Stat 366 A1 (Fall 2006) Midterm Solutions (October 23) page 1
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1 Stat 366 A1 Fall 6) Midterm Solutions October 3) page 1 1. The opening prices per share Y 1 and Y measured in dollars) of two similar stocks are independent random variables, each with a density function given by 1 y β fy) = α e α, y β,, elsewhere. On a given morning a stockholder is going to sell shares of whichever stock is more expensive. a) Find the probability density function for the price per stock that the person will sell. b) Find the expected cost per share that the stockholder will receive. c) What is the probability that the stockholder will be able to sell his most expensive stock at the price larger than $5 per share if α = and β =? d) Find the probability that Y 1 + Y will be smaller than $5 if α = and β =. Solution. a) Let s compute the cumulative distribution function for Y. Assume t > β: Let u = y β α, so that du = dy α : F Y t) = t β 1 y β α e α dy. F Y t) = t β α t β e u du = e u α = 1 e t β α, t β. To find the distribution of the order statistic Y ) = maxy 1, Y ), use the formula: F Y) y) = P Y ) y) = P Y 1 y, Y y) = P Y 1 y) P Y y) = F Y y) ) ), = 1 e y β α y β. Now differentiate it to get the density: exp 1 ) f Y) y) = α α y β) exp )) α y β), y β,, elsewhere.
2 Stat 366 A1 Fall 6) Midterm Solutions October 3) page b) To find the expected value, we just need to integrate: where v = α Notice that E[Y ) ] = = β 1 α y exp 1 ) α y β) dy ) αu + β e u du β αv + β ) e v dv, α y exp ) α y β) dy dy y β), so dv =, y = αv + β, and u is as defined before with y = αu + β). α ue u du = u de u = ue u + e u du = ue u e u = u + 1)e u. We have E[Y ) ] = α u e u du + β e u du α v e v dv β = 3 α u + 1) e u ) + β e u ) = 3 α + β. e v dv c) Notice that P Y ) > 5) = 1 F Y) 5) = 1 1 exp 1 )) α y β) ) = 1 1 e 3/ = e 3/ e 3 = 39.65%. d) Let s compute the double integral of the joint density over the triangle with vertices at,),,3), 3,): P Y 1 + Y < 5) = = 1 4 = y1 3 3 e e y 1/ 1 exp y 1 5 y1 e y / dy dy 1 y ) dy dy 1 e y 1/ e / e y 1 5)/ ) dy 1 = 1 3 = e 1 y1/ 1 3 e 1/ y 1 = e 1/ 3 e 1/ e 1/ = 1 3 e 1/ 9.%. e 1 y 1 / e 1/) dy 1
3 Stat 366 A1 Fall 6) Midterm Solutions October 3) page 3. Let X 1, X be i.i.d. random variables with probability function given by P X = 1) = P X = 1) = 1/. Write X 3 = X 1 X the product of X 1 and X ). Show that X 1, X and X 3 are not independent, but pairwise independent only. Hint: Prove the following results a) P X 3 = x X j = a) = P X 3 = x) for j = 1,, a = ±1 and for all x. b) P X 3 = x X 1 = a, X = b) P X 3 = x) for all x and for a, b = ±1. Solution. To show a), denote k = 3 j and consider the following: P X 3 = x X j = a) = P X 3 = x, X j = a) P X j = a) = P X k = x/a, X j = a) 1/ = P X k = x/a)p X j = a) = 1 1 = 1. This is exactly P X 3 = x) for any x = ±1. b) Similarly, P X 3 = x X 1 = a, X = b) = P X 3 = x, X 1 = a, X = b) P X 1 = a, X = b) 4 1 = 1, x = ab, = 4 4 =, x ab. = 4P X 3 = x, X 1 = a, X = b) We conclude that for different values of x = ±1 the latter is either or 1, but never 1 = P X 3 = x). So, X 3 is not independent from X 1 and X even though pairwise independence takes place.
