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Example 2.2.2 (Exponential mean) Suppose \( X \) has an exponential \( \left( \lambda \right) \) distribution, that is, it has pdf given by\n\n\[ \n{f}_{X}\left( x\right) = \frac{1}{\lambda }{e}^{-x/\lambda },\;0 \leq x < \infty ,\;\lambda > 0.\n\]\nThen \( \mathrm{E}X \) is given by\n\n\[ \n\mathrm{E}X = {\int }_{0}^{... | \[ \n= - {\left. x{e}^{-x/\lambda }\right| }_{0}^{\infty } + {\int }_{0}^{\infty }{e}^{-x/\lambda }{dx}\;\text{ (integration by parts) }\n\]\n\n\[ \n= {\int }_{0}^{\infty }{e}^{-x/\lambda }{dx} = \lambda\n\] | Yes |
If \( X \) has a binomial distribution, its pmf is given by\n\n\[ P\left( {X = x}\right) = \left( \begin{array}{l} n \\ x \end{array}\right) {p}^{x}{\left( 1 - p\right) }^{n - x},\;x = 0,1,\ldots, n, \]\n\nwhere \( n \) is a positive integer, \( 0 \leq p \leq 1 \), and for every fixed pair \( n \) and \( p \) the pmf s... | (the \( x = 0 \) term is 0 ). Using the identity \( x\left( \begin{array}{l} n \\ x \end{array}\right) = n\left( \begin{array}{l} n - 1 \\ x - 1 \end{array}\right) \), we have\n\n\[ \mathrm{E}X = \mathop{\sum }\limits_{{x = 1}}^{n}n\left( \begin{array}{l} n - 1 \\ x - 1 \end{array}\right) {p}^{x}{\left( 1 - p\right) }^... | Yes |
Example 2.2.4 (Cauchy mean) A classic example of a random variable whose expected value does not exist is a Cauchy random variable, that is, one with pdf\n\n\[ \n{f}_{X}\left( x\right) = \frac{1}{\pi }\frac{1}{1 + {x}^{2}},\; - \infty < x < \infty .\n\] | It is straightforward to check that \( {\int }_{-\infty }^{\infty }{f}_{X}\left( x\right) {dx} = 1 \), but \( \mathrm{E}\left| X\right| = \infty \) . Write\n\n\[ \n\mathrm{E}\left| X\right| = {\int }_{-\infty }^{\infty }\frac{\left| x\right| }{\pi }\frac{1}{1 + {x}^{2}}{dx} = \frac{2}{\pi }{\int }_{0}^{\infty }\frac{x}... | Yes |
Theorem 2.2.5 Let \( X \) be a random variable and let \( a, b \), and \( c \) be constants. Then for any functions \( {g}_{1}\left( x\right) \) and \( {g}_{2}\left( x\right) \) whose expectations exist,\n\na. \( \mathrm{E}\left( {a{g}_{1}\left( X\right) + b{g}_{2}\left( X\right) + c}\right) = a\mathrm{E}{g}_{1}\left( ... | Proof: We will give details for only the continuous case, the discrete case being similar. By definition,\n\n\[ \mathrm{E}\left( {a{g}_{1}\left( X\right) + b{g}_{2}\left( X\right) + c}\right) \]\n\n\[ = {\int }_{-\infty }^{\infty }\left( {a{g}_{1}\left( x\right) + b{g}_{2}\left( x\right) + c}\right) {f}_{X}\left( x\rig... | Yes |
Example 2.2.6 (Minimizing distance) The expected value of a random variable has another property, one that we can think of as relating to the interpretation of \( \\mathrm{E}X \) as a good guess at a value of \( X \) . Suppose we measure the distance between a random variable \( X \) and a constant \( b \) by \( {\\lef... | We could proceed with the minimization of \( \\mathrm{E}{\\left( X - b\\right) }^{2} \) by using calculus, but there is a simpler method. (See Exercise 2.19 for a calculus-based proof.) Using the belief that there is something special about \( \\mathrm{E}X \), we write\n\n\[ \n\\mathrm{E}{\\left( X - b\\right) }^{2} = ... | No |
Let \( X \) have the exponential \( \left( \lambda \right) \) distribution, defined in Example 2.2.2. There we calculated \( \mathrm{E}X = \lambda \), and we can now calculate the variance by | \[ \operatorname{Var}X = \mathrm{E}{\left( X - \lambda \right) }^{2} = {\int }_{0}^{\infty }{\left( x - \lambda \right) }^{2}\frac{1}{\lambda }{e}^{-x/\lambda }{dx} \] \[ = {\int }_{0}^{\infty }\left( {{x}^{2} - {2x\lambda } + {\lambda }^{2}}\right) \frac{1}{\lambda }{e}^{-x/\lambda }{dx} \] To complete the integration... | Yes |
Theorem 2.3.4 If \( X \) is a random variable with finite variance, then for any constants \( a \) and \( b \) ,\n\n\[ \operatorname{Var}\left( {{aX} + b}\right) = {a}^{2}\operatorname{Var}X \] | Proof: From the definition, we have\n\n\[ \operatorname{Var}\left( {{aX} + b}\right) = \mathrm{E}{\left( \left( aX + b\right) - \mathrm{E}\left( aX + b\right) \right) }^{2} \]\n\n\[ = \mathrm{E}{\left( aX - a\mathrm{E}X\right) }^{2}\;\left( {\mathrm{E}\left( {{aX} + b}\right) = a\mathrm{E}X + b}\right) \]\n\n\[ = {a}^{... | Yes |
Example 2.3.5 (Binomial variance) Let \( X \sim \operatorname{binomial}\left( {n, p}\right) \), that is,\n\n\[ P\left( {X = x}\right) = \left( \begin{array}{l} n \\ x \end{array}\right) {p}^{x}{\left( 1 - p\right) }^{n - x},\;x = 0,1,\ldots, n. \] | We have previously seen that \( \mathrm{E}X = {np} \) . To calculate \( \operatorname{Var}X \) we first calculate \( \mathrm{E}{X}^{2} \) . We have\n\n(2.3.2)\n\n\[ \mathrm{E}{X}^{2} = \mathop{\sum }\limits_{{x = 0}}^{n}{x}^{2}\left( \begin{array}{l} n \\ x \end{array}\right) {p}^{x}{\left( 1 - p\right) }^{n - x}. \]\n... | Yes |
Theorem 2.3.7 If \( X \) has \( {mgf}{M}_{X}\left( t\right) \), then\n\n\[ \n\mathrm{E}{X}^{n} = {M}_{X}^{\left( n\right) }\left( 0\right) \n\]\n\nwhere we define\n\n\[ \n{M}_{X}^{\left( n\right) }\left( 0\right) = {\left. \frac{{d}^{n}}{d{t}^{n}}{M}_{X}\left( t\right) \right| }_{t = 0}. \n\]\n\nThat is, the nth moment... | Proof: Assuming that we can differentiate under the integral sign (see the next section), we have\n\n\[ \n\frac{d}{dt}{M}_{X}\left( t\right) = \frac{d}{dt}{\int }_{-\infty }^{\infty }{e}^{tx}{f}_{X}\left( x\right) {dx} \n\]\n\n\[ \n= {\int }_{-\infty }^{\infty }\left( {\frac{d}{dt}{e}^{tx}}\right) {f}_{X}\left( x\right... | No |
Consider the two pdfs given by\n\n\[ \n{f}_{1}\left( x\right) = \frac{1}{\sqrt{2\pi }x}{e}^{-{\left( \log x\right) }^{2}/2},\;0 \leq x < \infty ,\n\]\n\n\[ \n{f}_{2}\left( x\right) = {f}_{1}\left( x\right) \left\lbrack {1 + \sin \left( {{2\pi }\log x}\right) }\right\rbrack ,\;0 \leq x < \infty .\n\]\n\n(The pdf \( {f}_... | However, the transformation \( y = \log x - r \) shows that this last integral is that of an odd function over \( \left( {-\infty ,\infty }\right) \) and hence is equal to 0 for \( r = 0,1,\ldots \) . Thus, even though \( {X}_{1} \) and \( {X}_{2} \) have distinct pdfs, they have the same moments for all \( r \) . The ... | Yes |
Lemma 2.3.14 Let \( {a}_{1},{a}_{2},\ldots \) be a sequence of numbers converging to a, that is, \( \mathop{\lim }\limits_{{n \rightarrow \infty }}{a}_{n} = a \) . Then\n\n\[ \mathop{\lim }\limits_{{n \rightarrow \infty }}{\left( 1 + \frac{{a}_{n}}{n}\right) }^{n} = {e}^{a} \] | Returning to the example, we have\n\n\[ {M}_{X}\left( t\right) = {\left\lbrack p{e}^{t} + \left( 1 - p\right) \right\rbrack }^{n} = {\left\lbrack 1 + \frac{1}{n}\left( {e}^{t} - 1\right) \left( np\right) \right\rbrack }^{n} = {\left\lbrack 1 + \frac{1}{n}\left( {e}^{t} - 1\right) \lambda \right\rbrack }^{n}, \]\n\nbeca... | No |
Example 2.4.5 (Interchanging integration and differentiation-I) Let \( X \) have the exponential \( \left( \lambda \right) \) pdf given by \( f\left( x\right) = \left( {1/\lambda }\right) {e}^{-x/\lambda },0 < x < \infty \), and suppose we want to calculate\n\n\[\n\frac{d}{d\lambda }\mathrm{E}{X}^{n} = \frac{d}{d\lambd... | (2.4.6)\n\n\[\n= {\int }_{0}^{\infty }\frac{{x}^{n}}{{\lambda }^{2}}\left( {\frac{x}{\lambda } - 1}\right) {e}^{-x/\lambda }{dx}\n\]\n\n\[\n= \frac{1}{{\lambda }^{2}}\mathrm{E}{X}^{n + 1} - \frac{1}{\lambda }\mathrm{E}{X}^{n}\n\]\n\nTo justify the interchange of integration and differentiation, we bound the derivative ... | Yes |
Example 2.4.7 (Interchanging summation and differentiation) Let \( X \) be a discrete random variable with the geometric distribution\n\n\[ P\left( {X = x}\right) = \theta {\left( 1 - \theta \right) }^{x},\;x = 0,1,\ldots ,\;0 < \theta < 1.\]\n\nWe have that \( \mathop{\sum }\limits_{{x = 0}}^{\infty }\theta {\left( 1 ... | \[ = \mathop{\sum }\limits_{{x = 0}}^{\infty }\left\lbrack {{\left( 1 - \theta \right) }^{x} - {\theta x}{\left( 1 - \theta \right) }^{x - 1}}\right\rbrack \]\n\n\[ = \frac{1}{\theta }\mathop{\sum }\limits_{{x = 0}}^{\infty }\theta {\left( 1 - \theta \right) }^{x} - \frac{1}{1 - \theta }\mathop{\sum }\limits_{{x = 0}}^... | Yes |
