Problem 4. Gaussian Random Variables

Question

Let X be a standard normal random variable. Let Y be a continuous random variable such that

fY|X(y|x)=12π−−√exp(−(y+2x)22).

Find E[Y|X=x] (as a function of x , in standard notation) and E[Y] .

E[Y|X=x]=
unanswered

E[Y]=
unanswered
Compute Cov(X,Y) .

Cov(X,Y)=
unanswered
The conditional PDF of X given Y=y is of the form

α(y)exp{−quadratic(x,y)}

By examining the coefficients of the quadratic function in the exponent, find E[X∣Y=y] and Var(X∣Y=y) .

E[X∣Y=y]=
unanswered

Var(X∣Y=y)=
unanswered

Answer 1

anyone else getting 0 for the Covariance?

Answer 2

no, I am getting Cov=1

Answer 3

E[Y|X=x] = -2x
E[Y] = 0
cov(X,Y)=0

Answer 4

sorry, made a mistake with my Covariance, it is not zero. Any one else get these numbers:

(1) -2*x
0
(2) -2
(3) (-2*y)/5
(4) (2*x)+(1/5)

Answer 5

Sorry, (4) is wrong above: Here are my latest answers: let me know if I am wrong.

E[Y|X=x] = -2*x
E[Y] = 0
Cov[X,Y]= -2
E[X|Y=y]= (-2*y)/5
Var[X|Y=y]= 1/5

Answer 6

How did you guys calculate cov(x,y)?

Answer 7

1. E[Y|X=x] = -2*x
2. E[Y] = 0
3. Cov[X,Y]= -2

Answer 8

I've got the same 4 answers. Not sure if they're right though

E[Y|X=x] = -2*x
E[Y] = 0
Cov[X,Y]= -2
E[X|Y=y]= (-2*y)/5
Var[X|Y=y]= 1/5