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

Question

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].

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)=

Answer 1

So if we look at the formula, fX|Y(x|y) ~ N(-2x,1), so
E[Y|X=x] = -2x
E[Y] = 0

Answer 2

The rest I'm not sure:
cov(X,Y) = 0
E[X|Y=y] = -y/2

Answer 3

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

Answer 4

E[Y|X=x] = -2x
E[Y] = 0
Cov(X,Y)= -2
E[X∣Y=y]=-y/2
Var(X∣Y=y)=1/16

Answer 5

why the cov(X,Y) = -2

Answer 6

E[Y|X=x] = -2x
E[Y] = 0
Cov(X,Y)= -2
E[X∣Y=y]=-y/2
Var(X∣Y=y)=1/4

Answer 7

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

Answer 8

what kind of math is this😵😵😵😵😵🤯🤯🤯🤯🤯