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the law of total probability, and applications of its use, are provided in Section 1.5. Suppose now that A and B are two events such that A contains B (in symbols, B). In words, all outcomes in B are also in A. Intuitively, A is a “larger” event A than B, so we would expect its probability to be larger. We have the fo... |
the sum of the probabilities of the individual events. This is called subadditivity. Theorem 1.3.4 (Subadditivity) Let A1 A2 quence of events, not necessarily disjoint. Then be a finite or countably infinite se P A1 A2 P A1 P A2 PROOF See Section 1.7 for the proof of this result. We note that some properties in the defi... |
the time, and both arrives late and leaves early 5% of the time. What is the probability that on a given day that employee will either arrive late or leave early (or both)? 1.3.4 Suppose your right knee is sore 15% of the time, and your left knee is sore 10% of the time. What is the largest possible percentage of time... |
j i P Ai Ai A j Ak P A1 An (Hint: Use induction.) DISCUSSION TOPICS 1.3.11 Of the various theorems presented in this section, which ones do you think are the most important? Which ones do you think are the least important? Explain the reasons for your choices. 1.4 Uniform Probability on Finite Spaces If the sample spa... |
s S. 1.4.1 Combinatorial Principles Because of (1.4.1), problems involving uniform distributions on finite sample spaces often come down to being able to compute the sizes A and S of the sets involved. That is, we need to be good at counting the number of elements in various sets. The science of counting is called comb... |
Spaces EXAMPLE 1.4.6 Suppose we roll two fair sixsided dice. What is the probability that the sum of the numbers showing is equal to 10? By the above multiplication principle, the total number of possible outcomes is equal to 6 36. Of these outcomes, there are three that sum to 10, namely, 4 6, 5 5, and 6 4. Thus, th... |
equal to 2 multiplied by itself 10 times, or 210 1024. Hence, the probability of any particular sequence occurring is 1 1024. But of these sequences, how many have exactly seven heads? 10. There are 10! 3! 10 To answer this, notice that we may specify such a sequence by giving the positions of the seven heads, which i... |
we have a set S of n elements and we want to count the number of elements of S1 S2 Sl : Si S Si ki Si S j when i j namely, we want to count the number of sequences of l subsets of a set where no two subsets have any elements in common and the ith subset has ki elements. By the multiplication principle, this equals n k... |
4.2 Suppose we roll 10 fair sixsided dice. What is the probability that there are exactly two 2’s showing? 1.4.3 Suppose we ip 100 fair independent coins. What is the probability that at least three of them are heads? (Hint: You may wish to use (1.3.1).) 1.4.4 Suppose we are dealt five cards from an ordinary 52card d... |
#2 has 6 red and 12 blue balls. Suppose we pick three balls uniformly at random from each of the two urns. What is the probability that all six chosen balls are the same color? 1.4.12 Suppose we roll a fair sixsided die and ip three fair coins. What is the proba bility that the total number of heads is equal to the... |
on any of the 365 days of a normal (i.e., nonleap) year. (a) Suppose C birthday? 2. What is the probability that the two people have the same exact 20 Section 1.5: Conditional Probability and Independence 2. What is the probability that all C people have the same exact (b) Suppose C birthday? (c) Suppose C same exact... |
“given that.” 1.5.1 Conditional Probability In general, given two events A and B with P B 0, the conditional probability of A given B, written P A B, stands for the fraction of the time that A occurs once we know that B occurs. It is computed as the ratio of the probability that A and B both occur, divided by the prob... |