Example 2.4.9 (Continuation of Example 2.4.7) To apply Theorem 2.4.8 we identify\n\n\\[ h\\left( {\\theta, x}\\right) = \\theta {\\left( 1 - \\theta \\right) }^{x} \\]\n\nand\n\n\\[ \\frac{\\partial }{\\partial \\theta }h\\left( {\\theta, x}\\right) = {\\left( 1 - \\theta \\right) }^{x} - {\\theta x}{\\left( 1 - \\thet... | Define \\( {S}_{n}\\left( \\theta \\right) \\) by\n\n\\[ {S}_{n}\\left( \\theta \\right) = \\mathop{\\sum }\\limits_{{x = 0}}^{n}\\left\\lbrack {{\\left( 1 - \\theta \\right) }^{x} - {\\theta x}{\\left( 1 - \\theta \\right) }^{x - 1}}\\right\\rbrack . \\]\n\nThe convergence will be uniform on \\( \\left\\lbrack {c, d}\... | Yes |
Theorem 3.2.2 (Binomial Theorem) For any real numbers \( x \) and \( y \) and integer \( n \geq 0 \)\n\n\[{\left( x + y\right) }^{n} = \mathop{\sum }\limits_{{i = 0}}^{n}\left( \begin{matrix} n \\ i \end{matrix}\right) {x}^{i}{y}^{n - i}.\] | Proof: Write\n\n\[{\left( x + y\right) }^{n} = \left( {x + y}\right) \left( {x + y}\right) \cdots \cdot \left( {x + y}\right) ,\]\n\nand consider how the right-hand side would be calculated. From each factor \( \left( {x + y}\right) \) we choose either an \( x \) or \( y \), and multiply together the \( n \) choices. F... | Yes |
Example 3.2.5 (Poisson approximation) A typesetter, on the average, makes one error in every 500 words typeset. A typical page contains 300 words. What is the probability that there will be no more than two errors in five pages? | If we assume that setting a word is a Bernoulli trial with success probability \( p = \frac{1}{500} \) (notice that we are labeling an error as a \ | No |
Example 3.4.3 (Binomial mean and variance) From Example 3.4.1 we have | \[ \frac{d}{dp}{w}_{1}\left( p\right) = \frac{d}{dp}\log \frac{p}{1 - p} = \frac{1}{p\left( {1 - p}\right) } \] \[ \frac{d}{dp}\log c\left( p\right) = \frac{d}{dp}n\log \left( {1 - p}\right) = \frac{-n}{1 - p} \] and thus from Theorem 3.4.2 we have \[ \mathrm{E}\left( {\frac{1}{p\left( {1 - p}\right) }X}\right) = \frac... | Yes |
Example 3.4.8 (A curved exponential family) The normal family of Example 3.4.4 is a full exponential family. However, if we assume that \( {\sigma }^{2} = {\mu }^{2} \), the family becomes curved. (Such a model might be used in the analysis of variance; see Exercises 11.1 and 11.2.) We then have\n\n\[ f\left( {x \mid \... | \[ = \frac{1}{\sqrt{{2\pi }{\mu }^{2}}}\exp \left( {-\frac{1}{2}}\right) \exp \left( {-\frac{{x}^{2}}{2{\mu }^{2}} + \frac{x}{\mu }}\right) . \] | Yes |
Example 3.5.3 (Exponential location family) Let \( f\left( x\right) = {e}^{-x}, x \geq 0 \), and \( f\left( x\right) = 0, x < 0 \) . To form a location family we replace \( x \) with \( x - \mu \) to obtain | \[ f\left( {x \mid \mu }\right) = \left\{ \begin{array}{ll} {e}^{-\left( {x - \mu }\right) } & x - \mu \geq 0 \\ 0 & x - \mu < 0 \end{array}\right. \] \[ = \left\{ \begin{array}{ll} {e}^{-\left( {x - \mu }\right) } & x \geq \mu \\ 0 & x < \mu \end{array}\right. \] Graphs of \( f\left( {x \mid \mu }\right) \) for variou... | Yes |
Theorem 3.5.6 Let \( f\left( \cdot \right) \) be any pdf. Let \( \mu \) be any real number, and let \( \sigma \) be any positive real number. Then \( X \) is a random variable with \( \operatorname{pdf}\left( {1/\sigma }\right) f\left( {\left( {x - \mu }\right) /\sigma }\right) \) if and only if there exists a random v... | Proof: To prove the \ | No |
Theorem 3.5.7 Let \( Z \) be a random variable with \( {pdf}\;f\left( z\right) \) . Suppose \( \mathrm{E}Z \) and \( \operatorname{Var}Z \) exist. If \( X \) is a random variable with \( \operatorname{pdf}\left( {1/\sigma }\right) f\left( {\left( {x - \mu }\right) /\sigma }\right) \), then\n\n\[ \mathrm{E}X = \sigma \m... | Proof: By Theorem 3.5.6, there is a random variable \( {Z}^{ * } \) with pdf \( f\left( z\right) \) and \( X = \) \( \sigma {Z}^{ * } + \mu \) . So \( \mathrm{E}X = \sigma \mathrm{E}{Z}^{ * } + \mu = \sigma \mathrm{E}Z + \mu \) and \( \operatorname{Var}X = {\sigma }^{2}\operatorname{Var}{Z}^{ * } = {\sigma }^{2}\operat... | Yes |
Theorem 3.6.1 (Chebychev’s Inequality) Let \( X \) be a random variable and let \( g\left( x\right) \) be a nonnegative function. Then, for any \( r > 0 \), \[ P\left( {g\left( X\right) \geq r}\right) \leq \frac{\mathrm{E}g\left( X\right) }{r}. \] | Proof: \[ \mathrm{E}g\left( X\right) = {\int }_{-\infty }^{\infty }g\left( x\right) {f}_{X}\left( x\right) {dx} \] \[ \geq {\int }_{\{ x : g\left( x\right) \geq r\} }g\left( x\right) {f}_{X}\left( x\right) {dx} \] ( \( g \) is nonnegative) \[ \geq r{\int }_{\{ x : g\left( x\right) \geq r\} }{f}_{X}\left( x\right) {dx} ... | Yes |
Example 3.6.2 (Illustrating Chebychev) The most widespread use of Chebychev’s Inequality involves means and variances. Let \( g\left( x\right) = {\left( x - \mu \right) }^{2}/{\sigma }^{2} \), where \( \mu = \mathrm{E}X \) and \( {\sigma }^{2} = \operatorname{Var}X \) . For convenience write \( r = {t}^{2} \) . Then | \[ P\left( {\frac{{\left( X - \mu \right) }^{2}}{{\sigma }^{2}} \geq {t}^{2}}\right) \leq \frac{1}{{t}^{2}}\mathrm{E}\frac{{\left( X - \mu \right) }^{2}}{{\sigma }^{2}} = \frac{1}{{t}^{2}}. \] Doing some obvious algebra, we get the inequality \[ P\left( {\left| {X - \mu }\right| \geq {t\sigma }}\right) \leq \frac{1}{{t... | Yes |
Theorem 3.6.4 Let \( {X}_{\alpha ,\beta } \) denote a gamma \( \left( {\alpha ,\beta }\right) \) random variable with pdf \( f\left( {x \mid \alpha ,\beta }\right) \) , where \( \alpha > 1 \) . Then for any constants a and \( b \) ,\n\n\[ P\left( {a < {X}_{\alpha ,\beta } < b}\right) = \beta \left( {f\left( {a \mid \al... | Proof: By definition,\n\n\[ P\left( {a < {X}_{\alpha ,\beta } < b}\right) = \frac{1}{\Gamma \left( \alpha \right) {\beta }^{\alpha }}{\int }_{a}^{b}{x}^{\alpha - 1}{e}^{-x/\beta }{dx} \]\n\n\[ = \frac{1}{\Gamma \left( \alpha \right) {\beta }^{\alpha }}\left\lbrack {-{\left. {x}^{\alpha - 1}\beta {e}^{-x/\beta }\right| ... | Yes |
Lemma 3.6.5 (Stein’s Lemma) Let \( X \sim \mathrm{n}\left( {\theta ,{\sigma }^{2}}\right) \), and let \( g \) be a differentiable function satisfying \( \mathrm{E}\left| {{g}^{\prime }\left( X\right) }\right| < \infty \) . Then\n\n\[ \mathrm{E}\left\lbrack {g\left( X\right) \left( {X - \theta }\right) }\right\rbrack = ... | Proof: The left-hand side is\n\n\[ \mathrm{E}\left\lbrack {g\left( X\right) \left( {X - \theta }\right) }\right\rbrack = \frac{1}{\sqrt{2\pi }\sigma }{\int }_{-\infty }^{\infty }g\left( x\right) \left( {x - \theta }\right) {e}^{-{\left( x - \theta \right) }^{2}/\left( {2{\sigma }^{2}}\right) }{dx}. \]\n\nUse integratio... | Yes |
Example 3.6.6 (Higher-order normal moments) Stein's Lemma makes calculation of higher-order moments quite easy. For example, if \( X \sim \mathrm{n}\left( {\theta ,{\sigma }^{2}}\right) \), then | \[ \mathrm{E}{X}^{3} = \mathrm{E}{X}^{2}\left( {X - \theta + \theta }\right) \] \[ = \mathrm{E}{X}^{2}\left( {X - \theta }\right) + \theta \mathrm{E}{X}^{2} \] \[ = 2{\sigma }^{2}\mathrm{E}X + \theta \mathrm{E}{X}^{2} \] \[ = 2{\sigma }^{2}\theta + \theta \left( {{\sigma }^{2} + {\theta }^{2}}\right) \] \[ = {3\theta }... | Yes |
