Math 3215 Intro. Probability & Statistics Summer 14. Homework 5: Due 7/3/14

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1 Math 325 Intro. Probability & Statistics Summer Homework 5: Due 7/3/. Let X and Y be continuous random variables with joint/marginal p.d.f. s f(x, y) 2, x y, f (x) 2( x), x, f 2 (y) 2y, y. Find the conditional p.d.f. h(y x) and the conditional probability P ( Y 3 2 X ). What is the expected value of Y when X? Solution: The conditional p.d.f h(y x) f(x, y)/f (x) is immediately seen to be h(y x) 2 2( x) x. To find P ( Y 3 X ) we integrate the conditional p.d.f. h(y ) on the interval 2 /2 y 3/, and we obtain P ( 2 Y 3 X ) 3/ /2 dy ( 3 Since expectation is linear, we have E(Y X ) E(/3) /3. ) Let X and Y be discrete random variables with joint p.m.f. f(x, y) x + y, x, 2, y, 2, 3,. 32 Find the marginal p.m.f. s of X and Y and the conditional p.m.f. s g(x y) and h(y x). Find P ( Y 3 X ) and P (Y 2 X 2). Finally, find E(Y X ) and find V ar(y X ). Solution: The marginal p.m.f. s of X and Y are immediately seen to be f (x) y,2,3, f 2 (y) x,2 f(x, y) x +, 32 f(x, y) 3 + 2y 32.

2 2 The conditional p.m.f. s are thus seen to be h(y x) g(x y) f(x, y) f (x) f(x, y) f 2 (y) (x + y)/32 (x + )/32 x + y x +, (x + y)/32 (3 + 2y)/32 x + y 3 + 2y. Also, we have P ( Y 3 X ) y,2,3 h(y ) Note, this can also be computed as h( ) 5 9. Next, we compute Finally, we have that E(Y X ) P (Y 2 X 2) y,2,3, y,2 h(y 2) y h(y ) () 2 + (2) 3 + (3) + () 8 7, and V ar(y X ) E(Y 2 X ) E(Y X ) 57 7 ( )

3 3 3. Let W equal the weight of a box of oranges which is supposed to weight -kg. Suppose that P (W < ).5 and P (W >.5).. Call a box of oranges light, good, or heavy depending on if W <, W.5, or W >.5, respectively. In n 5 independent observations of these boxes, let X equal the number of light boxes and Y the number of good boxes. Find the joint p.m.f. of X and Y. How is Y distributed? Name the distribution and state the values of the parameters associated to this distribution. Given X 3, how is Y distributed? Determine E(Y X 3) and find the correlation coefficient ρ of X and Y. Solution: The random variables are said to come from a trinomial distribution in this case since there are three exhaustive and mutually exclusive outcomes light, good, or heavy, having probabilities p.5, p 2.85, and p 3., respectively. It is easy to see that the trinomial p.m.f. in this case is 5! f(x, y) x!y!(5 x y)! px p y 2p 5 x y 3. That is, f(x, y) is exactly the joint p.m.f. of X and Y, the number of light boxes and good boxes, where p.5, p 2.85, and p 3. are the various probabilities of each of the three events: light, good, and heavy. The random variable is binomially distributed, but the parameters of the distribution depend on the value of the random variable X. We have that the random variable Y is (conditionally) p binomially distributed b(n x, 2 p ) since the marginal distributions of X, Y are b(n, p ), b(n, p 2 ) and the conditional p.m.f. of Y is thus, with n 5, h(y x) f(x, y) f (x) n! x!y!(n x y)! px p y 2p n x y n!x!(n x)! x!y!(n x y)!n! px (n x)! y!(n x y)! ( p2 p x p p y 2 3 x!(n x)! n! p n x y 3 ( p ) y ( p ) n x ( p ) y ) y ( ) n x y p3. p In this case, with the specified values of n, p, p 2, and p 3, we have for X 3 h(y 3) 7! y!(7 y)! (.897)y (.53) 7 y, p x ( p ) n x so that Y is conditionally b(7,.897) when X 3. Since µ np for binomial distribution, we have that E(Y X 3) (7)(.897) 2.5. It is not hard to see that in fact E(Y x) (n x) p 2 p in general, and that a similar formula holds for E(X y). The correlation coefficient is now found using the fact that since each of