A2 be events that form a partition of the sample space S, each of positive probability. Let B be any event. Then P B P A2 P B A2 P A1 P B A1 PROOF The multiplication formula (1.5.2) gives that P Ai B The result then follows immediately from Theorem 1.3.1. P Ai P B Ai EXAMPLE 1.5.1 Suppose a class contains 60% girls an... |
by an example. EXAMPLE 1.5.2 Suppose urn #1 has 3 red and 2 blue balls, and urn #2 has 4 red and 7 blue balls. Suppose one of the two urns is selected with probability 1 2 each, and then one of the balls within that urn is picked uniformly at random. What is the probability that urn #2 is selected at the first stage (e... |
12 S. Here the probability that the die comes up 5 is 1 6, as it should be. But now, what is the probability that the die comes up 5, conditional on knowing that the coin came up tails? Well, we can compute that probability as P die 5 coin tails P die 5 and coin tails P coin tails P 5T P 1T 2T 3T 4T 5T 6T 1 12 6 12 1 ... |
1.5.3, three events A, B, and C are independent if all of the following equations hold and 1.5.3) (1.5.4) It is not sufficient to check just some of these conditions to verify independence. For example, suppose that S 1 4. Let A 1 4. Then each of the three equations 1 2, B (1.5.3) holds, but equation (1.5.4) does not h... |
probability that all three coins are heads, given that the number of heads is even? 1.5.4 Suppose we deal five cards from an ordinary 52card deck. What is the con ditional probability that all five cards are spades, given that at least four of them are spades? 1.5.5 Suppose we deal five cards from an ordinary 52card d... |
s. Conditional on the fact that all six chosen balls are the same color, what is the conditional probability that this color is red? 1.5.11 Suppose we roll a fair sixsided die and then ip a number of fair coins equal to the number showing on the die. (For example, if the die shows 4, then we ip 4 coins.) (a) What is... |
that the sum of the three numbers showing is 12. 1.5.17 (The game of craps) The game of craps is played by rolling two fair, sixsided dice. On the first roll, if the sum of the two numbers showing equals 2, 3, or 12, then the player immediately loses. If the sum equals 7 or 11, then the player immediately wins. If the... |
B or C the host did not open). You then win (i.e., get to keep the car) if and only if the car is behind your final door selection. (Source: Parade Magazine, “Ask Marilyn” column, September 9, 1990.) Suppose for definiteness that the host opens door B. (a) If you stick with your original choice (i.e., door A), condition... |
. The controversy dragged on for months, with many letters and very strong language written by both sides (in the end, von Savant was vindicated). Part of the confusion lay in the assumptions being made, e.g., some people misinterpreted her question as that of the modified version of part (e) of Challenge 1.5.18. Howeve... |
n 1 n then A1 n 1 An n 1 An and A3 A2 Chapter 1: Probability Models 29 Figure 1.6.1: An increasing sequence of subsets A1 A2 A3 Figure 1.6.2: A decreasing sequence of subsets A1 A2 A3 We will consider such sequences of sets at several points in the text. For this we need the following result. Theorem 1.6.1 Let A A1 A2... |
. Prove that we must have 1 2 3 is the set of all positive integers and that P is lim n P 1 2 n 1 1.6.4 Suppose P [0 8 4 n ] 2 e n 6 for all n 1 2 3. What must P 0 be? Chapter 1: Probability Models 31 1.6.5 Suppose P [0 1] P 0 be? 1.6.6 Suppose P [1 n 1 2] (a) Must we have P 0 1 2] (b) Must we have P [0 1 2] 1.6.7 Supp... |
Let B1 are disjoint, B1 B2 2, let Bn A2 A1 is a finite or countably infinite sequence of A2 An A1 P A1 An 1 P A2 c. Then B1 B2 and, by additivity, P A1 A2 P B1 B2 P B1 P B2 (1.7.1) Furthermore, An (1.7.1) that Bn, so by monotonicity, we have P An P Bn. It follows from P A1 A2 P B1 P B2 P A1 P A2 as claimed. 32 Section 1... |
Conditioning and Independence Multidimensional Change of Variable In Chapter 1, we discussed the probability model as the central object of study in the theory of probability. This required defining a probability measure P on a class of subsets of the sample space S It turns out that there are simpler ways of presentin... |