Theorem 3.6.8 (Hwang) Let \( g\left( x\right) \) be a function with \( - \infty < \mathrm{E}g\left( X\right) < \infty \) and \( - \infty < g\left( {-1}\right) < \infty \) . Then:\n\na. If \( X \sim \operatorname{Poisson}\left( \lambda \right) \),\n\n\[ \mathrm{E}\left( {{\lambda g}\left( X\right) }\right) = \mathrm{E}\... | Proof: We will prove part (a), saving part (b) for Exercise 3.50. We have\n\n\[ \mathrm{E}\left( {{\lambda g}\left( X\right) }\right) = \mathop{\sum }\limits_{{x = 0}}^{\infty }{\lambda g}\left( x\right) \frac{{e}^{-\lambda }{\lambda }^{x}}{x!} \]\n\n\[ = \mathop{\sum }\limits_{{x = 0}}^{\infty }g\left( x\right) \frac{... | No |
Example 3.6.9 (Higher-order Poisson moments) For \( X \sim \operatorname{Poisson}\left( \lambda \right) \), take \( g\left( x\right) = {x}^{2} \) and use (3.6.5); | \[ \mathrm{E}\left( {\lambda {X}^{2}}\right) = \mathrm{E}\left( {X{\left( X - 1\right) }^{2}}\right) = \mathrm{E}\left( {{X}^{3} - 2{X}^{2} + X}\right) . \] Therefore, the third moment of a \( \operatorname{Poisson}\left( \lambda \right) \) is \[ \mathrm{E}{X}^{3} = \lambda \mathrm{E}{X}^{2} + 2\mathrm{E}{X}^{2} - \mat... | Yes |
Lemma 3.8.3 (Markov’s Inequality) If \( P\left( {Y \geq 0}\right) = 1 \) and \( P\left( {Y = 0}\right) < 1 \) , then, for any \( r > 0 \) , | \[ P\left( {Y \geq r}\right) \leq \frac{\mathrm{E}Y}{r} \] with equality if and only if \( P\left( {Y = r}\right) = p = 1 - P\left( {Y = 0}\right) ,0 < p \leq 1 \) . | Yes |
Theorem 3.8.5 (Vysochanskiĭ-Petunin Inequality) Let \( X \sim f \), where \( f \) is unimodal, and define \( {\xi }^{2} = \mathrm{E}{\left( X - \alpha \right) }^{2} \) for an arbitrary point \( \alpha \) . Then\n\n\[ \n\begin{array}{r} P\left( {\left| {X - \alpha }\right| > \varepsilon }\right) \leq \left\{ \begin{arra... | Pukelsheim points out that taking \( \alpha = \mu = \mathrm{E}\left( X\right) \) and \( \varepsilon = {3\sigma } \), where \( {\sigma }^{2} = \operatorname{Var}\left( X\right) \) , yields\n\n\[ \nP\left( {\left| {X - \mu }\right| > {3\sigma }}\right) \leq \frac{4}{81} < {.05} \n\]\n\nthe so-called three-sigma rule, tha... | Yes |
What is \( P\left( {X = 5\text{ and }Y = 3}\right) \) ? | You can verify that the only two sample points that yield \( X = 5 \) and \( Y = 3 \) are \( \left( {4,1}\right) \) and \( \left( {1,4}\right) \) . Thus the event \ | No |
For the \( \left( {X, Y}\right) \) whose joint pmf is given in Table 4.1.1, what is the average value of \( {XY} \) ? | Letting \( g\left( {x, y}\right) = {xy} \), we compute \( \mathrm{E}{XY} = \mathrm{E}g\left( {X, Y}\right) \) by computing \( {xyf}\left( {x, y}\right) \) for each of the 21 \( \left( {x, y}\right) \) points in Table 4.1.1 and summing these 21 terms. Thus,\n\n\[ \mathrm{E}{XY} = \left( 2\right) \left( 0\right) \frac{1}... | Yes |
Example 4.1.5 (Joint pmf for dice) Define \( f\left( {x, y}\right) \) by\n\n\[ f\left( {0,0}\right) = f\left( {0,1}\right) = \frac{1}{6} \]\n\n\[ f\left( {1,0}\right) = f\left( {1,1}\right) = \frac{1}{3} \]\n\n\[ f\left( {x, y}\right) = 0\text{ for any other }\left( {x, y}\right) . \]\n\nThen \( f\left( {x, y}\right) \... | All this can be done without reference to the sample space \( S \) . Indeed, there are many sample spaces and functions thereon that lead to this joint pmf for \( \left( {X, Y}\right) \) . Here is one. Let \( S \) be the 36-point sample space for the experiment of tossing two fair dice. Let \( X = 0 \) if the first die... | No |
Theorem 4.1.6 Let \( \left( {X, Y}\right) \) be a discrete bivariate random vector with joint pmf \( {f}_{X, Y}\left( {x, y}\right) \) . Then the marginal pmfs of \( X \) and \( Y,{f}_{X}\left( x\right) = P\left( {X = x}\right) \) and \( {f}_{Y}\left( y\right) = \) \( P\left( {Y = y}\right) \), are given by\n\n\[ \n{f}... | Proof: We will prove the result for \( {f}_{X}\left( x\right) \) . The proof for \( {f}_{Y}\left( y\right) \) is similar. For any \( x \in \Re \), let \( {A}_{x} = \{ \left( {x, y}\right) : - \infty < y < \infty \} \) . That is, \( {A}_{x} \) is the line in the plane with first coordinate equal to \( x \) . Then, for a... | Yes |
Using the result of Theorem 4.1.6, we can compute the marginal distributions for \( X \) and \( Y \) from the joint distribution given in Table 4.1.1. To compute the marginal pmf of \( Y \), for each possible value of \( Y \) we sum over the possible values of \( X \) . | \[ {f}_{Y}\left( 0\right) = {f}_{X, Y}\left( {2,0}\right) + {f}_{X, Y}\left( {4,0}\right) + {f}_{X, Y}\left( {6,0}\right) \] \[ \begin{matrix} & & + {f}_{X, Y}\left( {8,0}\right) + {f}_{X, Y}\left( {{10},0}\right) + {f}_{X, Y}\left( {{12},0}\right) \end{matrix} \] \[ = \frac{1}{6} \] Similarly, we obtain \[ {f}_{Y}\lef... | Yes |
Example 4.1.9 (Same marginals, different joint pmf) Define a joint pmf by | \[ \begin{array}{l} f\left( {0,0}\right) = \frac{1}{12},\;f\left( {1,0}\right) = \frac{5}{12},\;f\left( {0,1}\right) = f\left( {1,1}\right) = \frac{3}{12}, \end{array} \]\n\[ f\left( {x, y}\right) = 0\text{ for all other values. } \]\n\nThe marginal pmf of \( Y \) is \( {f}_{Y}\left( 0\right) = f\left( {0,0}\right) \) ... | Yes |
Define a joint pdf by\n\n\\[ \nf\\left( {x, y}\\right) = \\left\\{ \\begin{array}{ll} {6x}{y}^{2} & 0 < x < 1\\text{ and }0 < y < 1 \\\\ 0 & \\text{ otherwise. } \\end{array}\\right.\n\\]\n\n(Henceforth, it will be understood that \\( f\\left( {x, y}\\right) = 0 \\) for \\( \\left( {x, y}\\right) \\) values not specifi... | That \\( f\\left( {x, y}\\right) \\geq 0 \\) for all \\( \\left( {x, y}\\right) \\) in the defined range is fairly obvious. To compute the integral of \\( f\\left( {x, y}\\right) \\) over the whole plane, note that, since \\( f\\left( {x, y}\\right) \\) is 0 except on the unit square, the integral over the plane is the... | Yes |
Example 4.1.12 (Calculating joint probabilities-II) As another example of a joint pdf, let \( f\left( {x, y}\right) = {e}^{-y},0 < x < y < \infty \) . Although \( {e}^{y} \) does not depend on \( x \) , \( f\left( {x, y}\right) \) certainly is a function of \( x \) since the set where \( f\left( {x, y}\right) \) is non... | The integration is easier over the intersection of the set \( B = \{ \left( {x, y}\right) : x + y < 1\} \) and the set where \( f\left( {x, y}\right) \) is nonzero, the triangular region (darker shading in Figure 4.1.1) bounded by the lines \( x = y, x + y = 1 \), and \( x = 0 \) . Thus\n\n\[ P\left( {X + Y \geq 1}\rig... | Yes |
Define the joint pmf of \( \\left( {X, Y}\\right) \) by\n\n\[ \n\\begin{matrix} f\\left( {0,{10}}\\right) = f\\left( {0,{20}}\\right) = \\frac{2}{18},\\;f\\left( {1,{10}}\\right) = f\\left( {1,{30}}\\right) = \\frac{3}{18}, \\end{matrix} \n\]\n\n\[ \nf\\left( {1,{20}}\\right) = \\frac{4}{18},\\;\\text{ and }\\;f\\left(... | First, the marginal pmf of \( X \) is\n\n\[ \n{f}_{X}\\left( 0\\right) = f\\left( {0,{10}}\\right) + f\\left( {0,{20}}\\right) = \\frac{4}{18}, \n\]\n\n\[ \n{f}_{X}\\left( 1\\right) = f\\left( {1,{10}}\\right) + f\\left( {1,{20}}\\right) + f\\left( {1,{30}}\\right) = \\frac{10}{18}, \n\]\n\n\[ \n{f}_{X}\\left( 2\\right... | Yes |
Example 4.2.4 (Calculating conditional pdfs) As in Example 4.1.12, let the continuous random vector \( \left( {X, Y}\right) \) have joint pdf \( f\left( {x, y}\right) = {e}^{-y},0 < x < y < \infty \) . Suppose we wish to compute the conditional pdf of \( Y \) given \( X = x \) . | The marginal pdf of \( X \) is computed as follows. If \( x \leq 0, f\left( {x, y}\right) = 0 \) for all values of \( y \), so \( {f}_{X}\left( x\right) = 0 \) . If \( x > 0, f\left( {x, y}\right) > 0 \) only if \( y > x \) . Thus\n\n\[ \n{f}_{X}\left( x\right) = {\int }_{-\infty }^{\infty }f\left( {x, y}\right) {dy} =... | Yes |
