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Mathematics 426 Robert Gross Homework 7 Answers 1. Suppose that X is a Poisson random variable with parameter λ. Show that E[Xn ] = λE[(X + 1)n−1 ]. Use this result to compute E[X], E[X2 ], and E[X3 ]. Answer : We have E[Xn ] = ∞ X kn P(X = k) = k=0 ∞ X =λ ∞ X λk X n −λ λk X n−1 −λ λk = k e = k e k! k! (k − 1)! k=1 k=1 ∞ kn e−λ k=0 ∞ ∞ X λ λk =λ (k + 1)n−1 e−λ = λE[(X + 1)n−1 ] (k − 1)! k! k=0 k−1 k n−1 −λ e k=1 Now we have E[X] = λE[10 ] = λ, and then E[X2 ] = λE[(X + 1)1 ] = λE[X + 1] = λ(λ + 1) = λ2 + λ, and nally E[X3 = λE[(X + 1)2 ] = λE[X2 + 2X + 1] = λ((λ2 + λ) + 2λ + 1) = λ(λ2 + 3λ + 1) = λ3 + 3λ2 + λ. 2. (a ) Write the power series expansion for eλ + e−λ . (b ) Suppose that X is a Poisson random variable with parameter λ. Show that P(X is even) = (1 + e−2λ )/2. Answer : (a ) We have λ e +e = ∞ X λk k=0 (b ) We have P(X is even) = −λ ∞ X ∞ X λ2k + (−1) =2 k! k=0 k! (2k)! k=0 P(X = 2k) = e k=0 ∞ X −λ kλ k λ ∞ X e + e−λ 1 + e−2λ λ2k −λ =e = . (2k)! 2 2 k=0 3. Whenever Alice and Bob have lunch, they decide who will pay for the meal by each rolling a die. The one who rolls the lower number must pay. If there is a tie, they roll again, and continue rolling until one rolls a lower number than the other. What is the expected number of rolls needed to determine who will pay for lunch? Answer : The probability of a tie is 16 , so the probability that the game ends in a single roll is 56 . Because this problem describes a geometric distribution, we know that the expected value is p1 = 56 . 4. Suppose that X is a continuous random variable with probability density function f(x) = c(1 − x4 ) 0 −1 < x < 1 otherwise (a ) What is c? (b ) What is the cumulative distribution function for X? Answer : (a ) We compute Z1 x5 (1 − x ) dx = x − 5 −1 4 1 −1 8 = . 5 Therefore, c = . (b ) If we denote the cumulative distribution function as F(a), we know that F(a) = 0 for a < −1 and F(a) = 1 for a > 1. For −1 6 a 6 1, we have 5 8 5 F(a) = 8 Za 5 (1 − x ) dx = 8 −1 4 x5 x− 5 a −1 5 = 8 a5 4 a− + 5 5 5. Suppose that X is a binomial random variable with E[X] = 6 and Var(X) = 2.4. Compute P(X = 5). Answer : We have np = 6 and np(1 −p) = 2.4. Division yields 1 − p = 264 = 0.4, p = 0.6, and n = p6 = 10. Now, P(X = 5) = 10 0.65 0.45 ≈ 0.2007. 5 . 6. A lling station is provided with gasoline once each week. Suppose that the weekly volume of sales in thousands of gallons is a random variable with probability density function f(x) = 6(1 − x)5 0<x<1 otherwise What must the capacity of the lling station's tank be so that the probability of the station running out of gas in a given week is 0.01? Answer : One way to think about this problem is in terms of the cumulative distribution function F(a): we wish to nd a so that F(a) = 0.99. We know that for a 6 0 6 1, we have Za a F(a) = 6(1 − x)5 dx = −(1 − x)6 0 = 1 − (1 − a)6 . 0 0 We then must solve (1 − a)6 = 0.01, or 1 − a ≈ 0.4642, or a ≈ 0.5358. This is the capacity of the station's tank, in thousands of gallons. 