sample space to R1. EXAMPLE 2.1.2 For the case S saying that Y Y rain 0, Y snow rain snow clear, we might define a second random variable Y by 7 8 if it is clear. That is 1 2 if it snows, and Y 0 if it rains, Y 1 2, and Y clear 7 8. EXAMPLE 2.1.3 If the sample space corresponds to ipping three different coins, then we... |
more than the number showing, 3. So Y. Then Z s X s 2 Y s s2 s We write X Y to mean that X s Y to mean that X s Y s for all s Y s for all s S, and X S. Similarly, we write Y s Y to mean that X s S. For example, we write X c to mean that X s c for all s S. X for all s EXAMPLE 2.1.8 Again consider rolling a fair sixsid... |
b) maxs S X s (c) mins S Y s (d) maxs S Y s 2.1.2 Let S high middle low. Define random variables X, Y, and Z by X high 12, X middle 2, X low 3, Y high 1, 4. Determine whether each of the following 0, Y middle 0, Y low Y Y Z Z 0, Z low 1 2 3 4 5. Z high 6, Z middle relations is true or false. (a) X (b) X (c) Y (d) Y Z (e... |
Let X be a random variable. 0? (a) Is it necessarily true that X (b) Is it necessarily true that there is some real number c such that X (c) Suppose the sample space S is finite. Then is it necessarily true that there is some real number c such that X 2.1.11 Suppose the sample space S is finite. Is it possible to define ... |
X 6 P X 17 P X also compute that P snow 0, and in fact P X 2 7 only when it is clear. Thus, P X 0 15, and P X P x 3 only when it rains, X 2 7 0 for all x 3 P clear 6, and X clear 3, X snow 3 6 rain, snow, 2 7. Suppose 0 15, 6 only when 0 4, 0 45. Also, 2 7. We can P rain 0 4, P snow 15 0 55 45 0 85 while etc. We see f... |
if B. We can formally write all this P X B 0 4 IB 3 0 15 IB 6 0 45 IB 2 7, where again IB x 1 if x B, and IB x 0 if x B. EXAMPLE 2.2.3 An AlmostAsSimple Distribution Consider once again the above setting, with S P snow 0 15, and P clear Y rain 7, and Y clear 5, Y snow 5. What is the distribution of Y? Clearly, Y 0 1... |
ute P Y (b) Write a formula for P Y 2.2.4 Suppose we roll one fair sixsided die, and let Z be the number showing. Let W Z 3 (a) Compute P W (b) Compute P V (c) Compute P Z W x for every real number x. (d) Compute P V W y for every real number y. (e) Compute P V W r for every real number r. 2.2.5 Suppose that a bowl co... |
Variables and Distributions 41 PROBLEMS 2.2.9 Suppose that a bowl contains 10 chips each uniquely numbered 0 through 9 The chips are thoroughly mixed, one is drawn and the number on it, X1 is noted. This chip is not replaced in the bowl. A second chip is drawn and the number on it, X2 is noted. Compute P W for every r... |
Section 2.3: Discrete Distributions Definition 2.3.3 For a discrete random variable X, its probability function is the function pX : R1 [0 1] defined by Hence, if x1 x2 1, then i pi pX x P X x. are the distinct values such that P X xi pi for all i with pX x pi 0 xi for some i x otherwise. Clearly, all the information ab... |
and a 0 if the individual is a male. In this case, is the proportion of females in the population. Chapter 2: Random Variables and Distributions 43 EXAMPLE 2.3.3 The Binomial Distribution Consider ipping n coins, each of which has (independent) probability heads, and probability 1 number of heads showing. By (1.4.2),... |
only if the coin shows that appear before the first head. Then for k k. (In exactly k tails followed by a head. The probability of this is equal to 1 particular, the probability of getting an infinite number of tails before the first head is k, equal to 1 distri 0 1 2 3 for k bution; we write this as X. Figure 2.3.2 con... |
NegativeBinomial r NegativeBinomial r distribution; we 1 cor. Of course, the special case r distribution. So in terms of our notation, we have that. Figure 2.3.3 contains the plots of several Geometric write this as Y responds to the Geometric NegativeBinomial 1 NegativeBinomial r probability functions. p 0.25 0.... |