Example 4.2.6 (Checking independence-I) Consider the discrete bivariate random vector \( \\left( {X, Y}\\right) \), with joint pmf given by\n\n\[ \nf\\left( {{10},1}\\right) = f\\left( {{20},1}\\right) = f\\left( {{20},2}\\right) = \\frac{1}{10},\n\]\n\n\[ \n\\begin{matrix} f\\left( {{10},2}\\right) = f\\left( {{10},3}... | For example,\n\n\[ \nf\\left( {{10},3}\\right) = \\frac{1}{5} \\neq \\frac{1}{2}\\frac{1}{2} = {f}_{X}\\left( {10}\\right) {f}_{Y}\\left( 3\\right) .\n\]\n\nThe relationship (4.2.1) must hold for every choice of \( x \) and \( y \) if \( X \) and \( Y \) are to be independent. Note that \( f\\left( {{10},1}\\right) = \... | Yes |
Lemma 4.2.7 Let \( \left( {X, Y}\right) \) be a bivariate random vector with joint pdf or pmf \( f\left( {x, y}\right) \) . Then \( X \) and \( Y \) are independent random variables if and only if there exist functions \( g\left( x\right) \) and \( h\left( y\right) \) such that, for every \( x \in \Re \) and \( y \in \... | Proof: The \ | No |
Consider the joint pdf \( f\left( {x, y}\right) = \frac{1}{384}{x}^{2}{y}^{4}{e}^{-y - \left( {x/2}\right) }, x > 0 \) and \( y > 0 \). If we define \( g\left( x\right) = \left\{ {\begin{array}{ll} {x}^{2}{e}^{-x/2} & x > 0 \\ 0 & x \leq 0 \end{array}\;\text{ and }\;h\left( y\right) = \left\{ \begin{array}{ll} {y}^{4}{... | We do not have to compute marginal pdfs. | Yes |
Theorem 4.2.10 Let \( X \) and \( Y \) be independent random variables.\n\na. For any \( A \subset \Re \) and \( B \subset \Re, P\left( {X \in A, Y \in B}\right) = P\left( {X \in A}\right) P\left( {Y \in B}\right) \) ; that is, the events \( \{ X \in A\} \) and \( \{ Y \in B\} \) are independent events.\n\nb. Let \( g\... | Proof: For continuous random variables, part (b) is proved by noting that\n\n\[ \mathrm{E}\left( {g\left( X\right) h\left( Y\right) }\right) = {\int }_{-\infty }^{\infty }{\int }_{-\infty }^{\infty }g\left( x\right) h\left( y\right) f\left( {x, y}\right) {dxdy} \]\n\n\[ = {\int }_{-\infty }^{\infty }{\int }_{-\infty }^... | Yes |
Let \( X \) and \( Y \) be independent exponential(1) random variables. From Theorem 4.2.10 we have\n\n\[ P\left( {X \geq 4, Y < 3}\right) = P\left( {X \geq 4}\right) P\left( {Y < 3}\right) = {e}^{-4}\left( {1 - {e}^{-3}}\right) . \] | Letting \( g\left( x\right) = {x}^{2} \) and \( h\left( y\right) = y \), we see that\n\n\[ \mathrm{E}\left( {{X}^{2}Y}\right) = \left( {\mathrm{E}{X}^{2}}\right) \left( {\mathrm{E}Y}\right) = \left( {\operatorname{Var}X + {\left( \mathrm{E}X\right) }^{2}}\right) \mathrm{E}Y = \left( {1 + {1}^{2}}\right) 1 = 2. \] | Yes |
Theorem 4.2.12 Let \( X \) and \( Y \) be independent random variables with moment generating functions \( {M}_{X}\left( t\right) \) and \( {M}_{Y}\left( t\right) \). Then the moment generating function of the random variable \( Z = X + Y \) is given by \[ {M}_{Z}\left( t\right) = {M}_{X}\left( t\right) {M}_{Y}\left( t... | Proof: Using the definition of the mgf and Theorem 4.2.10, we have \[ {M}_{Z}\left( t\right) = \mathrm{E}{e}^{tZ} = \mathrm{E}{e}^{t\left( {X + Y}\right) } = \mathrm{E}\left( {{e}^{tX}{e}^{tY}}\right) = \left( {\mathrm{E}{e}^{tX}}\right) \left( {\mathrm{E}{e}^{tY}}\right) = {M}_{X}\left( t\right) {M}_{Y}\left( t\right)... | Yes |
Example 4.3.1 (Distribution of the sum of Poisson variables) Let \( X \) and \( Y \) be independent Poisson random variables with parameters \( \theta \) and \( \lambda \), respectively. Thus the joint pmf of \( \left( {X, Y}\right) \) is\n\n\[ \n{f}_{X, Y}\left( {x, y}\right) = \frac{{\theta }^{x}{e}^{-\theta }}{x!}\f... | In this example it is interesting to compute the marginal pmf of \( U \) . For any fixed nonnegative integer \( u,{f}_{U, V}\left( {u, v}\right) > 0 \) only for \( v = 0,1,\ldots, u \) . This gives the set of \( v \) values to sum over to obtain the marginal pmf of \( U \) . It is\n\n\[ \n{f}_{U}\left( u\right) = \math... | Yes |
Example 4.3.4 (Sum and difference of normal variables) Let \( X \) and \( Y \) be independent, standard normal random variables. Consider the transformation \( U = \) \( X + Y \) and \( V = X - Y \) . In the notation used above, \( U = {g}_{1}\left( {X, Y}\right) \) where \( {g}_{1}\left( {x, y}\right) = x + y \) and \... | (4.3.4)\n\n\[ u = x + y\\text{ and }v = x - y \]\n\n take on as \( \\left( {x, y}\\right) \) range over the set \( \\mathcal{A} = {\\Re }^{2} \) . But we can set \( u \) to be any number and \( v \) to be any number and uniquely solve equations (4.3.4) for \( x \) and \( y \) to obtain\n\n(4.3.5)\n\n\[ x = {h}_{1}\\lef... | Yes |
Example 4.3.6 (Distribution of the ratio of normal variables) Let \( X \) and \( Y \) be independent \( \mathrm{n}\left( {0,1}\right) \) random variables. Consider the transformation \( U = X/Y \) and \( V = \left| Y\right| \) . ( \( U \) and \( V \) can be defined to be any value, say \( \left( {1,1}\right) \), if \( ... | \[ {A}_{1} = \{ \left( {x, y}\right) : y > 0\} ,\;{A}_{2} = \{ \left( {x, y}\right) : y < 0\} ,\;\text{ and }\;{A}_{0} = \{ \left( {x, y}\right) : y = 0\} . \] \( {A}_{0},{A}_{1} \), and \( {A}_{2} \) form a partition of \( \mathcal{A} = {\Re }^{2} \) and \( P\left( {\left( {X, Y}\right) \in {A}_{0}}\right) = P\left( {... | Yes |
Example 4.4.1 (Binomial-Poisson hierarchy) Perhaps the most classic hierarchical model is the following. An insect lays a large number of eggs, each surviving with probability \( p \) . On the average, how many eggs will survive? | The \ | No |
The random variable of interest, \( X = \) number of survivors, has the distribution given by\n\n\[ P\left( {X = x}\right) = \mathop{\sum }\limits_{{y = 0}}^{\infty }P\left( {X = x, Y = y}\right) \] | \[ = \mathop{\sum }\limits_{{y = 0}}^{\infty }P\left( {X = x \mid Y = y}\right) P\left( {Y = y}\right) \;\left( \begin{matrix} \text{ definition of } \\ \text{ conditional probability } \end{matrix}\right) \]\n\n\[ = \mathop{\sum }\limits_{{y = x}}^{\infty }\left\lbrack {\left( \begin{array}{l} y \\ x \end{array}\right... | Yes |
Theorem 4.4.3 If \( X \) and \( Y \) are any two random variables, then\n\n\[ \mathrm{E}X = \mathrm{E}\left( {\mathrm{E}\left( {X \mid Y}\right) }\right) \]\n\nprovided that the expectations exist. | Proof: Let \( f\left( {x, y}\right) \) denote the joint pdf of \( X \) and \( Y \) . By definition, we have\n\n\[ \mathrm{E}X = \iint {xf}\left( {x, y}\right) {dxdy} = \int \left\lbrack {\int {xf}\left( {x \mid y}\right) {dx}}\right\rbrack {f}_{Y}\left( y\right) {dy} \]\n\nwhere \( f\left( {x \mid y}\right) \) and \( {... | Yes |
Consider a generalization of Example 4.4.1, where instead of one mother insect there are a large number of mothers and one mother is chosen at random. We are still interested in knowing the average number of survivors, but it is no longer clear that the number of eggs laid follows the same Poisson distribution for each... | The mean of \( X \) can easily be calculated as\n\n\[ \mathrm{E}X = \mathrm{E}\left( {\mathrm{E}\left( {X \mid Y}\right) }\right) \]\n\n\[ = \mathrm{E}\left( {pY}\right) \]\n\n(as before)\n\n\[ = \mathrm{E}\left( {\mathrm{E}\left( {{pY}|\Lambda }\right) }\right) \]\n\n\[ = \mathrm{E}\left( {p\Lambda }\right) \]\n\n\[ =... | Yes |
Example 4.4.6 (Beta-binomial hierarchy) One generalization of the binomial distribution is to allow the success probability to vary according to a distribution. A standard model for this situation is\n\n\[ \nX \mid P \sim \operatorname{binomial}\left( P\right) ,\;i = 1,\ldots, n, \n\]\n\n\[ \nP \sim \operatorname{beta}... | By iterating the expectation, we calculate the mean of \( X \) as\n\n\[ \n\mathrm{E}X = \mathrm{E}\left\lbrack {\mathrm{E}\left( {X \mid P}\right) }\right\rbrack = \mathrm{E}\left\lbrack {nP}\right\rbrack = n\frac{\alpha }{\alpha + \beta }. \n\] | Yes |
To calculate the variance of \( X \), we have from (4.4.4), | \n\[
\operatorname{Var}X = \operatorname{Var}\left( {\mathrm{E}\left( {X \mid P}\right) }\right) + \mathrm{E}\left( {\operatorname{Var}\left( {X \mid P}\right) }\right) .
