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An) = 1 − exp(−λn). Define N = (An; n ≥ 1) independent exp sn. are ∞ n − λk 1 (b) Find E(s N ) and E(N ), when λn = a + log n. The probability of obtaining heads when a certain coin is tossed is p. The coin is tossed repeatedly until a sequence of three heads is obtained. If pn is the probability that this event occurs ... |
) denote for 1 ≤ n ≤ K − 1 the probability that he loses all his money, and let p(0) = 1, p(K ) = 0. Show that p(n) = 1 2 ( p(n − 1) + p(n + 1)); (1 ≤ n ≤ K − 1). 20 21 G(s) = k−1 n=0 p(n)sn then, provided s = 1, G(s) = 1 (1 − s)2 (1 − (2 − p(1))s + p(K − 1)s K +1). 22 Hence, or otherwise, show that p(1) = 1 − 1/K, p(K... |
X )/ log n → 1. Find the probability generating function of the distribution P(X = k) = λ λ(λ + 1... (λ + k − 1) (1 + a)kk! ; a 1 + a k > 0,P(X = 0) = λ. a 1 + a Let X and Y be independent Poisson random variables with parameters λ and µ respectively. Find the joint probability generating function of X − Y and X + Y. F... |
Y0) = (0, 0). Define T = min{n; X n +Yn = m}. Find the probability generating function of T. In Problem 30, if α1 = β1 and α2 = β2, show that E(X 2 n Also, in Problem 30, if α1 = β1 = α2 = β2 = 1 4, (a) Show that E(T ) = ∞. (b) Show that the point at which the walk hits x + y = m is a proper random variable. (c) Find i... |
; r ≥ 1. Show that κ1 = E(X ), κ2 = var (X ), and κ3 = µ3 − 3µ1µ2 + 2µ3 1. Show that the joint probability mass function. Show that X + Y has r th cumulant and κ (Y ). r r f (x, y λx µy; x ≥ 0, y ≥ 1 has joint p.g.f. G(s, t) = (1 − λ − µ)t 1 − λs − µt. What is cov (X, Y )? Let X m have generating function ( p/(1 − qs)... |
qEX k for k ≥ 1. If X has p.g.f. G(s), show that T (s) = EX (X − 1)... (X − k + 1) = kT (k−1)(1). s snP(X > n) = (1 − G(s))/(1 − s). Deduce that n 43 44 7 Continuous Random Variables He talks at random: sure the man is mad. W. Shakespeare, Henry VI 7.1 Density and Distribution √ Hitherto, we have assumed that a random... |
s of possible values of X. Happily, we already have such a function; recall the following: 287 288 7 Continuous Random Variables (1) (2) (3) Definition The distribution function F of the random variable X is the function F(x) = P(Ax ), where Ax is the event Ax = {ω : X (ω) ≤ x}, xR. We usually write (2) as F(x) = P(X ≤ ... |
be the perpendicular distance from Q to the base. What is the distribution function FX (x)? [See figure 7.2 for a sketch of the triangle]. Solution inside the triangle ABC. For reasons of symmetry, Let the height of the triangle AP be h. The event X > x occurs when Q lies P(Q ABC) = (area of ABC)/a = h − x h 2. Hence, ... |
F(x + hk) = lim k→∞ P(A(k)) A(k)) = P(Ax ) = F(x), as required. Finally, for (9), lim x→∞ F(x) = lim n→∞ P(An) = P() by (1.5.4). The last part is proved similarly. Although the distribution function has not played a very active role so far in this book, it now assumes a greater importance. One reason for this is Theor... |
4 Revisited: Uniform Density uniformly at random in [0, 1], then It follows that X has a density FX (x) = f X (x) = In Example 4, we found that if X was chosen. 0 < x < 1 otherwise. Figures 7.3 to 7.6 illustrate the density and distribution of random variables Y and Z, which are uniformly distributed on (a, b) a... |
−λx 0 if x > 0, x ≤ 0. F(x) = 1 − e−λx = x 0 λe−λvdv = x ∞ f (v)dv, and, for x < 0, F(x) = 0 = x −∞ f (v)dv. Hence, f (x) is a density of X. See Figures 7.7 and 7.8. Notice that F (0) does not exist, and also that the function f (x) = λe−λx x ≥ 0 x < 0 0 FZ(z) 1 0 a b c d z Figure 7.6 The distribution function of a ran... |
does not matter very much. 2 and write f (x) = 1 2 λe−λ|x| for all x. The Finally, we note the obvious facts that for any density f, (18) (19) and if P(|X | < ∞) = 1, then f (x) ≥ 0, ∞ −∞ f (v)dv = 1. It is straightforward to see that any function with these properties is the density of some random variable, and so an... |