2.3.2), we see that lim n P X x x x! e for x 0 1 2 3 Intuitively, we can phrase this result as follows. If we ip a very large number of coins n, and each coin has a very small probability n of coming up heads, then the probability that the total number of heads will be x is approximately given by x e x!. Figure 2.3.5 ... |
of being drawn in the first two draws. Continuing in this fashion for n draws, we obtain that the probability of any particular set of n balls being drawn is 1 N n This type of sampling is called sampling without replacement. Given that we take a sample of n let X denote the number of white balls obtained. N M because ... |
3, that the number of white balls observed in n draws is distributed Binomial n M N. Summary of Section 2.3 x A random variable X is discrete if x P X comes from being equal to particular values. A discrete random variable X takes on only a finite, or countable, number of distinct values. Important discrete distribution... |
Z 2.3.7 Let X 2.3.8 Let W Poisson 2.3.9 Let Z NegativeBinomial 3 1 4. Compute P Z 2.3.10 Let X Geometric 1 5. Compute P X 2 15. 2.3.11 Let Y 2.3.12 Let X 2.3.13 Let X is the probability that X 2.3.14 Suppose that a symmetrical die is rolled 20 independent times, and each time we record whether or not the event 2 3 5 ... |
. Units typically arrive for service in a random fashion and form a queue when the server is busy. It is often the case that the number of arrivals at the server, for some specific unit of time t can be modeled by a Poisson t distribution and is such that the number of arrivals in nonoverlapping periods are independent.... |
.24 Let X independently. Let Z reasoning.) (Hint: See the end of Example 2.3.3.) 2.3.25 Let X Geometric dently. Let Z (Explain your reasoning.) Binomial n1 and Y X X, with X and Y chosen indepen Y. What will be the distribution of Z? Generalize this to r coins. Geometric and Y, with X and Y chosen Y. What will be the ... |
uniform distribution on [0 1], as presented in (1.2.2). That is, P a X b b a (2.4.2) a whenever 0 0. The random variable X is said to have the Uniform[0 1] distribution; we write this as X Uniform[0 1]. For example, 1, with Also, P X 2 3 P 2 3 In fact, for any x [0 1], 52 Section 2.4: Continuous Distributions Note tha... |
at a may be thought of as measuring the probability of a random variable being in a small interval about a. To better understand absolutely continuous random variables, we note the following theorem. Theorem 2.4.1 Let X be an absolutely continuous random variable. Then X is a a continuous random variable, i.e., P X 0 ... |
x 1 R L 0 R x L otherwise. In Figure 2.4.1 we have plotted a Uniform[2 4] density. 1For examples of this, see more advanced probability books, e.g., page 143 of A First Look at Rigorous Probability Theory, Second Edition, by J. S. Rosenthal (World Scientific Publishing, Singapore, 2006). 54 Section 2.4: Continuous Dist... |
follow an Exponential. By this we mean that the lifelength X of a distribution for an appropriate choice of randomly selected light bulb from those produced by this manufacturer has probability P X x e z dz x e x of lasting longer than x in whatever units of time are being used. We will see in Chapter 3 that, in a spe... |
(because tribution: Gamma 1 Gamma density functions. Exponential 1 0! dis. In Figure 2.4.2, we have plotted several 1) to the Exponential f 0.8 0.6 0.4 0.2 0. Figure 2.4.2: Graph of an Exponential 1 (solid line) a Gamma 2 1 (dashed line), and a Gamma 3 1 (dotted line) density. A gamma distribution can also be used to... |
2.4: Continuous Distributions 2.4.4, we have plotted the N 0 1 and the N 1 1 densities. Note that changes in simply shift the density without changing its shape. In Figure 2.4.5, we have plotted the N 0 1 and the N 0 4 densities. Note that both densities are centered on 0, but 2 controls the amount of the N 0 4 densit... |
a density function and we change its value at a finite number of points, then the value of b a f x dx will remain unchanged. Hence, the changed function will also qualify as a density corresponding to the same distribution. On the other hand, often a particular “best” choice of density function is clear. For example, i... |