\]
\nNow \( \mathrm{E}\left( {X \mid P}\right) = {nP} \), and since \( P \sim \operatorname{beta}\left( {\alpha ,\beta }\right) \) ,\n\n\[
\operatorn... | Yes |
Theorem 4.5.3 For any random variables \( X \) and \( Y \) ,\n\n\[ \operatorname{Cov}\left( {X, Y}\right) = \mathrm{E}{XY} - {\mu }_{X}{\mu }_{Y} \] | Proof: \( \operatorname{Cov}\left( {X, Y}\right) = \mathrm{E}\left( {X - {\mu }_{X}}\right) \left( {Y - {\mu }_{Y}}\right) \)\n\n\[ = \mathrm{E}\left( {{XY} - {\mu }_{X}Y - {\mu }_{Y}X + {\mu }_{X}{\mu }_{Y}}\right) \;\text{ (expanding the product) } \]\n\n\[ = \mathrm{E}{XY} - {\mu }_{X}\mathrm{E}Y - {\mu }_{Y}\mathrm... | Yes |
Let the joint pdf of \( \left( {X, Y}\right) \) be \( f\left( {x, y}\right) = 1,0 < \) \( x < 1, x < y < x + 1 \) . See Figure 4.5.1. The marginal distribution of \( X \) is uniform \( \left( {0,1}\right) \) so \( {\mu }_{X} = \frac{1}{2} \) and \( {\sigma }_{X}^{2} = \frac{1}{12} \) . The marginal pdf of \( Y \) is \(... | \[ \mathrm{E}{XY} = {\int }_{0}^{1}{\int }_{x}^{x + 1}{xydydx} = {\int }_{0}^{1}{\left. \frac{1}{2}x{y}^{2}\right| }_{x}^{x + 1}{dx} = {\int }_{0}^{1}\left( {{x}^{2} + \frac{1}{2}x}\right) {dx} = \frac{7}{12}. \] Using Theorem 4.5.3, we have \( \operatorname{Cov}\left( {X, Y}\right) = \frac{7}{12} - \left( \frac{1}{2}\... | Yes |
Theorem 4.5.5 If \( X \) and \( Y \) are independent random variables, then \( \operatorname{Cov}\left( {X, Y}\right) = 0 \) and \( {\rho }_{XY} = 0 \) . | Proof: Since \( X \) and \( Y \) are independent, from Theorem 4.2.10 we have \( \mathrm{E}{XY} = \) \( \left( {\mathrm{E}X}\right) \left( {\mathrm{E}Y}\right) \) . Thus\n\n\[ \operatorname{Cov}\left( {X, Y}\right) = \mathrm{E}{XY} - \left( {\mathrm{E}X}\right) \left( {\mathrm{E}Y}\right) = \left( {\mathrm{E}X}\right) ... | Yes |
Theorem 4.5.6 If \( X \) and \( Y \) are any two random variables and \( a \) and \( b \) are any two constants, then\n\n\[ \operatorname{Var}\left( {{aX} + {bY}}\right) = {a}^{2}\operatorname{Var}X + {b}^{2}\operatorname{Var}Y + {2ab}\operatorname{Cov}\left( {X, Y}\right) . \] | Proof: The mean of \( {aX} + {bY} \) is \( \mathrm{E}\left( {{aX} + {bY}}\right) = a\mathrm{E}X + b\mathrm{E}Y = a{\mu }_{X} + b{\mu }_{Y} \) . Thus\n\n\[ {Var}\left( {{aX} + {bY}}\right) = E{\left( \left( aX + bY\right) - \left( a{\mu }_{X} + b{\mu }_{Y}\right) \right) }^{2} \]\n\n\[ = E{\left( a\left( X - {\mu }_{X}\... | Yes |
Theorem 4.5.7 For any random variables \( X \) and \( Y \), a. \( - 1 \leq {\rho }_{XY} \leq 1 \). b. \( \left| {\rho }_{XY}\right| = 1 \) if and only if there exist numbers \( a \neq 0 \) and \( b \) such that \( P(Y = {aX} + b) = 1 \). If \( {\rho }_{XY} = 1 \), then \( a > 0 \), and if \( {\rho }_{XY} = - 1 \), then... | Proof: Consider the function \( h\left( t\right) \) defined by \[ h\left( t\right) = \mathrm{E}{\left( \left( X - {\mu }_{X}\right) t + \left( Y - {\mu }_{Y}\right) \right) }^{2}. \] Expanding this expression, we obtain \[ h\left( t\right) = {t}^{2}\mathrm{E}{\left( X - {\mu }_{X}\right) }^{2} + {2t}\mathrm{E}\left( {X... | Yes |
Example 4.5.8 (Correlation-II) This example is similar to Example 4.5.4, but we develop it differently to illustrate other model building and computational techniques. Let \( X \) have a uniform \( \left( {0,1}\right) \) distribution and \( Z \) have a uniform \( \left( {0,\frac{1}{10}}\right) \) distribution. Suppose ... | \[ f\left( {x, y}\right) = {10},\;0 < x < 1,\;x < y < x + \frac{1}{10}. \] Rather than using the formal techniques of Section 4.3, we can justify this as follows. Given \( X = x, Y = x + Z \) . The conditional distribution of \( Z \) given \( X = x \) is just uniform \( \left( {0,\frac{1}{10}}\right) \) since \( X \) a... | Yes |
In this example, let \( X \) have a uniform \( \left( {-1,1}\right) \) distribution and let \( Z \) have a uniform \( \left( {0,\frac{1}{10}}\right) \) distribution. Let \( X \) and \( Z \) be independent. Let \( Y = {X}^{2} + Z \) and consider the random vector \( \left( {X, Y}\right) \). As in Example 4.5.8, given \(... | Since \( X \) has a uniform \( \left( {-1,1}\right) \) distribution, \( \mathrm{E}X = \mathrm{E}{X}^{3} = 0 \), and since \( X \) and \( Z \) are independent, \( \mathrm{E}{XZ} = \left( {\mathrm{E}X}\right) \left( {\mathrm{E}Z}\right) \). Thus,\n\n\[ \operatorname{Cov}\left( {X, Y}\right) = \mathrm{E}\left( {X\left( {{... | Yes |
Example 4.6.1 (Multivariate pdfs) Let \( n = 4 \) and\n\n\[ f\left( {{x}_{1},{x}_{2},{x}_{3},{x}_{4}}\right) = \left\{ \begin{array}{ll} \frac{3}{4}\left( {{x}_{1}^{2} + {x}_{2}^{2} + {x}_{3}^{2} + {x}_{4}^{2}}\right) & 0 < {x}_{i} < 1, i = 1,2,3,4 \\ 0 & \text{ otherwise. } \end{array}\right. \]\n\nThis nonnegative fu... | \[ = {\int }_{0}^{1}{\int }_{0}^{1}{\int }_{0}^{1}{\int }_{0}^{1}\frac{3}{4}\left( {{x}_{1}^{2} + {x}_{2}^{2} + {x}_{3}^{2} + {x}_{4}^{2}}\right) d{x}_{1}d{x}_{2}d{x}_{3}d{x}_{4} \]\n\n\[ = 1\text{.} \] | Yes |
Theorem 4.6.4 (Multinomial Theorem) Let \( m \) and \( n \) be positive integers. Let \( \mathcal{A} \) be the set of vectors \( \mathbf{x} = \left( {{x}_{1},\ldots ,{x}_{n}}\right) \) such that each \( {x}_{i} \) is a nonnegative integer and \( \mathop{\sum }\limits_{{i = 1}}^{n}{x}_{i} = m \) . Then, for any real num... | \[ {\left( {p}_{1} + \cdots + {p}_{n}\right) }^{m} = \mathop{\sum }\limits_{{\mathbf{x} \in \mathcal{A}}}\frac{m!}{{x}_{1}!\cdots \cdots {x}_{n}!}{p}_{1}^{{x}_{1}}\cdots \cdots {p}_{n}^{{x}_{n}}. \] | Yes |
Theorem 4.6.7 (Generalization of Theorem 4.2.12) Let \( {X}_{1},\ldots ,{X}_{n} \) be mutu- \( \textit{ally independent random variables with mgfs}\;{M}_{{X}_{1}}\left( t\right) ,\ldots ,{M}_{{X}_{n}}\left( t\right) .\textit{ Let}\;Z = {X}_{1} + \cdots + \) \( {X}_{n} \) . Then the mgf of \( Z \) is | \[ {M}_{Z}\left( t\right) = {M}_{{X}_{1}}\left( t\right) \cdots \cdots {M}_{{X}_{n}}\left( t\right) \] In particular, if \( {X}_{1},\ldots ,{X}_{n} \) all have the same distribution with \( {mgf}{M}_{X}\left( t\right) \), then \[ {M}_{Z}\left( t\right) = {\left( {M}_{X}\left( t\right) \right) }^{n}. \] | Yes |
Example 4.6.8 (Mgf of a sum of gamma variables) Suppose \( {X}_{1},\ldots ,{X}_{n} \) are mutually independent random variables, and the distribution of \( {X}_{i} \) is \( \operatorname{gamma}\left( {{\alpha }_{i},\beta }\right) \) . From Example 2.3.8, the \( \mathrm{{mgf}} \) of a \( \operatorname{gamma}\left( {\alp... | \[ {M}_{Z}\left( t\right) = {M}_{{X}_{1}}\left( t\right) \cdots \cdots {M}_{{X}_{n}}\left( t\right) = {\left( 1 - \beta t\right) }^{-{\alpha }_{1}}\cdots \cdots {\left( 1 - \beta t\right) }^{-{\alpha }_{n}} = {\left( 1 - \beta t\right) }^{-\left( {{\alpha }_{1} + \cdots + {\alpha }_{n}}\right) }.\] | Yes |