find a constant b such that for all n > 0, + 0 g(v)dv < b < ∞, then n ∞ 0 exists by monotone convergence. g(v)dv = lim n→∞ n 0 g(v)dv (21) Example: Normal Density Show that − 1 2 x 2 for all x R f = c exp can be a density. Solution For any n > 1, n exp − 1 2 v2 −n + ∞ −∞ exp(− 1 2 dv < 2 1 0 dv + n 1 e−vdv < 2(1 + e−1)... |
−1 = gration by parts gives c−1 = (α − 1)! λα x α−1e−λx d x exists, and if α is a positive integer then repeated inte- λe−λvdv = (α − 1)! + ∞ 0 This density is known as the gamma density with parameters α and λ. When α is not an integer, the integral above defines a function of α known as the gamma function, and denoted... |
= P(g(X ) ≤ y) = f X (v)dv, C where C = {v : g(v) ≤ y}. Then, if F(y) is continuous and differentiable, we can go on to find the density of Y, if it exists. Here are some simple examples of this idea in practice. (2) Example Let X be uniformly distributed on (0, 1) with density f (x) = 1 if 0 < x < 1 otherwise. 0 If Y ... |
y ≤ 1. 7.2 Functions of Random Variables 299 Finally, if r = 0, then X r = 1, FY (y) is not continuous (having a jump from 0 to 1 at y = 1) and so Y does not have a density in this case. Obviously, Y is discrete, with P(Y = 1) = 1. (4) Example Let X have the standard normal distribution with density f (x) = (2π)− 1 2 ... |
case of (7.1.28). (8) Example: Inverse Functions Let X have distribution function F(x), where F(x) is continuous and strictly increasing. Let g(x) be a function satisfying F(g) = x. Because 300 7 Continuous Random Variables F(x) is continuous and strictly increasing, this defines g(x) uniquely for every x in (0, 1). Th... |
) is, in some sense, a rough approximation to the continuous random variable X. It is easy to get much better approximations as follows. (14) Example: Discrete Approximation As usual, X has density f (x); suppose also that X > 0. For fixed n, with 0 ≤ r ≤ 2n − 1 and k ≥ 0, define Sn(X ) = k + r 2−n if k + r 2−n ≤ X < k ... |
, 1). The word randomly appears in quotations because each ui is not really random. Because the machine was programmed to produce it, the outcome is known in advance, but such numbers behave for many practical purposes as though they were random. They are called pseudorandom numbers. Now if we have a pseudorandom numbe... |
. Because of the ready availability of large numbers of uniform pseudorandom numbers, interest is concentrated on finding transformations that then yield random variables of arbitrary type. We have seen several in Section 7.2. Here is another idea. Example: Composition (0, 1). Show how to simulate a random variable with... |
2π) 1 σ (2π) µ σ (2π) 1 2 + v exp(−(v − µ)2/(2σ 2))dv (v − µ) exp −∞ ∞ − 1 2 v − µ σ 2 2 dv − 1 2 v − µ σ u2 du + −∞ exp u exp 1 2 ∞ −∞ − 1 2 = 1 (2π) 1 2 dv µ (2π ) 1 2 ∞ exp −∞ − 1 2 u2 du on making the substitution u = (v − µ)/σ in both integrands. The first integrand is an odd function, so the integral over R is zer... |
� v=−∞ v f (v), in the discrete case. Of course, Definition 1 is much more than just a plausible analogy, but a complete account of expectation is well beyond our scope. However, we can use Example 7.2.14 to give a little more justification for Definition 1. Let k + r 2−n = a(k, r, n). Recall from (7.2.15) that |Sn − X | ... |
What is E(Y )? If we know the density of X, then we may be able to find E(Y ) by first discovering fY (y), if it exists. This is often an unattractive procedure. We may do much better to use the following theorem, which we state without proof. (9) (10) (11) (12) (13) 7.4 Expectation 305 Theorem f (x). Then Y has an expe... |
useful special case of Theorem 9. (14) Example Let the nonnegative random variable X have density f, and let g(X ) ≥ 0. Show that E(g(X )) = + ∞ 0 g(v) f (v)dv. Solution E(g(X )) = ∞ 0 ∞ P(g(X )) ≥ v) dv by (12) ∞ = 0 x:g(x)≥v f (x)d xdv = f (x) 0 0 g(x) ∞ dvd x = f (x)g(x)d x, 0 306 7 Continuous Random Variables as r... |
Then: (i) If g(X ) and h(X ) have finite mean, then E(g(X ) + h(X )) = E(g(X )) + E(h(X )). (ii) If P(a ≤ X ≤ b) = 1, then a ≤ E(X ) ≤ b. (iii) If h is nonnegative, then for a > 0, P(h(X ) ≥ a) ≤ E(h(X )/a). (iv) Jensen’s inequality If g is convex then E(g(X )) ≥ g(E(X )). Proof The proof is an exercise for you. When h... |