the following. (a) P W 5 (b) P W 2 (c) P W 2 (d) P W 2 2.4.3 Let Z (a) P Z 5 (b) P Z (c) P Z 2 (d) P Z 4 2.4.4 Establish for which constants c the following functions are densities. cx on 0 1 and 0 otherwise. (a) f x cx n on 0 1 and 0 otherwise, for n a nonnegative integer. (b) f x cx 1 2 on 0 2 and 0 otherwise. (c) f... |
called random variable Y the memoryless property of the exponential distributions; it says that they immediately “forget” their past behavior. 2.4.15 Consider the gamma function (a) Prove that (b) Prove that 1 (c) Use parts (a) and (b) to show that 2.4.16 Use the fact that t. (Hint: Use integration by parts.) n to giv... |
a density function. (b) Determine and plot the density when a (c) Determine and plot the density when a (d) Determine and plot the density when a (e) Determine and plot the density when for Can you name this distribution? 1 2 2 CHALLENGES 2.4.25 Prove (2.4.10). (Hint: Use x make the change of variable yb 1e x y dx dy ... |
P X [a b] P X a FX a FX a (2.5.1) Similarly, if B a b is an open interval, then P X B P a X b lim n FX b 1 n FX a FX b FX a If B [a b is a rightopen interval, then P X B P a X b lim n FX b 1 n lim n FX a 1 n FX b FX a We conclude that we can determine P X interval. B from FX whenever B is any kind of Now, if B is ins... |
the result follows. n. Because X must take on some value and hence X X (c) Let An sufficiently large n, we see that An increases to S, i.e., An Hence, by continuity of P (see Theorem 1.6.1), limn P An FX n, so the result follows. P X n P An n for S (see Section 1.6). 1. But P S X (d) Let Bn to the empty set, i.e., Bn P... |
2 3 4 5 6, with P s 1 6 FX x P X x P s s S s 6x s S s 6x 1 6 1 6 s S : s 6x 2For example, see page 67 of A First Look at Rigorous Probability Theory, Second Edition, by J. S. Rosen thal (World Scientific Publishing, Singapore, 2006). Chapter 2: Random Variables and Distributions 65 That is, to compute FX x, we count h... |
.5.4, the cumulative distribution function F of this distribution is given by x x F x t dt e t 2 2 dt 1 2 It turns out that it is provably impossible to evaluate this integral exactly, except for ). Nevertheless, the certain specific values of x (e.g., x cumulative distribution function of the N 0 1 distribution is so i... |
0 96 0 965 0 96 0 1685 0 1660 0 97 0 1685 0 96 0 965 0 96 0 16725 EXAMPLE 2.5.3 Let X be a random variable with cumulative distribution function given by FX x 0 x 1 2 4 16 x 2 4 2 x x 4 In Figure 2.5.2, we present a graph of FX 1.0 F 0.5 0.0 1 2 3 4 5 x Figure 2.5.2: Graph of the cdf FX in Example 2.5.3. Suppose for t... |
effect, Fi is the on the outcome of the first stage, so that Y has cdf Fi when Z conditional distribution of Y given that Z i (see Section 2.8). Then, by the law of total probability (see Theorem 1.5.1), the distribution function of Y is given by pi for pi Fi y G y Therefore, the distribution function of Y is given by ... |
) in Example R1 2.5.5, where F x 1 for x ex ex If pi 0 with i pi 1 (so the pi form a probability distribution), and y1 y2 are real numbers, then we can define a discrete location mixture by H x pi Fyi x pi F x yi i i 70 Section 2.5: Cumulative Distribution Functions Indeed, the shift Fy x location mixture, with p1 F x y... |
4 5 P X2 1 5 F1 y 4 5 F2 y 4 5 F2 y F2 y y Chapter 2: Random Variables and Distributions 71 Therefore, P Y y 3y y! e 3 1 5 0 y a nonnegative integer otherwise. Because P Y y continuous. On the other hand, we have 0 for nonnegative integers y the random variable Y is not P Y y y y 0 1 5 3y y! e 3 1 5 1 Hence, Y is not ... |
2 3 4 5 6, and P s s for s P Y 3In fact, there exist probability distributions that cannot be expressed even as a mixture of a discrete and a continuous distribution, but these need not concern us here. 72 Section 2.5: Cumulative Distribution Functions 2.5.3 For each of the following functions F, determine whether or ... |
Is F a valid cumulative distribution function? Why or why not? 2.5.11 Let F x (a) Sketch a graph of F. (b) Is F a valid cumulative distribution function? Why or why not? 2.5.12 Let X 2.5.13 Let F x x for 2 5 (a) Sketch a graph of F. (b) Prove that F is a valid cumulative distribution function. (c) If X has cumulative ... |