4 Stat 366 A1 Fall 6) Midterm Solutions October 3) page 4 3. Let X, Y ) be jointly distributed with density given by k x + y), e x 3y, x, y f X,Y x, y) =, elsewhere. a) Find k. b) Find f y ) X, the conditional distribution of Y given X. c) Find E [ Y ] X, the conditional expectation of Y. Solution. a) Integrating the would-be density ignoring k yields: 1 k = = 1 3 x,y x + y) e x 3y dxdy = xe x dx + xe x e 3y dydx + ye 3y dy = 1 3 e x x + 1) + 9 = 5 9. ye 3y e x dxdy The aside computations were: e 3y dy = 1 3 e 3y = 1 3, e x dx = 1, ye 3y dy = 1 3 y de 3y = 1 3 ye 3y e 3y dy = 1 9 e 3y = 1 9. One can also look at these integrals from the probabilistic point of view. Let Z be an exponentially distributed random variable with mean 1. We have e 3y dy = 1 3 e 3y d3y) = 1 3 P Z ) = 1 3. ye 3y dy = 1 9 3y) e 3y d3y) = 1 9 E[Z] = 1 9. So, we conclude that k = 9 5. b) By definition, f y X = x ) = fx, y) f 1 x). So, we basically need to find the marginal density of X given the joint density. Let s integrate: 9 f 1 x) = 5 x + y) e x 3y dy = 9 5 xe x e 3y dy e x ye 3y dy = xe x + 5 e x = 3x + e x, x >. 5
5 Stat 366 A1 Fall 6) Midterm Solutions October 3) page 5 Now, we can write the answer: f y X = x ) = 9 x + y 3x + e 3y, x, y. c) Now that we know the conditional density, the conditional expectation E[Y X] is easy to find: E [ Y X = x ] = 9 x + y 3x + y e 3y dy = 9x 3x + ye 3y dy x + y e 3y dy. Let s do some auxiliary calculations let Z be exponentially distributed with mean 1): Or, alternatively: y e 3y dy = 1 7 3y) e 3y d3y) = 1 7 E[Z ] = 1 ) V [Z] + E[Z]) = 7 7. y e 3y dy = 1 3 y de 3y = 1 3 y e 3y e 3y dy = 3 ye 3y dy = 7. So, we finally have E [ Y X = x ] = 9x 3x x + 7 = 1 3 3x + 4 3x +, x >. 4. Let Y 1, Y be i.i.d. random variables distributed as normal with mean µ and variance σ. Using the m.g.f. method, obtain the probability density function of i) U 1 = Y 1 + Y. ii) U = Y 1 Y /. Are U 1 and U independent? Justify your answer. Solution. We are given that Y 1, Y Nµ, σ ). Recall that the moment generating function of the normal random variable is m Y t) = E[e Y t ] = exp µt + σ t ).
6 Stat 366 A1 Fall 6) Midterm Solutions October 3) page 6 Let s derive the m.g.f. of Z = Y 1 + ay, where a is a real number recall that Y 1 and Y are independent): m Z t) = E[e Y 1+aY )t ] = E[e Y1t ]E[e ayt ] = m Y1 t) m Y at) = exp µt + σ t ) exp aµt + σ a t ) ) = exp 1 + a)µt a ) σ t = exp µ t + σ t ), for µ = 1 + a)µ and σ = 1 + a )σ. We see that Z Nµ, σ ). Take a = for U 1 and a = 1 for U : U 1 N3µ, 5σ ), U N 1 µ, 5 4 σ) To check the independence of U 1 and U we only need to find the covariance because in the case of normal variables as we have for U 1 and U ) a covariance of zero means independence, and vice versa). We have Cov[U 1, U ] = Cov[Y 1 + Y, Y 1 Y /] = V [Y 1 ] /frac1cov[y 1, Y ] + Cov[Y, Y 1 ] V [Y ] = σ σ =. We see that Cov[U 1, U ] =, and therefore U 1 and U are independent.
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