Problem 4. Gaussian Random Variables
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
How did you guys calculate cov(x,y)?
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
1. E[Y|X=x] = -2*x
2. E[Y] = 0
3. Cov[X,Y]= -2
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
To find E[Y|X=x], we need to integrate over the possible values of Y while keeping X=x.
1. Start by writing out the integral: E[Y|X=x] = ∫ y * fY|X(y|x) dy, where fY|X(y|x) is the conditional PDF given in the problem statement.
2. Substitute the given expression for fY|X(y|x) into the integral: E[Y|X=x] = ∫ y * (12π)^(-1/2) * exp(-(y+2x)^2/2) dy.
3. Solve the integral by recognizing that the integrand is proportional to the PDF of a normal distribution. The integral evaluates to ∫ y * (12π)^(-1/2) * exp(-(y+2x)^2/2) dy = x.
Therefore, E[Y|X=x] = x.
To find E[Y], we need to compute the unconditional expectation of Y, which means integrating over the possible values of X as well.
1. Start by writing out the integral: E[Y] = ∫ E[Y|X=x] * fX(x) dx, where fX(x) is the standard normal PDF.
2. Substitute the expression for E[Y|X=x] from the previous step: E[Y] = ∫ x * fX(x) dx.
3. Recognize that the integrand is proportional to the PDF of a normal random variable. The integral evaluates to ∫ x * fX(x) dx = 0.
Therefore, E[Y] = 0.
To compute Cov(X,Y), we need to integrate the joint PDF of X and Y multiplied by the deviations of X and Y from their means.
1. Start by writing out the integral: Cov(X,Y) = ∫ ∫ (X - E[X]) * (Y - E[Y]) * fX,Y(x,y) dx dy.
2. Substitute the given expression for fY|X(y|x) into fX,Y(x,y): Cov(X,Y) = ∫ ∫ (X - E[X]) * (Y - E[Y]) * (12π)^(-1/2) * exp(-(y+2x)^2/2) dx dy.
3. Solve the double integral by recognizing the integrand as a bivariate normal distribution. The integral evaluates to Cov(X,Y) = ∫ (X - E[X]) * (Y - E[Y]) * fX(x) dx.
4. Recognize that the integrand is proportional to the PDF of a normal random variable. The integral evaluates to Cov(X,Y) = ∫ (X - E[X]) * (Y - E[Y]) * fX(x) dx = ∫ x * 0 * fX(x) dx = 0.
Therefore, Cov(X,Y) = 0.
Now, let's move on to finding E[X|Y=y] and Var(X|Y=y) using the conditional PDF of X given Y=y.
By examining the coefficients of the quadratic function in the exponent, we can determine the mean and variance of X given Y=y.
1. The expression α(y)exp{-quadratic(x,y)} contains the quadratic term -(y+2x)^2/2. From this, we can deduce that the mean of X given Y=y is -2y.
2. The quadratic term also allows us to determine the variance of X given Y=y. In this case, the variance is 1/2.
Therefore, E[X|Y=y] = -2y and Var(X|Y=y) = 1/2.
anyone else getting 0 for the Covariance?
no, I am getting Cov=1
E[Y|X=x] = -2x
E[Y] = 0
cov(X,Y)=0
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)