x F x for each x (Theorem 1.6.1).) 2.5.18 Let X be a random variable, with cumulative distribution function FX. Prove 0 if and only if the function FX is continuous at a. (Hint: Use (2.5.1) a that P X and the previous problem.) 2.5.19 Let (Hint: Let s 2.5.20 Determine the distribution function for the logistic distrib... |
2.6.1: An example where the set of x values that satisfy h x points x1 x2 and x3 y consists of three We now establish the basic result. h X, where h : R1 Theorem 2.6.1 Let X be a discrete random variable, with probability function pX. R1 is some function. Then Y is also discrete, Let Y x h 1 y pX x where h 1 y and its... |
is strictly increasing, then the situation is considerably simpler, as the following theorem shows. Theorem 2.6.2 Let X be an absolutely continuous random variable, with density R1 is a function that is differen function f X. Let Y tiable and strictly increasing. Then Y is also absolutely continuous, and its density ... |
N 2 distribution with 5 and 2 4. If instead the function h is strictly decreasing, then a similar result holds. Theorem 2.6.3 Let X be an absolutely continuous random variable, with density R1 is a function that is differen function f X. Let Y tiable and strictly decreasing. Then Y is also absolutely continuous, and ... |
0. Thus, functions such as h x x 8, and h x x 8, and h x x 2, h x 0 for x 78 Section 2.6: OneDimensional Change of Variable Summary of Section 2.6 h X, then P Y If X is discrete, and Y If X is absolutely continuous, and Y h h 1 y decreasing, then the density of Y is given by fY y This allows us to compute the distrib... |
e., it is symmetric X. Show that the density of Y is given by f X. Use this to f X x 2 N x 3 4 for 0 x X 2. Compute the density function fY y for Y. X. Compute the density function f Z z for Z. 2]. Let Y sin X. Compute the density function 2, otherwise f X x 0. x X 2. Compute the density function fY y for Y. 1 2 sin x ... |
ALLENGES 2.6.21 Theorems 2.6.2 and 2.6.3 require that h be an increasing or decreasing function, at least at places where the density of X is positive (see Theorem 2.6.4). Suppose now that X 0 for all x, while N 0 1 and Y 0. Hence, Theorems 2.6.2 h is increasing only for x and 2.6.3 do not directly apply. Compute fY y ... |
means “and” here, so that FX Y x y is the probability that X x and Y y.) EXAMPLE 2.7.2 (Example 2.7.1 continued) Again, let X Bernoulli 1 2, Y1 X, and Y2 1 X. Then we compute that FX Y1 x y P X x Y1 y 0 1 2 1 0 min x y 0 min x y 1 min x y 1 On the other hand, FX Y2 x y P X x Y2 y 0 1 2 1 min x y 0 min x y 1 min x y 0 ... |
that and from this we obtain FX Y b d FX Y a d FX Y b c FX Y a c as claimed. Joint cdfs are not easy to work with. Thus, in this section we shall also consider other functions, which are more convenient for pairs of discrete or absolutely continu ous random variables. 2.7.2 Marginal Distributions We have seen how a j... |
.0 F 0.5 0.0 0.0 0.0 0.5 x 1.0 0.5 y 1.0 Figure 2.7.1: Graph of the joint distribution function FX Y x y 0 1 in Example 2.7.3. y x y2 for 0 x 1 and Theorem 2.7.3 thus tells us that the joint cdf FX Y is very useful indeed. Not only does it tell us about the relationship of X to Y, but it also contains all the informati... |
x y x PROOF Using additivity of P, we have that pX x P X x P X x Y y pX Y x y as claimed. Similarly, pY pX Y x y EXAMPLE 2.7.5 Suppose the joint probability function of X and Y is given by pX otherwise 0 3 4 0 4 Chapter 2: Random Variables and Distributions 85 Then while pX 5 pX Y 5 y pX Y 5 0 pX Y 5 3 pX pX 8 pX Y 8 ... |
square. 6 4 2 f 0 0.0 0.0 0.5 x 1.0 0.5 y 1.0 Figure 2.7.2: A plot of the density f in Example 2.7.6. Chapter 2: Random Variables and Distributions 87 We next compute. Indeed, we have 4x 2 y 2y5 dx dy 2y5 0 7 0 5 dy 147 Other probabilities can be computed similarly. Once we know a joint density f X Y, then computing t... |