α > 0 and λ > 0 by (5) Hence, (α) = ∞ 0 x α−1λαe−λx d x. f (x) = λα (α) x α−1e−λx, x ≥ 0, is the density of a random variable x. Find E(etX ). Where does it exist? Solution E(etX ) = ∞ 0 etv λα (α) vα−1e−λvdv = λα (α) ∞ 0 vα−1e−(λ−t)vdv. The integral exists if λ > t, and then making the substitution (λ − t)v = u gives... |
MX (t)? After all, the probability generating function uniquely determines the corresponding mass function. Unfortunately, the answer is no in general because densities not uniquely determined by their moments do exist. However, none appear here; every density in this book is uniquely determined by its moment generati... |
n 3 2 = O(n2) with c = 2. Observe that this is an abuse of notation (= being the abused symbol) because it does not follow from these two examples that log n = n 3 2. Also, if h(n) = O(g(n)) and k(n) = O(g(n)), then h(n) + k(n) = O(g(n)). A similar definition holds for small values of the argument. Definition x → 0, if ... |
p) (npq) n 1 2 *. * n Next we expand the two exponential terms in (12) to give E(etYn ) = 1 + t 2 2n n− 3 2 + O n. Now we recall the useful result that says that, for constant a, 1 + a n lim n→∞ + o(n−1) n = ea. Applying this to (13) shows that E(etYn ) = e 1 2 t 2, lim n→∞ which is the m.g.f. of the standard normal d... |
, c). More generally, it is easy to see that a uniform random variable, constrained to lie in any subset A of its range, is uniformly distributed over the subset A. Because P(X ≤ x|A) is a distribution, it may have an expectation. For example, suppose that X has density f, and A is given by (2). Then, by (1), P(X ≤ x|A... |
) dv = 1 λ − s eλs − 1. 0 7.7 Ageing and Survival Many classic examples of continuous random variables arise as waiting times or survival times. For instance, the time until the cathode-ray tube in your television fails, the time until you are bitten by a mosquito after disembarking in the tropics, the time until a st... |
given that it has not failed by time t. Now 1 s FT |At (s + t) = (1 − F(t))−1 lim s→0 F(t + s) − F(t) s = f (t) 1 − F(t) 1 − F(t) = r (t). lim s→0. (5) Thus, r (t) may be thought of as the “intensity” of the probability that a device aged t will fail. (6) Example: Exponential Life If T has an exponential density, then... |
i) If r (t) increases, then T is (or has) increasing failure rate, denoted by IFR. (ii) If H (t) (iii) If for all s ≥ 0, t ≥ 0, increases, then T is (or has) increasing failure rate average, denoted by IFRA. t H (s + t) ≥ H (s) + H (t), then T is new better than used, denoted by NBU. 314 7 Continuous Random Variables (... |
FY (x) for all x, then X is said to be stochastically larger than Y. Now we can supply a connection with the ideas of the preceding section (7.7). (4) Example If T is a random variable with residual life R(s), s > 0, show that T has increasing failure rate if and only if R(s) is stochastically larger than R(t) for all... |
��ning ourselves to two dimensions for definiteness, suppose a point Q is picked at random in a region R of area |R|. Then it is natural to let the probability P(S), that Q lies in a set S ⊆ R, be given by (1) P(S) = |S| |R|, where, now, |S| denotes the area of S. It follows from the properties of area that P(.) has the... |
points are picked independently, then X n is a binomial random variable with parameters n and P(A1), and we have shown that as n → ∞, for > 0, + 1 P |n−1 X n − 1 0 f (x)d x| > → 0. + 1 0 f (x)d x. In practice, one would This therefore offers a method for evaluating the integral be unlikely to use this method in one di... |
, surrounded by an annulus A of width h. Then, if Q1 and Q2 are dropped at random on to the disc of radius x + h, we have (using independence and the properties of the uniform density) that (10) P(Q1 ∈ D ∩ Q2 ∈ D) = Also, 2 π x 2 π(x + h)2 = 1 − 4h x + o(h). P(Q1 ∈ D ∩ Q2 ∈ A) = = 2h x and P(Q1 ∈ A ∩ Q2 ∈ A) = o(h). He... |