4 the conditional expectations E(Y x) (n x) p 2 p and E(X y) (n y) p p 2 is linear, then the square of the correlation coefficient ρ 2 is equal to the product of the respective coefficients of x and y in the conditional expectations. ( ) ( ) ρ 2 p2 p p p 2 p p 2 ( p )( p 2 ), from which it follows that p p 2 ρ ( p )( p 2 ).89. The fact that the correlation coefficient is negative follows from the fact that, for example, E(Y x) µ Y + ρ σ Y σ X (x µ X ), and noting that the coefficient of x in E(Y x) is seen to be negative (and also σ Y, σ X > ).. Let X have the uniform distribution U(, 2) and let the conditional distribution of Y, given that X x, be U(, x). Find the joint p.d.f. f(x, y) of X and Y, and be sure to state the domain of f(x, y). Find E(Y x). Solution: We have that f(x, y) y, < x 2, y x. x Now, E(Y x) x y y x dx 3x y3 x x2 3.

5 5 5. The support of a random variable X is the set of x-values such that f(x). Given that X has p.d.f. f(x) x 2 /3, < x < 2, what is the support of X 2? Find the p.m.f. of the random variable Y X 2. Solution: The p.d.f. g(y) of Y X 2 is obtained as follows. We note that the possible y-values that can be obtained are in the range y, so the support of g(y) needs to be the interval [, ]. Now, on the interval [, ], there is a one-to-one transformation represented by x y. We first find G(y) P (Y y) P (X 2 y) P (X y) for X x in [, 2], corresponding to Y y in [, ]. We have G(y) y f(x) dx y x 2 3 dx, and in particular g(y) G (y) ( y) 2 ( y), from the chain rule and the Fundamental 3 Theorem of Calculus. Simplifying, we have g(y) y, < y <. 6 In order to find g(y) on < y <, we need to work a little harder. For x in the interval < x < there is a two-to-one transformation given by x y for < x <, and x y for < x <. We then calculate G(y) for < y < (i.e., < x < ) as before, but now using two integrals G(y) P (Y y) P (X 2 y) P ( y < X < ) + P ( < X < y), so for y in the interval < y < we have G(y) y f(x) dx + y f(x) dx. Again, from the chain rule and the Fundamental Theorem of Calculus we have g(y) G (y) f( y) ( y) + f( y) ( y) y 3 2 y + y 3 2 y y 3. So, g(y) { y/3 if < y <, y/6 if < y <. There is no problem defining g() g(), or even just leaving the p.d.f. undefined at the points y and y.

6 6 6. Let X, X 2 denote two independent random variables each with the χ 2 (2) distribution. Find the joint p.d.f. of Y X and Y 2 X + X 2. What is the support of Y, Y 2 (i.e., what is the domain of the joint p.d.f., where f(y, y 2 ) )? Are Y and Y 2 independent? Solution: We have that X, X 2 have the same p.d.f. h(x) 2 e x/2, x <, corresponding to the χ 2 (r) distribution with r 2 degrees of freedom. By the way, this is the same as saying that X, X 2 follow exponential distributions with θ 2. Since X, X 2 are independent, the joint p.d.f. of X and X 2 is f(x, x 2 ) h(x )h(x 2 ) e (x +x 2 )/2. The change of variables formula is g(y, y 2 ) J f(v (y, y 2 ), v 2 (y, y 2 )) using the determinant of the Jacobian v (x, x 2 ) v (x, x 2 ) y y 2 J v 2 (x, x 2 ) v 2 (x, x 2 ) y y 2 where v i (y, y 2 ) is the inverse of u i, Y i u i (X, X 2 ), i, 2. In this case, Y u (X, X 2 ) X, so X v (Y, Y 2 ) Y, and Y 2 u 2 (X, X 2 ) X + X 2, so X 2 v 2 (Y, Y 2 ) Y 2 Y. Hence, J, so the joint p.d.f. of Y and Y 2 is g(y, y 2 ) f(y, y 2 y ) e y 2/2, y y 2 <. To determine if Y and Y 2 are independent we compute the marginal p.d.f. s. We have, and g 2 (y 2 ) g (y ) y2 y e y 2/2 dy 2 e y 2/2 dy y 2 e y 2/2, 2 e y 2/2 Since, g(y, y 2 ) g (y )g 2 (y 2 ), Y and Y 2 are dependent. y 2 e y /2.