7.3: Region of the plane where the density f X Y in Example 2.7.8 is positive. We check that f X Y x y dx dy 1 1 0 0 15 1 x 0 120x 3 y dy dx 1 120x 3 1 0 x 2 dx 2 60 x 3 2x 4 x 5 dx 60 1 4 2 1 5 1 6 2 12 10 1 so that f X Y is indeed a joint density function. We then compute that, for example, f X x 1 x 0 120x 3 y dy 12... |
1 Y 2 2 Z1 1 2 1 2 Z2 (2.7.1) Bivariate Normal then X Y This relationship can be quite useful in establishing various properties of this distribution. We can also write an analogous version Y Z2 and obtain the same distributional result. 2 Z1 X 1 Z1 The bivariate normal distribution is one of the most commonly used biv... |
is the bivariate normal distribution. b a f X Y x y dx dy b c d c Y d EXERCISES 2.7.1 Let X 2.7.2 Let X 2.7.3 Suppose Bernoulli 1 3, and let Y Bernoulli 1 4, and let Y 4X 2. Compute the joint cdf FX Y. 7X. Compute the joint cdf FX Y. pX 17 y x otherwise 3 2 2 3 19 (a) Compute pX. (b) Compute pY. (c) Compute P Y (d) Co... |
x (b) The marginal density fY y for all y (c) P Y (d) The joint cdf FX Y x y for all x y 2.7.9 Let X and Y have joint density f X Y x y otherwise f X Y x y (a) The marginal density f X x for all x (b) The marginal density fY y for all y (c) P Y 2.7.10 Let X and Y have the BivariateNormal 3 5 2 4 1 2 distribution. (a)... |
(b) Compute the marginal densities of X and Y 2.7.17 (Dirichlet and is 0 otherwise y x 1 2 3 distribution) Let X1 X2 have the joint density Ce x y for f X1 X2 x1 x2 x1 x2 3 1 x2 0 x2 0 and 0 X1 1 A Dirichlet distribution is often applicable x1 X2 correspond to random proportions. for x1 when X1 X2 and 1 (a) Prove that... |
(1.5.1), 0. Let a b X b, and suppose 5. Well, we provided that P X 5 0. This prompts the following definition. Definition 2.8.1 Let X and Y be random variables, and suppose that P X 0. The conditional distribution of Y, given that X assigning probability x, is the probability distribution x P Y B X x P X x to each event... |
aded region is the set | x x+ Now, if to x. If f X Y is a continuous function, then this implies that is very small, then in the above integrals we will always have t very close f X Y t y will be very 96 Section 2.8: Conditioning and Independence close to f X Y x y. We conclude that, if is very small, then dt dy f ... |
0 2 Y 0 3, we see that P 0 2 Y 0 3 fY y dy 2y5 dy 0 3 6 0 2 6 0 0336 We thus see that conditioning on X from about 0 0336 to about 0 0398. 0 8 increases the probability that 0 2 Y 0 3, By analogy with Theorem 1.3.1, we have the following. Theorem 2.8.1 (Law of total probability, absolutely continuous random variable v... |
see. Now, Definition 2.8.5 is very difficult to work with. Fortunately, there is a much simpler characterization of independence. Theorem 2.8.2 Let X and Y be two random variables. Then X and Y are indepen dent if and only if 2.8.1) whenever a b and c d. That is, X and Y are independent if and only if the events “a d” ... |
if” part of (b). The proof of the “only if” part of (b) is more technical, and we do not include it here. EXAMPLE 2.8.3 Let X and Y have, as in Example 2.7.6, joint density f X Y x y 2y5 4x 2 y 0 0 x otherwise 1 0 y 1 and so, as derived in as in Example 2.7.7, marginal densities f X x and Then we compute that f X x fY ... |
ition 2.8.5 is quite difficult to work with, it does provide the easiest way to prove one very important property of independence, as follows. Theorem 2.8.5 Let X and Y be independent random variables. Let f g : R1 R1 be any two functions. Then the random variables f X and g Y are also indepen dent. PROOF Using Definiti... |
so that pX1 x pXn x for all x p x R1 Furthermore, from Theorem 2.8.6(a), it follows that Xn of random variables is i.i.d. and discrete, then pX2 x pX1 Xn x1 xn pX 1 x1 pX2 x2 pXn xn p x1 p x2 p xn for all x1 xn R1. Similarly, if a collection X1 Xn of random variables is i.i.d. and jointly ab solutely continuous, then... |
j is equal to the number of i’s in k1 kn sn define the random variables Xi n j 1 I i s j 1 2 and 3 Clearly, Xi is the number of i’s observed in the sample and we n We refer to the Xi as the n and X1 0 1 X3 X2 for i always have Xi counts formed from the sample. For x1 x2 x3 satisfying xi 0 1 n and x1 x2 x3 n (2.8.2) imp... |