can find the density of L. The next natural step is to pick lines (or other objects) at random and ask how they divide up the region R in random tessellations or coverings. This is well beyond our scope, but the trivial Example 7.18 illustrates some of the problems. 7.10 Review and Checklist for Chapter 7 We introduced... |
X > x)d x. 0 If random variables X and Y are such that Y = g(X ) and X is continuous, then Y has an expected value if + ∞ −∞ |g(x)| f X (x)d x < ∞ and ∞ EY = Eg(X ) = g(x) f X (x)d x. −∞ Moments: In particular, if g(X ) = X r, this yields the r th moment µr of X. When X > 0, ∞ EX r = 0 r x r −1P(X > x)d x. When g(X ) =... |
−a) λ/(λ − t) exp (µt + 1 2 σ 2t 2) r λ λ−t λ2 λ2−t 2 Table 7.1 gives some useful continuous random variables with their elementary properties. Checklist of Terms for Chapter 7 7.1 distribution function density standard normal density φ(x) mixture 7.2 functions inverse function 7.3 simulation composition 7.4 expected v... |
st part or by inspection, we see that g1(U ) = 1 2 (1 − U 2) is a random variable with density f1(x). (To see this, just make the simple calculation P(g1(U ) ≤ x) = P (1 − U 2) ≤ x = P(U ≥ (1 − 2x) 1 2 ) = 1 − (1 − 2x) 1 2, 1 2 and differentiate to get the density f1.) Likewise, f2(x) = is a density function, and 2 = |... |
�� (2U ) 1 2 2 − (2 − 2U ) 1 2 if U < 1 2 if U ≥ 1 2. Show that X has a triangular density on [0, 2]. Worked Examples and Exercises 323 Find the densities of: (7) Exercise (a) tan(πU ). (b) tan( π 2 U ). (1) (2) (3) (4) (5) Let 7.12 Example: Normal Distribution φ(x) = (2π)− 1 2 e−x 2/2; (x) = x −∞ φ(u)du. (a) Define the... |
φ(u) u − ∞ x φ(u) u φ(u) u2 du = = φ(x) x φ(x) x ≥ φ(x) + − 1 x x φ(u) u3 du ∞ +! x φ(x) x 3 − 1 x 3 x. du by (4) on integrating by parts, by (4), 3φ(u) u4 du on integrating by parts, Remark For large x, these bounds are clearly tight. (7) Exercise Show that they are orthogonal with respect to φ(x) over R, which is to... |
2) = d 2/a2. Worked Examples and Exercises 325 Figure 7.9 Bertrand’s paradox. In this case, X < √ 3a because O P > 1 2 a. √ Now X > of the disc. This occurs (see Figure 7.9) if and only if OP has length less than 1 3a if and only if the chord R Q subtends an angle greater than 2π 3 at the centre 2 a. Hence, by (1), P(X... |
If the shop sells every bit before the weekend, then further customers that week are supplied by post at the end of the week; this costs p per bit, due to postage, packing, paperwork, and other penalties, and p > c. (iv) The demand Z for bits each week is a random variable with density f (z) and distri- bution F(z) wh... |
− cx, and to deliver ˆy − x when x < ˆy and Now, if we set g(x) = λ(x) + cx, we have λ(x) > k + c ˆy + λ( ˆy) − cx. and g(x) = c − p + (h + p)F(x) g(x) = (h + p)F(x) ≥ 0. Because g(0) < 0 and g( ˆy) = 0, it follows that there is a unique point ˆx such that g( ˆx) = λ( ˆx) + c ˆx = k + c ˆy + λ( ˆy). Hence, the optimal... |
not answered by then. Show that to minimize the expected time you spend listening to the ringing tone, you should choose s to be the unique positive root s0 of log s = (s + 1)(s − 2). Solution not, we have (a) Let R be the ringing time. Conditioning on whether the clerk is there or (2) P(R > s) = 1 2 P(R > s| absent) ... |
+ s−1 log(1 + s)). 7.16 Example: Pirates Expensive patented (or trade marked) manufactures are often copied and the copies sold as genuine. You are replacing part of your car; with probability p you buy a pirate part, with probability 1 − p a genuine part. In each case, lifetimes are exponential, pirate parts with par... |
FR. Exercise random variable with density f (λ). Let M(t) be the continuous mixture Let T" be a family of random variables indexed by a parameter ", where " is a ∞ M(t) = P(T" ≤ t) = FTλ (t) f (λ)dλ. 0 330 7 Continuous Random Variables Show that if FTλ (t) is DFR for all λ, then M(t) is DFR. [Hint: The Cauchy–Schwarz i... |
FR? A point P is chosen at random along a rod of length l. 7.18 Example: Triangles (a) The rod is bent at P to form a right angle, thus forming the two shorter sides of a right-angled triangle. Let! be the smallest angle in this triangle. Find E(tan!) and E(cot!). (b) The rod is now cut into two pieces at P. A piece is... |
