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

Question

f_{Y|X}(y|x)=\frac{1}{\sqrt{2\pi }}\exp \left(-\frac{(y+2x)^2}{2}\right).

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

\mathbf{E}[Y|X=x]=\quad


\mathbf{E}[Y]=\quad

Compute {\rm Cov}(X,Y).

{\rm Cov}(X,Y)=\quad

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

\alpha (y) \exp \left\{ -\text {quadratic}(x,y)\right\}

By examining the coefficients of the quadratic function in the exponent, find {\bf E}[X\mid Y=y] and {\rm Var}(X\mid Y=y).

{\bf E}[X\mid Y=y]=\quad


{\rm Var}(X\mid Y=y)=\quad

Answer 1

To find E[Y|X=x], we can use the definition of conditional expectation:

E[Y|X=x] = ∫y * f_{Y|X}(y|x) dy

Plug in the given expression for f_{Y|X}(y|x):

E[Y|X=x] = ∫y * (1/√(2π)) * exp(-(y+2x)^2/2) dy

This integral can be difficult to solve analytically, so we will leave it as is.

Next, to find E[Y], we can use the law of total expectation:

E[Y] = E[E[Y|X]]

Since Y is a continuous random variable, we can integrate over the entire range of X to compute the inner expectation:

E[Y] = ∫E[Y|X=x] * f_X(x) dx

Since X is a standard normal random variable, its probability density function is:

f_X(x) = (1/√(2π)) * exp(-x^2/2)

Plugging this into the equation for E[Y], we get:

E[Y] = ∫E[Y|X=x] * (1/√(2π)) * exp(-x^2/2) dx

Again, this integral may be difficult to solve analytically, so we will leave it as is.

To compute Cov(X,Y), we can use the definition of covariance:

Cov(X,Y) = E[(X - E[X]) * (Y - E[Y])]

Since E[X] = 0 (since X is a standard normal random variable), this simplifies to:

Cov(X,Y) = E[X * (Y - E[Y])]

Again, we can use the law of total expectation to compute this expectation:

Cov(X,Y) = ∫E[X * (Y - E[Y])|X=x] * f_X(x) dx

But E[X * (Y - E[Y])|X=x] = x * (E[Y|X=x] - E[Y]), so we have:

Cov(X,Y) = ∫x * (E[Y|X=x] - E[Y]) * f_X(x) dx

Note that E[Y] was already computed in the previous step, so we can use that value in this equation.

To find E[X|Y=y] and Var(X|Y=y), we need to find the conditional PDF of X given Y=y. The given expression for f_{Y|X}(y|x) implies:

f_{X|Y}(x|y) = α(y) * exp(-(x + 2y)^2/2)

We can normalize this PDF by integrating it over its entire range and setting it equal to 1:

∫α(y) * exp(-(x + 2y)^2/2) dx = 1

To find α(y), we can divide the PDF by its integral:

α(y) = (1 / ∫exp(-(x + 2y)^2/2) dx)

Again, this integral may be difficult to solve analytically, so we will leave α(y) as is.

Given this conditional PDF, we can find E[X|Y=y] and Var(X|Y=y) using the definitions:

E[X|Y=y] = ∫x * f_{X|Y}(x|y) dx

Var(X|Y=y) = ∫(x - E[X|Y=y])^2 * f_{X|Y}(x|y) dx