1. Suppose X and Y are jointly distributed with pdf fx,y (x, y), where fx,y(x, y) = — e-(x−y)² e-x². TT Note that fx (x) = f e-(x-y)² e-x² dy. TT a. By making the substitution t = y - x, simplify the expression for fx (x) by 1 e-((t-μ)²/(20²)) evaluating the integral above. You may use the fact that is the probability density for a normal random variable with mean μ and variance o2, so that 8 Love e-((t-μ)²/(20²)) dt = 1 1 V2πσ2 for any values μ and o b. Find f(y|X = x) (y). (Note that your answer will depend on both x and y. This is because for each value of x, there is a different marginal distribution for Y which is a function of y.) c. Find E[Y|X = x]. (Hint: for part b., you should have gotten a normal distribution. Note your answer will be a function of x only.) d. Find Var E[YIX]]. (Hint: In your answer for c., replace x with X.)

MATLAB: An Introduction with Applications
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Chapter1: Starting With Matlab
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1. Suppose X and Y are jointly distributed with pdf fx,y (x, y), where
fx,y (x, y) = ¹⁄e-(x-y)² e-x².
Note that
TT
fx(x) = √ ² e-(x-y)² e-x² dy.
-00 T
a. By making the substitution t = y - x, simplify the expression for fx(x) by
evaluating the integral above. You may use the fact that
e-((t-μ)²/(20²))
1
V2πσ2
is the probability density for a normal random variable with mean u and
variance o2, so that
8
1
√2² e-
for
J18
e-((t-μ)²/(20²)) dt = 1
any
values u and o
b. Find f(Y|X = x)(y). (Note that your answer will depend on both x and y.
This is because for each value of x, there is a different marginal distribution for Y
which is a function of y.)
c. Find E[Y|X = x]. (Hint: for part b., you should have gotten a normal
distribution. Note your answer will be a function of x only.)
d. Find Var[E[Y|X]]. (Hint: In your answer for c., replace x with X.)
Transcribed Image Text:1. Suppose X and Y are jointly distributed with pdf fx,y (x, y), where fx,y (x, y) = ¹⁄e-(x-y)² e-x². Note that TT fx(x) = √ ² e-(x-y)² e-x² dy. -00 T a. By making the substitution t = y - x, simplify the expression for fx(x) by evaluating the integral above. You may use the fact that e-((t-μ)²/(20²)) 1 V2πσ2 is the probability density for a normal random variable with mean u and variance o2, so that 8 1 √2² e- for J18 e-((t-μ)²/(20²)) dt = 1 any values u and o b. Find f(Y|X = x)(y). (Note that your answer will depend on both x and y. This is because for each value of x, there is a different marginal distribution for Y which is a function of y.) c. Find E[Y|X = x]. (Hint: for part b., you should have gotten a normal distribution. Note your answer will be a function of x only.) d. Find Var[E[Y|X]]. (Hint: In your answer for c., replace x with X.)
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