2 (1 − X ) √ 2 − 1. 332 Hence, 7 Continuous Random Variables P (no obtuse angle | the triangle exists) = P {X < √ 2 − 1) = 2). = P(X < P X < 1 2 (3) (4) Exercise What is the distribution of the length X ∧ (1 − X ) of the shortest side of the triangle? Exercise The longest side is X ∨ (1 − X ). Show that E {X ∧ (1 − X ... |
(8) Worked Examples and Exercises 333 (b) Let the integrand in (2) be f (x, u). Then for x − 1) log(1 + ux 1 1 log f (x, u) = −ux → − 1 2 u2 as x → ∞. u2 − ux − 1 2 + O(x −1) Now, if we were justified in saying that lim x→∞ f (x, u)du = lim x→∞ f (x, u) du, then (3) would follow from (2), (4), and (5). However, it is a... |
n. Show that if n → ∞ and b → ∞ in 2, then Pb → e−y2. such a way that b = yn 1 Exercise: de Moivre–Laplace Theorem Let Sn be binomial with parameters n and p. Define 2 = (2π) 1 2. Yn = Sn − np (npq) 1 2, q = 1 − p, and yk = k − np (npq) 1 2. Show that as n → ∞ P(Sn = k) = n 2πk(n − k) 1 2 k np k nq n − k n−k (1 + o(1)).... |
) can f be a density function? Let X have distribution F(x). Show that P(X = x) > 0 if and only if F(x) is discontinuous at x. + va−1(1 − v)b−1dv; a > 0, b > 0. The beta disThe beta function B(a, b) is given by B(a, b) = tribution has density 1 0 f (x) = 1 B(a, b) x a−1(1 − x)b−1 for 0 < x < 1. If X has the beta distri... |
What is the density and expectation of the period? Let X have density f (x). Construct a simple random variable Sn(X ) such that given > 0, P(|Sn(X ) − X | > ) < 2−n. (Assume X is proper.) If X is exponentially distributed find the m.g.f. of X, E(et X ). A point Q is chosen at random inside an equilateral triangle of u... |
∂b √ π 23 24 25 What is the moment generating function of the two-sided exponential density? Where is it defined? Let U be uniform on (0, 1). Show that, if [a] denotes the integer part of a, and 0 < p < 1, 26 X = 1 + " # log U log(1 − p) 27 28 29 30 has a geometric distribution. Let U be uniform on (0, 1). Show how to ... |
I (X > x) to show that, for a random variable X that is nonnegative, (a) EX = P(X > x) d x. ∞ ∞ 0 (b) EX r = 0 r x r −1P(X > x) d x. ∞ (c) Eeθ X = 1 + θ (d) When X ≥ 0 is integer valued, eθ x P(X > x) d x. 0 ∞ k=0 skP(X > k) = 1 − G X (s) 1 − s. 36 Let X be a standard normal random variable with density φ(x) and distr... |
. Let F(x, y) be a joint distribution. Suppose that ∂ 2 F Definition ∂ x∂ y exists and is nonnegative, except possibly on a finite collection of lines in R2. Suppose further that the function f (x, y) defined by (1) (2) f (x, y∂ y 0 where this exists elsewhere, 337 338 8 Jointly Continuous Random Variables satisfies (3) F(... |
< x < a, 0 < y < b otherwise satisfies (3), and is the density of X and Y. It is uniformly distributed over the rectangle R. Furthermore, if A is a subset of R with area |A|, then using (7.8.1) (and a theorem about double integrals), we have P((X, Y ) ∈ A) = |A| ab = f (x, y) d xd y. (x,y)∈A In fact, a version of the u... |
( 1 2 (1, 1). Hence,, 1 2 ), (1, 0), P(X + Y > 1) = 8 1 x 1 2 1−x x y dy d x = 5 6. Finally, F(x, y) = y x 0 v 8uv du dv = 2x 2 y2 − y4. The geometric problems of Section 7.9 can now be reformulated and generalized in this new framework. Obviously, “picking a point Q at random in some region R” is what we would now de... |
dy = 1. cos θ = X 2 + (2 − X − Y )2 − Y 2 2X (2 − X − Y ) < 0, if θ is an obtuse angle. Hence, in this case, Y > X 2 − 2X + 2 2 − X = g(X ), say. Now g(x) ≥ 1 − x (with equality only at x = 0). Hence, p0 is given by P(θ is obtuse) = P(Y > g(X )) = 1 1 0 g(x) f (x, y) d yd x = c 1 0 (1 − g(x)). (e) When a = 0, p0 = 2x ... |
F(x, y). (12) (13) (14) (15) (16) Here are some examples to illustrate these properties. Note that in future we will specify f (x, y) only where it is nonzero. (17) Example Verify that the function f (x, y) = 8x y for 0 < y < x < 1 is a density. For what value of c is f (x, y) = cx y for 0 < x < y < 1, a density? Find... |
�2) 2 y τ − ρx σ + x 2 σ 2 − ρ2x 2 σ 2 * dy Now setting y τ − = u, and recalling that ρx σ ∞ −∞ yields exp − u2 2(1 − ρ2) τ du = (2π (1 − ρ2)) 1 2 τ f X (x) = 1 (2π) 1 2 σ exp − x 2 2σ 2. This is the N (0, σ 2) density, and so f satisfies (12) and is nonnegative. It is therefore a density. Interchanging the roles of x a... |