Theorem 4.6.11 (Generalization of Lemma 4.2.7) Let \( {\mathbf{X}}_{1},\ldots ,{\mathbf{X}}_{n} \) be random vectors. Then \( {\mathbf{X}}_{1},\ldots ,{\mathbf{X}}_{n} \) are mutually independent random vectors if and only if there exist functions \( {g}_{i}\left( {\mathbf{x}}_{i}\right), i = 1,\ldots, n \), such that ... | \[ f\left( {{\mathbf{x}}_{1},\ldots ,{\mathbf{x}}_{n}}\right) = {g}_{1}\left( {\mathbf{x}}_{1}\right) \cdots \cdots {g}_{n}\left( {\mathbf{x}}_{n}\right) . \] | Yes |
Example 4.6.13 (Multivariate change of variables) Let \( \\left( {{X}_{1},{X}_{2},{X}_{3},{X}_{4}}\\right) \) have joint pdf\n\n\[ \n{f}_{\\mathbf{X}}\\left( {{x}_{1},{x}_{2},{x}_{3},{x}_{4}}\\right) = {24}{e}^{-{x}_{1} - {x}_{2} - {x}_{3} - {x}_{4}},\\;0 < {x}_{1} < {x}_{2} < {x}_{3} < {x}_{4} < \\infty .\n\]\n\nConsi... | This transformation maps the set \( \\mathcal{A} \) onto the set \( \\mathcal{B} = \\left\\{ {\\mathbf{u} : 0 < {u}_{i} < \\infty, i = 1,2,3,4}\\right\\} \) . The transformation is one-to-one, so \( k = 1 \), and the inverse is\n\n\[ \n{X}_{1} = {U}_{1},\\;{X}_{2} = {U}_{1} + {U}_{2},\\;{X}_{3} = {U}_{1} + {U}_{2} + {U... | Yes |
Lemma 4.7.1 Let \( a \) and \( b \) be any positive numbers, and let \( p \) and \( q \) be any positive numbers (necessarily greater than 1) satisfying\n\n(4.7.1)\n\n\[ \frac{1}{p} + \frac{1}{q} = 1 \]\n\nThen\n\n(4.7.2)\n\n\[ \frac{1}{p}{a}^{p} + \frac{1}{q}{b}^{q} \geq {ab} \]\n\nwith equality if and only if \( {a}^... | Proof: Fix \( b \), and consider the function\n\n\[ g\left( a\right) = \frac{1}{p}{a}^{p} + \frac{1}{q}{b}^{q} - {ab}. \]\n\nTo minimize \( g\left( a\right) \), differentiate and set equal to 0 :\n\n\[ \frac{d}{da}g\left( a\right) = 0 \Rightarrow {a}^{p - 1} - b = 0 \Rightarrow b = {a}^{p - 1}. \]\n\nA check of the sec... | Yes |
Theorem 4.7.2 (Hölder’s Inequality) Let \( X \) and \( Y \) be any two random variables, and let \( p \) and \( q \) satisfy (4.7.1). Then\n\n\[ \left| {\mathrm{E}{XY}}\right| \leq \mathrm{E}\left| {XY}\right| \leq {\left( \mathrm{E}{\left| X\right| }^{p}\right) }^{1/p}{\left( \mathrm{E}{\left| Y\right| }^{q}\right) }^... | Proof: The first inequality follows from \( - \left| {XY}\right| \leq {XY} \leq \left| {XY}\right| \) and Theorem 2.2.5. To prove the second inequality, define\n\n\[ a = \frac{\left| X\right| }{{\left( \mathrm{E}{\left| X\right| }^{p}\right) }^{1/p}}\;\text{ and }\;b = \frac{\left| Y\right| }{{\left( \mathrm{E}{\left| ... | Yes |
Theorem 4.7.3 (Cauchy–Schwarz Inequality) For any two random variables \( X \) and \( Y \) , | \[ \left| {\mathrm{E}{XY}}\right| \leq \mathrm{E}\left| {XY}\right| \leq {\left( \mathrm{E}{\left| X\right| }^{2}\right) }^{1/2}{\left( \mathrm{E}{\left| Y\right| }^{2}\right) }^{1/2}. \] | Yes |
Example 4.7.4 (Covariance inequality) If \( X \) and \( Y \) have means \( {\mu }_{X} \) and \( {\mu }_{Y} \) and variances \( {\sigma }_{X}^{2} \) and \( {\sigma }_{Y}^{2} \), respectively, we can apply the Cauchy-Schwarz Inequality to get | \[ \mathrm{E}\left| {\left( {X - {\mu }_{X}}\right) \left( {Y - {\mu }_{Y}}\right) }\right| \leq {\left\{ \mathrm{E}{\left( X - {\mu }_{X}\right) }^{2}\right\} }^{1/2}{\left\{ \mathrm{E}{\left( Y - {\mu }_{Y}\right) }^{2}\right\} }^{1/2}. \]\n\nSquaring both sides and using statistical notation, we have\n\n\[ {\left( \... | Yes |
Theorem 4.7.5 (Minkowski’s Inequality) Let \( X \) and \( Y \) be any two random variables. Then for \( 1 \leq p < \infty \) ,\n\n\[{\left\lbrack \mathrm{E}{\left| X + Y\right| }^{p}\right\rbrack }^{1/p} \leq {\left\lbrack \mathrm{E}{\left| X\right| }^{p}\right\rbrack }^{1/p} + {\left\lbrack \mathrm{E}{\left| Y\right| ... | Proof: Write\n\n\[ \mathrm{E}{\left| X + Y\right| }^{p} = \mathrm{E}\left( {\left| {X + Y}\right| {\left| X + Y\right| }^{p - 1}}\right) \]\n\n\[ \leq \mathrm{E}\left( {\left| X\right| {\left| X + Y\right| }^{p - 1}}\right) + \mathrm{E}\left( {\left| Y\right| {\left| X + Y\right| }^{p - 1}}\right) ,\]\n\nwhere we have ... | Yes |
Theorem 4.7.7 (Jensen’s Inequality) For any random variable \( X \), if \( g\left( x\right) \) is a convex function, then\n\n\[ \mathrm{E}g\left( X\right) \geq g\left( {\mathrm{E}X}\right) \]\n\nEquality holds if and only if, for every line \( a + {bx} \) that is tangent to \( g\left( x\right) \) at \( x = {EX} \) , \(... | Proof: To establish the inequality, let \( l\left( x\right) \) be a tangent line to \( g\left( x\right) \) at the point \( g\left( {\mathrm{E}X}\right) \) . (Recall that \( \mathrm{E}X \) is a constant.) Write \( l\left( x\right) = a + {bx} \) for some \( a \) and \( b \) . The situation is illustrated in Figure 4.7.2.... | Yes |
Example 5.1.2 (Sample pdf-exponential) Let \( {X}_{1},\ldots ,{X}_{n} \) be a random sample from an exponential \( \left( \beta \right) \) population. Specifically, \( {X}_{1},\ldots ,{X}_{n} \) might correspond to the times until failure (measured in years) for \( n \) identical circuit boards that are put on test and... | \[ = {\int }_{2}^{\infty }\ldots {\int }_{2}^{\infty }\mathop{\prod }\limits_{{i = 1}}^{n}\frac{1}{\beta }{e}^{-{x}_{i}/\beta }d{x}_{1}\cdots d{x}_{n} \]\n\n\[ = {e}^{-2/\beta }{\int }_{2}^{\infty }\cdots {\int }_{2}^{\infty }\mathop{\prod }\limits_{{i = 2}}^{n}\frac{1}{\beta }{e}^{-{x}_{i}/\beta }d{x}_{2}\cdots d{x}_{... | Yes |
Example 5.1.3 (Finite population model) As an example of an approximate calculation using independence, suppose \( \{ 1,\ldots ,{1000}\} \) is the finite population, so \( N = {1000} \) . A sample of size \( n = {10} \) is drawn without replacement. What is the probability that all ten sample values are greater than 20... | If \( {X}_{1},\ldots ,{X}_{10} \) were mutually independent we would have\n\n\[ P\left( {{X}_{1} > {200},\ldots ,{X}_{10} > {200}}\right) = P\left( {{X}_{1} > {200}}\right) \cdots \cdot P\left( {{X}_{10} > {200}}\right) \]\n\n\[ = {\left( \frac{800}{1000}\right) }^{10} = {.107374} \]\n\nTo calculate this probability ex... | Yes |
Theorem 5.2.4 Let \( {x}_{1},\ldots ,{x}_{n} \) be any numbers and \( \bar{x} = \left( {{x}_{1} + \cdots + {x}_{n}}\right) /n \) . Then a. \( \mathop{\min }\limits_{a}\mathop{\sum }\limits_{{i = 1}}^{n}{\left( {x}_{i} - a\right) }^{2} = \mathop{\sum }\limits_{{i = 1}}^{n}{\left( {x}_{i} - \bar{x}\right) }^{2} \) , b. \... | Proof: To prove part (a), add and subtract \( \bar{x} \) to get\n\n\[ \mathop{\sum }\limits_{{i = 1}}^{n}{\left( {x}_{i} - a\right) }^{2} = \mathop{\sum }\limits_{{i = 1}}^{n}{\left( {x}_{i} - \bar{x} + \bar{x} - a\right) }^{2} \]\n\n\[ = \mathop{\sum }\limits_{{i = 1}}^{n}{\left( {x}_{i} - \bar{x}\right) }^{2} + 2\mat... | Yes |