7. Remember that a poker hand contains 5 random cards. Let X be the number of suits in a hand. Find P(X = k) for k = 1, 2, 3, and 4. Hint : See Ch. 2, self-help prob. 11, for a discussion of P(X = 4). Answer : The simplest to compute is P(X = 1): pick a suit, and then the cards in that suit. 4 · 13 33 P(X = 1) = 525 = ≈ 0.0020. 5 16660 For X = 2, we can either use Inclusion{Exclusion, or else rely on the fact that a two-suited hand must have either 4-1 distribution or else 3-2 distribution. For Inclusion{Exclusion, we compute the probability that the hand contains no more than 2 suits, and then subtract the one-suited hands: 26 5 52 5 4 · P(X = 2) = 2 13 5 52 5 4 · −2 2 = 143 ≈ 0.1459. 980 Alternatively, we can compute the probability of the two dierent distributions mentioned above and add: 13 13 13 13 4·3 P(X = 2) = + 2 3 1 52 5 4 = 143 980 Inclusion{Exclusion is trickier when X = 3 (see if you can work it out on your own), but fortunately, there are again only 2 possible distributions: 2-2-1 and 3-1-1: P(X = 3) = 13 2 6·2 13 2 13 + 1 52 5 13 3 13 1 13 1 = 4901 ≈ 0.5884. 8330 Finally, for X = 4, the simplest approach is one that the text does not oer. The only possible distribution is 2-1-1-1: P(X = 4) = 4 13 2 13 13 1 1 52 5 13 1 = 2197 ≈ 0.2637. 8330 The 4 probabilities sum to 1, as they must. 8. Let Y be a random variable. Show that Z∞ E[Y] = Z∞ P(Y > y) dy − P(Y < −y) dy 0 by showing that 0 Z∞ Z∞ P(Y > y) dy = 0 Z∞ xfY (x) dx 0 Z0 xfY (x) dx. P(Y < −y) dy = − 0 −∞ In these formul, fY (x) is the probability density function for Y . Answer : I prefer to work in the other direction, assuming that the probability density function for Y is f(y) and starting with the formula for E[Y]: Z∞ E[Y] = Z0 Z∞ xf(x) dx = −∞ Z∞ Zx = xf(x) dx + −∞ 0 xf(x) dx = 0 Z0 Z0 dy f(x) dx − 0 Z∞ Z0 xf(x) dx − 0 (−x)f(x) dx −∞ dy f(x) dx −∞ x In the rst integral, we have 0 6 y 6 x 6 ∞, while in the second integral we have −∞ < x < y < 0. Changing the order of integration: Z∞ Z∞ = Z0 Zy f(x) dx dy − 0 y Z∞ f(x) dx dy = −∞ −∞ Z0 P(Y > y) dy − 0 P(Y < y) dy −∞ One more step: let y = −z in the second integral, with dy = −dz: Z∞ = Z0 P(Y > y) dy + Z∞ P(Y < −z) dz = ∞ 0 Z∞ P(Y > y) dy − 0 P(Y < −z) dz. 0 9. If X is a random variable with density function f(x), show that Z∞ g(x)f(x) dx. E[g(X)] = −∞ Hint : The previous exercise Zshows that Z∞ ∞ E[g(X)] = P(g(X) > y) dy − 0 P(g(X) < −y) dy. 0 Now proceed as we did in class. Answer : Write f(x) as the probability density function for X. We have Z∞ E[g(X)] = Z0∞ = 0 Z∞ P(g(X) > y) dy − P(g(X) < −y) dy 0 Z Z∞ Z f(x) dx dy − f(x) dx dy x:g(x)>y 0 x:g(x)<−y Again, we switch the order of integration in each of the integrals, noting in the second integral that if g(x) < −y, then −g(x) > y: Z Z g(x) Z Z −g(x) dy f(x) dx − dy f(x) dx x:g(x)<0 0 Z Z = g(x)f(x) dx − (−g(x))f(x) dx x:g(x)>0 x:g(x)<0 Z Z = g(x)f(x) dx + g(x)f(x) dx x:g(x)>0 x:g(x)<0 Z = g(x)f(x) dx = x:g(x)>0 0 x:g(x)6=0 Finally, notice that Z g(x)f(x) dx = 0, because the integrand is 0: x:g(x)=0 Z Z = Z∞ g(x)f(x) dx + x:g(x)6=0 g(x)f(x) dx = x:g(x)=0 g(x)f(x) dx −∞