T given by maps C one–one onto D, with inverse T −1 given by T (x, y) = (u(x, y), v(x, y)) T −1(u, v) = (x(u, v), y(u, v)), which maps D one–one onto C. We define the so-called Jacobian J as J (u, v) = ∂ x ∂u ∂ y ∂v − ∂ x ∂v ∂ y ∂u, where the derivatives are required to exist and be continuous in D. Then we have the fo... |
θ) is not uniform, as was f (x, y). 344 8 Jointly Continuous Random Variables (5) Example satisfying Let Q = (X, Y ) be uniformly distributed over the ellipse C with boundary of area |C|. What is P(X > Y, X > −Y )? x 2 a2 + y2 b2 = 1, Here the transformation x = ar cos θ and y = br sin θ maps the ellipse one– Solution... |
U and V are independent. To see this, just let A = (x: g(x) ≤ u) and B = (g: h(y) ≤ v), and the independence follows from (4) and (2). An important and useful converse is the following. 8.3 Independence 345 (6) Theorem If X and Y have density f (x, y), and for all x and y it is true that then X and Y are independent. ... |
Jointly Continuous Random Variables 2π f R(r ) = f (r, θ) dθ = 2r ; 0 ≤ r ≤ 1. 0 Hence, f (r, θ) = f!(θ) f R(r ), and so R and! are independent. Example: Bertrand’s Paradox Again Suppose we choose a random chord of a circle C radius a, as follows. A point P is picked at random (uniformly) inside C. Then a line through... |
are independent, they have joint density f (x, y) = k2 exp − 1 2 (x 2 + y2). Make the change of variables to polar coordinates, so that by Theorem 8.2.1 the random variables R = (X 2 + Y 2) 2 and! = tan−1(Y/ X ) have joint density 1 f (r, θ) = k2r exp − 1 2 r 2 for 0 ≤ r < ∞, 0 < θ ≤ 2π. Hence, R has density and! has ... |
Y are independent, then ∞ f Z (z) = −∞ f X (u) fY (z − u) du. Proof and Y are independent. Turning to the proof of (1), we give two methods of solution. First notice that by (8.3.3) the result (2) follows immediately from (1) when X I Let A be the region in which u + v ≤ z. Then P(Z ≤ z) = = (u,v)∈A ∞ z −∞ −∞ f (u, v)... |
the exponent in the integrand we have, after a little manipulation, −1 2(1 − ρ2) u2 a2σ 2 = −1 2(1 − ρ2) − 2ρu(z − u) abσ z − u)2 b2τ 2 − (1 − ρ2) a2b2σ 2τ 2 + z2 α, where α = 1 a2σ 2 + 2ρ abσ τ + 1 b2τ 2, and β = ρ abσ τ + 1 b2τ 2. Setting u = v + β α z in the integrand, we evaluate ∞ exp − −∞ αv2 2(1 − ρ2) dv = 2π (... |
X Y have joint density J (u, z) = = u−18) f (u, z) = 1 |u| The result (7) follows immediately as it is the marginal density of Z obtained from f (u, z). Alternatively, it is possible to derive the result directly by the usual plod, as follows: u, z u f. P(X Y ≤ z) = P 0 X > 0, Y ≥ z X ∞ z/u = = = −∞ z/u 0 z −∞ z −∞ ∞ ... |
. 8.5 Expectation 351 Solution orem 6 takes the special form When X and Y are independent, we have f (x, y) = f X (x) fY (y), and The- f (z) = ∞ 1 |u| −∞ ∞ f X (u) fY z u du = 1 |u| u>z ue− u2 2 π −1 − 1 2 1 − z2 u2 du = 1 π z 2 e− u2 (u2 − z2) u du. 1 2 Now we make the substitution u2 = z2 + v2 to find that f (z) = 1 π... |
discovered above, finding the density of g(X, Y ) may not be a trivial matter. Fortunately, this task is rendered unnecessary by the following result, which we state without proof. 352 8 Jointly Continuous Random Variables (1) Theorem then (2) If X and Y have joint density f (x, y) and ∞ ∞ |g(u, v)| f (u, v) dudv < ∞, ... |
var (Y )) 1 2. Remark When X and Y are independent, then it follows from Corollary 3(v) that cov (X, Y ) = ρ(X, Y ) = 0, but not conversely. There is an important exception to this, in that bivariate normal random variables are independent if and only if ρ(X, Y ) = 0. See Examples 8.11 and 8.20 for details (4) Example ... |
t=0 Furthermore, just as joint p.g.f.s factorize for independent discrete random variables, it is the case that joint m.g.f.s factorize for independent continuous random variables. That is to say, if and only if X and Y are independent. MX,Y (s, t) = MX (s)MY (t) We offer no proofs for the above statements, as a proper... |