Lemma 5.2.5 Let \( {X}_{1},\ldots ,{X}_{n} \) be a random sample from a population and let \( g\left( x\right) \) be a function such that \( \mathrm{E}g\left( {X}_{1}\right) \) and \( \operatorname{Var}g\left( {X}_{1}\right) \) exist. Then\n\n\( \left( {5.2.1}\right) \)\n\n\[ \mathrm{E}\left( {\mathop{\sum }\limits_{{i... | Proof: To prove (5.2.1), note that\n\n\[ \mathrm{E}\left( {\mathop{\sum }\limits_{{i = 1}}^{n}g\left( {X}_{i}\right) }\right) = \mathop{\sum }\limits_{{i = 1}}^{n}\mathrm{E}g\left( {X}_{i}\right) = n\left( {\mathrm{E}g\left( {X}_{1}\right) }\right) .\n\nSince the \( {X}_{i} \) s are identically distributed, the second ... | Yes |
Theorem 5.2.6 Let \( {X}_{1},\ldots ,{X}_{n} \) be a random sample from a population with mean \( \mu \) and variance \( {\sigma }^{2} < \infty \) . Then\n\na. \( \mathrm{E}\bar{X} = \mu \) ,\n\nb. Var \( \bar{X} = \frac{{\sigma }^{2}}{n} \)\n\nc. \( \mathrm{E}{S}^{2} = {\sigma }^{2} \) . | Proof: To prove (a), let \( g\left( {X}_{i}\right) = {X}_{i}/n \), so \( \mathrm{E}g\left( {X}_{i}\right) = \mu /n \) . Then, by Lemma 5.2.5,\n\n\[ \mathrm{E}\bar{X} = \mathrm{E}\left( {\frac{1}{n}\mathop{\sum }\limits_{{i = 1}}^{n}{X}_{i}}\right) = \frac{1}{n}\mathrm{E}\left( {\mathop{\sum }\limits_{{i = 1}}^{n}{X}_{i... | Yes |
Example 5.2.8 (Distribution of the mean) Let \( {X}_{1},\ldots ,{X}_{n} \) be a random sample from a \( \mathrm{n}\left( {\mu ,{\sigma }^{2}}\right) \) population. Then the mgf of the sample mean is | \[ {M}_{\bar{X}}\left( t\right) = {\left\lbrack \exp \left( \mu \frac{t}{n} + \frac{{\sigma }^{2}{\left( t/n\right) }^{2}}{2}\right) \right\rbrack }^{n} \] \[ = \exp \left( {n\left( {\mu \frac{t}{n} + \frac{{\sigma }^{2}{\left( t/n\right) }^{2}}{2}}\right) }\right) = \exp \left( {{\mu t} + \frac{\left( {{\sigma }^{2}/n... | Yes |
Theorem 5.2.9 If \( X \) and \( Y \) are independent continuous random variables with pdfs \( {f}_{X}\left( x\right) \) and \( {f}_{Y}\left( y\right) \), then the pdf of \( Z = X + Y \) is\n\n\[ \left( {5.2.3}\right) \]\n\n\[ {f}_{Z}\left( z\right) = {\int }_{-\infty }^{\infty }{f}_{X}\left( w\right) {f}_{Y}\left( {z -... | Proof: Let \( W = X \) . The Jacobian of the transformation from \( \left( {X, Y}\right) \) to \( \left( {Z, W}\right) \) is 1 . So using (4.3.2), we obtain the joint pdf of \( \left( {Z, W}\right) \) as\n\n\[ {f}_{Z, W}\left( {z, w}\right) = {f}_{X, Y}\left( {w, z - w}\right) = {f}_{X}\left( w\right) {f}_{Y}\left( {z ... | Yes |
Let \( U \) and \( V \) be independent Cauchy random variables, \( U \sim \operatorname{Cauchy}\left( {0,\sigma }\right) \) and \( V \sim \operatorname{Cauchy}\left( {0,\tau }\right) \) ; that is,\n\n\[ \n{f}_{U}\left( u\right) = \frac{1}{\pi \sigma }\frac{1}{1 + {\left( u/\sigma \right) }^{2}},\;{f}_{V}\left( v\right)... | This integral is somewhat involved but can be solved by a partial fraction decomposition and some careful antidifferentiation (see Exercise 5.7). The result is \n\n\( \left( {5.2.5}\right) \)\n\n\[ \n{f}_{Z}\left( z\right) = \frac{1}{\pi \left( {\sigma + \tau }\right) }\frac{1}{1 + {\left( z/\left( \sigma + \tau \right... | No |
Theorem 5.3.1 Let \( {X}_{1},\ldots ,{X}_{n} \) be a random sample from a \( \mathrm{n}\left( {\mu ,{\sigma }^{2}}\right) \) distribution, and let \( \dot{X} = \left( {1/n}\right) \mathop{\sum }\limits_{{i = 1}}^{n}{X}_{i} \) and \( {S}^{2} = \left\lbrack {1/\left( {n - 1}\right) }\right\rbrack \mathop{\sum }\limits_{{... | Proof: First note that, from Section 3.5 on location-scale families, we can assume, without loss of generality, that \( \mu = 0 \) and \( \sigma = 1 \) . (Also see the discussion preceding Theorem 5.2.11.) Furthermore, part (b) has already been established in Example 5.2.8, leaving us to prove parts (a) and (c).\n\nTo ... | Yes |
Lemma 5.3.2 (Facts about chi squared random variables) We use the notation \( {\chi }_{p}^{2} \) to denote a chi squared random variable with \( p \) degrees of freedom.\n\na. If \( Z \) is a \( n\left( {0,1}\right) \) random variable, then \( {Z}^{2} \sim {\chi }_{1}^{2} \) ; that is, the square of a standard normal r... | Proof: We have encountered these facts already. Part (a) was established in Example 2.1.7. Part (b) is a special case of Example 4.6.8, which has to do with sums of independent gamma random variables. Since a \( {\chi }_{p}^{2} \) random variable is a gamma \( \left( {p/2,2}\right) \) , application of the example gives... | Yes |
Lemma 5.3.3 Let \( {X}_{j} \sim \mathrm{n}\left( {{\mu }_{j},{\sigma }_{j}^{2}}\right), j = 1,\ldots, n \), independent. For constants \( {a}_{ij} \) and \( {b}_{rj}\left( {j = 1,\ldots, n;i = 1,\ldots, k;r = 1,\ldots, m}\right) \), where \( k + m \leq n \), define\n\n\[ \n{U}_{i} = \mathop{\sum }\limits_{{j = 1}}^{n}{... | Proof: It is sufficient to prove the lemma for \( {\mu }_{i} = 0 \) and \( {\sigma }_{i}^{2} = 1 \), since the general statement of the lemma then follows quickly. Furthermore, the implication from independence to 0 covariance is immediate (Theorem 4.5.5) and the expression for the covariance is easily verified (Exerci... | Yes |
Let \( {X}_{1},\ldots ,{X}_{n} \) be a random sample from a \( \mathrm{n}\left( {{\mu }_{X},{\sigma }_{X}^{2}}\right) \) population, and let \( {Y}_{1},\ldots ,{Y}_{m} \) be a random sample from an independent \( n\left( {{\mu }_{Y},{\sigma }_{Y}^{2}}\right) \) population. If we were interested in comparing the variabi... | \[ \frac{{S}_{X}^{2}/{S}_{Y}^{2}}{{\sigma }_{X}^{2}/{\sigma }_{Y}^{2}} = \frac{{S}_{X}^{2}/{\sigma }_{X}^{2}}{{S}_{Y}^{2}/{\sigma }_{Y}^{2}} \] Examination of (5.3.8) shows us how the \( F \) distribution is derived. The ratios \( {S}_{X}^{2}/{\sigma }_{X}^{2} \) and \( {S}_{Y}^{2}/{\sigma }_{Y}^{2} \) are each scaled ... | Yes |
To see how the \( F \) distribution may be used for inference about the true ratio of population variances, consider the following. The quantity \( \left( {{S}_{X}^{2}/{\sigma }_{X}^{2}}\right) /\left( {{S}_{Y}^{2}/{\sigma }_{Y}^{2}}\right) \) has an \( {F}_{n - 1, m - 1} \) distribution. (In general, we use the notati... | We can calculate\n\n\[ \mathrm{E}{F}_{n - 1, m - 1} = \mathrm{E}\left( \frac{{\chi }_{n - 1}^{2}/\left( {n - 1}\right) }{{\chi }_{m - 1}^{2}/\left( {m - 1}\right) }\right) \;\text{ (by definition) } \]\n\n\[ = \mathrm{E}\left( \frac{{\chi }_{n - 1}^{2}}{n - 1}\right) \mathrm{E}\left( \frac{m - 1}{{\chi }_{m - 1}^{2}}\r... | Yes |
Theorem 5.4.3 Let \( {X}_{1},\ldots ,{X}_{n} \) be a random sample from a discrete distribution with \( {pmf}{f}_{X}\left( {x}_{i}\right) = {p}_{i} \), where \( {x}_{1} < {x}_{2} < \cdots \) are the possible values of \( X \) in ascending order. Define\n\n\[ \n{P}_{0} = 0 \]\n\n\[ \n{P}_{1} = {p}_{1} \]\n\n\[ \n{P}_{2}... | Proof: Fix \( i \), and let \( Y \) be a random variable that counts the number of \( {X}_{1},\ldots ,{X}_{n} \)\n\nthat are less than or equal to \( {x}_{i} \) . For each of \( {X}_{1},\ldots ,{X}_{n} \), call the event \( \left\{ {{X}_{j} \leq {x}_{i}}\right\} \) a \ | No |
