# Math

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].

E[Y|X=x]=

Compute Cov(X,Y).

The conditional PDF of X given Y=y is of the form

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]=

Var(X∣Y=y)=

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

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

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

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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

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5. why the cov(X,Y) = -2

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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

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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

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