Let X be a standard normal random variable. Let Y be a continuous random variable such that
fY|X(y|x)=1/√2π*exp(−(y+2x)^2/2).
Find E[Y|X=x] (as a function of x, in standard notation) and E[Y].
unanswered
Compute Cov(X,Y).
unanswered
Continuing from above, find the conditional expectation E[X∣Y=y] and conditional variance Var(X∣Y=y).
Hint: The conditional PDF of X given Y=y is of the form
h(y)exp(−g(x,y))
for some function h(y) and some quadratic function g(x,y). Rewriting the function g(x,y) in a familiar form will give E[X∣Y=y] and Var(X∣Y=y) without much computation.
E[X∣Y=y]= unanswered
Var(X∣Y=y)= unanswered
Does someone have the answer please?
jiskha.com/questions/1798909/problem-4-gaussian-random-variables-let-x-be-a-standard-normal-random-variable-let-y-be
use this thread instead ^
-2x; 0; -2; -2y/5 and 1/5
I got 2y/5 for (4).
To find the conditional expectation E[Y|X=x], we can use the definition of conditional expectation:
E[Y|X=x] = ∫y fY|X(y|x) dy
where fY|X(y|x) is the conditional probability density function (PDF) of Y given X=x.
In this case, the conditional PDF is given by:
fY|X(y|x) = 1/√(2π) * exp(-(y+2x)^(2)/2)
To find E[Y|X=x], we substitute this conditional PDF into the formula:
E[Y|X=x] = ∫y (1/√(2π)) * exp(-(y+2x)^(2)/2) dy
This integral can be evaluated using standard techniques of integration, such as completing the square and making a change of variables.
To find E[Y], we need to find the marginal expectation of Y. This is given by integrating the joint PDF of X and Y over all possible values of Y:
E[Y] = ∫∫y fX,Y(x, y) dx dy
where fX,Y(x, y) is the joint probability density function of X and Y.
To find Cov(X,Y), we use the formula for covariance:
Cov(X,Y) = E[XY] - E[X]E[Y]
where E[XY] is the joint expectation of X and Y.
To find E[X|Y=y] and Var(X|Y=y), we can use the hint given. The conditional PDF of X given Y=y is given by:
fX|Y(x|y) = h(y) * exp(-g(x,y))
where h(y) is a function of y and g(x,y) is a quadratic function of x and y.
By rewriting g(x,y) in a familiar form, we can determine the conditional expectation E[X|Y=y] and conditional variance Var(X|Y=y) without much computation.
To summarize:
E[Y|X=x] = ∫y (1/√(2π)) * exp(-(y+2x)^(2)/2) dy
E[Y] = ∫∫y fX,Y(x, y) dx dy
Cov(X,Y) = E[XY] - E[X]E[Y]
E[X|Y=y] = ?
Var(X|Y=y) = ?