Theorem 5.4.4 Let \( {X}_{\left( 1\right) },\ldots ,{X}_{\left( n\right) } \) denote the order statistics of a random sample, \( {X}_{1},\ldots ,{X}_{n} \), from a continuous population with cdf \( {F}_{X}\left( x\right) \) and pdf \( {f}_{X}\left( x\right) \). Then the pdf of \( {X}_{\left( j\right) } \) is\n\n\[ \n{f... | Proof: We first find the cdf of \( {X}_{\left( j\right) } \) and then differentiate it to obtain the pdf. As in Theorem 5.4.3, let \( Y \) be a random variable that counts the number of \( {X}_{1},\ldots ,{X}_{n} \) less than or equal to \( x \). Then, defining a \ | Yes |
Theorem 5.4.6 Let \( {X}_{\left( 1\right) },\ldots ,{X}_{\left( n\right) } \) denote the order statistics of a random sample, \( {X}_{1},\ldots ,{X}_{n} \), from a continuous population with cdf \( {F}_{X}\left( x\right) \) and pdf \( {f}_{X}\left( x\right) \) . Then the joint pdf of \( {X}_{\left( i\right) } \) and \(... | \[ {f}_{{X}_{\left( i\right) },{X}_{\left( j\right) }}\left( {u, v}\right) = \frac{n!}{\left( {i - 1}\right) !\left( {j - 1 - i}\right) !\left( {n - j}\right) !}{f}_{X}\left( u\right) {f}_{X}\left( v\right) {\left\lbrack {F}_{X}\left( u\right) \right\rbrack }^{i - 1} \times {\left\lbrack {F}_{X}\left( v\right) - {F}_{X... | Yes |
Let \( {X}_{1},\ldots ,{X}_{n} \) be iid uniform \( \left( {0, a}\right) \) and let \( {X}_{\left( 1\right) },\ldots ,{X}_{\left( n\right) } \) denote the order statistics. The range was earlier defined as \( R = {X}_{\left( n\right) } - {X}_{\left( 1\right) } \). The midrange, a measure of location like the sample med... | From (5.4.7) we have that\n\n\[ \n{f}_{{X}_{\left( 1\right) },{X}_{\left( n\right) }}\left( {{x}_{1},{x}_{n}}\right) = \frac{n\left( {n - 1}\right) }{{a}^{2}}{\left( \frac{{x}_{n}}{a} - \frac{{x}_{1}}{a}\right) }^{n - 2} \n\]\n\n\[ \n= \frac{n\left( {n - 1}\right) {\left( {x}_{n} - {x}_{1}\right) }^{n - 2}}{{a}^{n}},\;... | Yes |
Theorem 5.5.2 (Weak Law of Large Numbers) Let \( {X}_{1},{X}_{2},\ldots \) be iid random variables with \( \mathrm{E}{X}_{i} = \mu \) and \( \operatorname{Var}{X}_{i} = {\sigma }^{2} < \infty \) . Define \( {\bar{X}}_{n} = \left( {1/n}\right) \mathop{\sum }\limits_{{i = 1}}^{n}{X}_{i} \) . Then, for every \( \epsilon >... | Proof: The proof is quite simple, being a straightforward application of Chebychev's Inequality. We have, for every \( \epsilon > 0 \) , \[ P\left( {\left| {{\bar{X}}_{n} - \mu }\right| \geq \epsilon }\right) = P\left( {{\left( {\bar{X}}_{n} - \mu \right) }^{2} \geq {\epsilon }^{2}}\right) \leq \frac{\mathrm{E}{\left( ... | Yes |
Example 5.5.3 (Consistency of \( {S}^{2} \) ) Suppose we have a sequence \( {X}_{1},{X}_{2},\ldots \) of iid random variables with \( \mathrm{E}{X}_{i} = \mu \) and Var \( {X}_{i} = {\sigma }^{2} < \infty \) . If we define\n\n\[ \n{S}_{n}^{2} = \frac{1}{n - 1}\mathop{\sum }\limits_{{i = 1}}^{n}{\left( {X}_{i} - {\bar{X... | Using Chebychev’s Inequality, we have\n\n\[ \nP\left( {\left| {{S}_{n}^{2} - {\sigma }^{2}}\right| \geq \epsilon }\right) \leq \frac{\mathrm{E}{\left( {S}_{n}^{2} - {\sigma }^{2}\right) }^{2}}{{\epsilon }^{2}} = \frac{\operatorname{Var}{S}_{n}^{2}}{{\epsilon }^{2}}\n\]\n\nand thus, a sufficient condition that \( {S}_{n... | Yes |
If \( {S}_{n}^{2} \) is a consistent estimator of \( {\sigma }^{2} \), then by Theorem 5.5.4, the sample standard deviation \( {S}_{n} = \sqrt{{S}_{n}^{2}} = h\left( {S}_{n}^{2}\right) \) is a consistent estimator of \( \sigma \) . | Note that \( {S}_{n} \) is, in fact, a biased estimator of \( \sigma \) (see Exercise 5.11), but the bias disappears asymptotically. | No |
Example 5.5.7 (Almost sure convergence) Let the sample space \( S \) be the closed interval \( \left\lbrack {0,1}\right\rbrack \) with the uniform probability distribution. Define random variables \( {X}_{n}\left( s\right) = s + {s}^{n} \) and \( X\left( s\right) = s \) . For every \( s \in \lbrack 0,1),{s}^{n} \righta... | But since the convergence occurs on the set \( \lbrack 0,1) \) and \( P\left( {\lbrack 0,1}\right) ) = 1,{X}_{n} \) converges to \( X \) almost surely. | Yes |
If \( {X}_{1},{X}_{2},\ldots \) are iid uniform \( \left( {0,1}\right) \) and \( {X}_{\left( n\right) } = \mathop{\max }\limits_{{1 \leq i \leq n}}{X}_{i} \), let us examine if (and to where) \( {X}_{\left( n\right) } \) converges in distribution. | As \( n \rightarrow \infty \), we expect \( {X}_{\left( n\right) } \) to get close to 1 and, as \( {X}_{\left( n\right) } \) must necessarily be less than 1, we have for any \( \varepsilon > 0 \), \[ P\left( {\left| {{X}_{\left( n\right) } - 1}\right| \geq \varepsilon }\right) = P\left( {{X}_{\left( n\right) } \geq 1 +... | Yes |
Theorem 5.5.12 If the sequence of random variables, \( {X}_{1},{X}_{2},\ldots \) , converges in probability to a random variable \( X \) , the sequence also converges in distribution to \( X \) . | See Exercise 5.40 for a proof. | No |
Theorem 5.5.14 (Central Limit Theorem) Let \( {X}_{1},{X}_{2},\ldots \) be a sequence of iid random variables whose mgfs exist in a neighborhood of 0 (that is, \( {M}_{{X}_{i}}\left( t\right) \) exists for \( \left| t\right| < h \), for some positive \( h \) ). Let \( \mathrm{E}{X}_{i} = \mu \) and \( \operatorname{Var... | Proof of Theorem 5.5.14: We will show that, for \( \left| t\right| < h \), the mgf of \( \sqrt{n}\left( {{\bar{X}}_{n} - \mu }\right) /\sigma \) converges to \( {e}^{{t}^{2}/2} \), the mgf of a \( \mathrm{n}\left( {0,1}\right) \) random variable.\n\nDefine \( {Y}_{i} = \left( {{X}_{i} - \mu }\right) /\sigma \), and let... | Yes |
Suppose we observe \( {X}_{1},{X}_{2},\ldots ,{X}_{n} \) independent Bernoulli \( \left( p\right) \) random variables. The typical parameter of interest is \( p \), the success probability, but another popular parameter is \( \frac{p}{1 - p} \), the odds. For example, if the data represent the outcomes of a medical tre... | Intuition abandons us, and exact calculation is relatively hopeless, so we have to rely on an approximation. The Delta Method will allow us to obtain reasonable, approximate answers to our questions.\n\nOne method of proceeding is based on using a Taylor series approximation, which allows us to approximate the mean and... | Yes |
Theorem 5.5.21 (Taylor) If \( {g}^{\left( r\right) }\left( a\right) = {\left. \frac{{d}^{r}}{d{x}^{r}}g\left( x\right) \right| }_{x = a} \) exists, then\n\n\[ \mathop{\lim }\limits_{{x \rightarrow a}}\frac{g\left( x\right) - {T}_{r}\left( x\right) }{{\left( x - a\right) }^{r}} = 0. \] | In general, we will not be concerned with the explicit form of the remainder. Since we are interested in approximations, we are just going to ignore the remainder. There are, however, many explicit forms, one useful one being\n\n\[ g\left( x\right) - {T}_{r}\left( x\right) = {\int }_{a}^{x}\frac{{g}^{\left( r + 1\right... | No |
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