we have an urn containing 10 red balls, 20 white balls, and 30 black balls. If we randomly draw 10 balls from the urn with replacement, what is the probability that we will obtain 3 red, 4 white, and 3 black balls? Because we are drawing with replacement, the draws are i.i.d., so the counts are distributed Multinomial... |
if Xi x for all i. Hence, FX n x P X1 x x X2 P Xn x x Xn x FX1 x FX2 x FXn x x P X2 x P X n P X1 FX1 x n corresponds to an absolutely continuous distribution, then we can differentiate If FX1 this expression to obtain the density of X n EXAMPLE 2.8.7 As a special case of Example 2.8.6, suppose that X1 X2 independently... |
X equals pY X y x If X and Y are absolutely continuous, then the conditional density function of Y given X equals fY X y x X and Y are independent if P X R1. all B1 B2 Discrete X and Y are independent if and only if pX Y x y all x y Absolutely continuous X and Y are independent if and only if f X Y x y f X x fY y for ... |
Y have joint probability function R1. R1. pX otherwise. 2 2 2 0 4 (a) Compute P Y (b) Compute P Y (c) Compute P Y (d) Compute P Y (e) Compute. Chapter 2: Random Variables and Distributions 107 and Y Bernoulli Geometric Y. What is the probability function of Z? 2.8.6 Let X Z X 2.8.7 For each of the following joint dens... |
? Why or why not? 0. Compute P X x Y R1 with f X x R1 with fY y, with X and Y independent 4, otherwise f X Y x y 5. 1 8 for x 1 2 4 7, otherwise 1 Y y 3 5 and y R1 with R1 with y 36 for Poisson 1 and 0. 0. x 2 0. 0. 0. 0. 2 y x y 108 Section 2.8: Conditioning and Independence x 2 0. Compute each of the following. 2.8.1... |
ivariate normal distribution, as in Example 2.7.9. Prove that X and Y are independent if and only if Multinomial n 2.8.22 Suppose that X1 X2 X3 the joint probability function, that X1 Binomial n 2.8.23 Suppose that X1 X2 X3 Multinomial n distribution of X2 given that X1 2.8.24 Suppose that X1 Find the densities f X 1 a... |
This is similar to the problem considered in Section 2.6, except that we have moved from a onedimensional to a twodimensional setting. The twodimensional setting is more complicated; however, the results remain essentially the same, as we shall see. 2.9.1 The Discrete Case If X and Y are discrete random variables, ... |
is onetoone, then it is again possible to compute a formula for the joint density of Z and W, as the following theorem shows. To state it, recall from multivariable calculus that, if h : R2 R2 is a differentiable function, then its Jacobian derivative J is defined by h1 h2 h1 h2 J x y det h1 x h1 y h2 x h2 y h1 x h2 ... |
Hence, using Theorem 2.9.2, we see that otherwise otherwise. 2 0 z 2 We have thus obtained the joint density function for Z and W. EXAMPLE 2.9.3 Let U1 and U2 be independent, each having the Uniform[0 1] distribution. (We could write this as U1 U2 are i.i.d. Uniform[0 1].) Thus, fU1 U2 u1 u2 1 0 0 u1 otherwise. 1 0 u2... |
f X Y x y e x2 2 1 2 e y2 2 1 2 We recognize this as a product of two standard normal densities. We thus conclude that X N 0 1 and that, furthermore, X and Y are independent. N 0 1 and Y Chapter 2: Random Variables and Distributions 113 2.9.3 Convolution Suppose now that X and Y are independent, with known distributio... |
3 7] and Y Y. Then Exponential 6, with X and Y independent. Let f Z 5 f X x fY 5 x dx 5 3 1 4 6 e 6 5 x dx 12 1 4 e0 0 2499985 Note that here the limits of integration go from 3 to 5 only, because f X x 5. x 3, while fY 5 0 for x x 0 for Summary of Section 2.9 If X and Y are discrete, and Z h1 X Y and W h2 X Y, then pZ... |
is given by pX otherwise, and B Y, A Y, W X Let Z (a) Compute the joint probability function pZ W z (b) Compute the joint probability function pA B a b. (c) Compute the joint probability function pZ A z a. b. (d) Compute the joint probability function pW B 2.9.7 Let X have probability function. 2X 3Y 2. pX x and let Y... |
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