− M), and so M is independent of V. Remark This remarkable property of the normal distribution extends to any independent collection (Xi ; 1 ≤ i ≤ n) of N (µ, σ 2) random variables, and is known as the independence of sample mean and sample variance property. (11) Example Let X and Y be independent and identically dis... |
view to using moment generating functions, we first find ∞ E(etX 2 1 ) = −∞ 1 (2π) 1 2 exp tx 1 − 2t) 1 2. 1))n E(etY ) = (E(etX 2 1 (1 − 2t) = n 2 by independence and by (7.1.28) and (7.5.6), this is the m.g.f. of the χ 2(n) density. Hence, by Theorem 7.5.9, Y has a χ 2(n) density. Many results about sums of random var... |
), and we have the Key Rule P(X ∈ A|Y = y) = f X |Y (x|y) d x. x∈A (4) Example Let (X, Y ) be the coordinates of the point Q uniformly distributed on a circular disc of unit radius. What is fY |X (y|x)? Solution definition, Recall that for the marginal density f X (x) = (2/π)(1 − x 2) 1 2. Hence, by fY |X (y|x) = f (x, ... |
enough is called conditional expectation. (6) Definition given Y = y is given by If + R |x| f X |Y (x|y) d x < ∞, then the conditional expectation of X E(X |Y = y) = x f X |Y (x|y)d x. R (7) Example (5) Revisited If X and Y are independent and exponential, then we showed that the density of X given X + Y = v is uniform... |
, we make the important observation that by writing (10) ψ(y) = E(X |Y = y) y fY,V (y, v) dy −∞ we emphasize the fact that the conditional expectation of X given Y is a function of Y. If the value of Y is left unspecified, we write ψ(Y ) = E(X |Y ) on the understanding that when Y = y, ψ(Y ) takes the value E(X |Y = y) ... |
orem 11 can also be used to calculate probabilities by the simple device of letting X be the indicator of the event of interest. (13) Example Let U and Y have density f (u, y). What is P(U < Y )? Solution Let X be the indicator of U < Y. Then P(U < Y ) = E(X ) = E(E(X |Y )) by Theorem 11 ∞ y = E f (u|Y ) du = f (u, y) ... |
|Y (tY )) " # = E exp σ 2(1 − ρ2)t 2 Y 2 ρσ τ t + 1 2 = 1 (1 − 2ρσ τ t − σ 2τ 2(1 − ρ2)t 2) 1 2 using Example 8.5.13, and so X 1Y1 + X 2Y2 has moment generating function, (16) (17) M(t) = (1 − 2ρσ τ t − σ 2τ 2(1 − ρ2)t 2)−1 − ρ)t Hence, Z = X 1Y1 + X 2Y2 has an asymmetric bilateral exponential density, 1 1 − σ τ (1 + ... |
�(Y ))2]. Proof Using (17), we have (20) Hence, E[(X − ψ)(ψ − g)] = E[(ψ − g)E(X − ψ|Y )] = 0. E[(X − g)2] = E[(X − ψ + ψ − g)2] = E[(X − ψ)2] + E[(ψ − g)2] by (20) ≥ E[(X − ψ)2]. 8.7 Transformations: Order Statistics 361 We conclude this section by recording one more useful property of conditional densities, which may... |
, Yn) is given by (3) fY (y1,..., yn) = 1 |det A| f X (x1(y1,..., yn)... xn(y1,..., yn)) = |det B| f X (x1,..., xn). (4) Example: Normal Sample Let (X 1,..., X n) be independent N (0, 1) random vari- ables, and define (5) Y j = n j=1 x j ai j for 1 ≤ j ≤ n, where the matrix A = (ai j ) is an orthogonal rotation with det... |
X n i=1 X 2 i 2 − nX Hence, s2 is independent of X, by the independence of Y1 and (Y2,..., Yn). Finally, because each Yi is N (0, 1), (n − 1)s2 has a χ 2(n − 1) density by Example 8.5.13. See Problem 8.43 for another way to do this. 8.7 Transformations: Order Statistics 363 A particularly important linear transformatio... |
., X (n) is Now we observe that X 1, X 2,..., X n lies in just one of the n! regions Rπ ; hence, the order statistics have joint density n (8) n! f (yi ) i=1 for y1 < y2 <... < yn. Here are some applications of this useful result. (9) Example Let (X 1,..., X n) be independently and uniformly distributed on (0, a). Then... |
toll booth, meteorites fall from the sky, or you get stung by a wasp. You can think of many more such examples yourself, and it is clear that it would be desirable to have a general theory of such processes. This is beyond our scope, but we can now consider one exceptionally important special case of such processes. T... |
−λv dv = e−λt (λt)k k! after an integration by parts. As an alternative, we could argue straight from (5) and (6.1.7) that 1 − sE(s N (t)) 1 − s = ∞ 0 skP(N (t) ≥ k) = 1 + ∞ k=1 sk t 0 λkvk−1 (k − 1)! e−λvdv = 1 + sλ t 0 = 1 + s 1 − s eλvs−λvdv = 1 − s − seλt(s−1) 1 − s 1 − s = 1 − seλt(s−1) 1 − s, [eλv(s−1)]t 0 and th... |
2, the random variables Tn = n 1 Xi ; 1 ≤ n ≤ k + 1 have joint density f (t1,..., tk+1) = λk+1e−λtk+1; 0 < t1 <... < tk+1. (9) Now P(0 < T1 < t1 < T2 <... < Tk < tk; N (t) = k) = P(0 < T1 < t1 <... < Tk < tk < t < Tk+1) = λkt1(t2 − t1)... (tk − tk−1)e−λt on integrating the density (9). Hence, the conditional distributi... |
inomial distribution with (13) E(wW z Z |N (v) = k. Hence, combining (12) and (13) and Theorem (3) gives (14) E(wW z Z ) = exp (λw(t − s) + λz(v − u) + λ(s + u − t) − λv) = exp (λ(t − s)(w − 1) + λ(v − u)(z − 1)) = E(wW )E(z Z ), as required. We may also observe from (14) that because E(wW ) = exp (λ(t − s)(w − 1)), it... |
shorthand; we cannot condition on an infinite number of values of X (u). A stopping time for X (t) is a random variable T taking values in [0, ∞), such that the event (T ≤ t) depends only on values of X (s) for s ≤ t. There are many technical details that should properly be dealt with here; we merely note that they can... |
− λt. (c) W (t) = exp {−θ N (t) + λt(1 − e−θ )}, θ ∈ R. To see (a), we calculate E(U (t)|N (u); 0 ≤ u ≤ s) = E(N (t) − N (s) + N (s)|N (u), 0 ≤ u ≤ s) = U (s) + E[N (t) − N (s)] − λ(t − s) = U (s), where we used the independent increments property. The proof of (b) proceeds likewise: E(V (t)|N (u), 0 ≤ u ≤ s) = = E[N ... |
that, as n increases, the sequence ¯x n undergoes smaller and smaller fluctuations, and indeed exhibits behaviour of the kind we call convergent. A special case of such measurements arises when each xi takes the value 1 or 0 according to whether some event A occurs. Then ¯x n is the proportion of times that A occurs in... |
independent and identically distributed random variTheorem ables with mean µ, variance σ 2 < ∞, and moment generating function MX (t), |t| < a. Then we have: (i) Weak Law of Large Numbers $ $ $ $ $ P (4) For ε > 0, as n → ∞, 1 n n (Xi − µ) i=. (ii) Central Limit Theorem As n → ∞, (5) P 1 √ n σ (Xi − µ) ≤ x → (x) = x −... |
we consider the joint behaviour of collections of continuous random variables having joint density functions. We also introduce the joint distribution function, and show how these yield the marginal densities and distributions. The change of variable technique is given and used to study important functions of sets of ... |
J (u, v) = ∂ x ∂u ∂ y ∂v − ∂ y ∂u ∂ x ∂v, where the derivatives are continuous in the domain of (u, v). Then the random variables (U, V ) = (u(X, Y ), v(X, Y )) are jointly continuous with density fU,V (u, v) = f X,Y (x(u, v), y(u, v))|J (u, v)|. Independence: X and Y are independent, if and only if FX,Y (x, y) = FX (... |
� f (x, y) fY (y) 0,, 0 < fY (y) < ∞ otherwise. P(X ∈ A|Y = y) = f X |Y (x|y) d x, x∈A the continuous partition rule is f X (x) = R f X |Y (x|y) fY (y) dy, and the conditional distribution function is FX |Y (x|y) = x −∞ f X |Y (x|y)d x = P(X < x|Y = y). |x| f X |Y (x|y)d x < ∞, then the conditional expectation of X g... |
for all choices of a = (a1,..., an). It follows that the if multinormal distribution is determined by the means and covariances of (X 1,..., X n). 374 8 Jointly Continuous Random Variables To see this, we simply calculate the joint moment generating function of X: MX(t) = E exp tr Xr. n 1 This is easy because is norma... |
normal sample 8.6 conditional density conditional distribution conditional expectation 8.7 normal sample order statistics 8.8 Poisson process conditional property Worked Examples and Exercises 375 independent increments martingales optional stopping theorem 8.9 weak law of large numbers central limit theorem.11 Exampl... |
. 376 8 Jointly Continuous Random Variables Find the conditional m.g.f. of Y given X. Use (3) to find E(es X +tY ). Show that ρ(X, Y ) = cov (X, Y ) = ρ. Deduce that X and Exercise Exercise Y are independent if and only if cov (X, Y ) = 0. Exercise and Z = Exercise Let X 1, X 2,..., X n be independent standard normal va... |
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