Let X be a standard normal random variable. Let Y be a continuous random variable such that
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
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
To find the conditional expectation E[Y|X=x], we need to integrate Y times the conditional PDF f_{Y|X}(y|x) over all possible values of Y.
E[Y|X=x] = ∫ Y * f_{Y|X}(y|x) dy
Plugging in the given conditional PDF, we get:
E[Y|X=x] = ∫ Y * (1/√(2π)) * exp(-(y+2x)^2/2) dy
To find the expectation E[Y], we need to take the average of E[Y|X=x] over all possible values of X. Since X is a standard normal random variable, its PDF is given by:
f_X(x) = (1/√(2π)) * exp(-x^2/2)
We can use this PDF to calculate E[Y] as the integral of E[Y|X=x] multiplied by f_X(x) over all possible values of X.
E[Y] = ∫ E[Y|X=x] * f_X(x) dx
To compute Cov(X,Y), we need to first find E[XY] and then subtract E[X] * E[Y].
E[XY] = ∫ ∫ X * Y * f_{X,Y}(x,y) dx dy, where f_{X,Y}(x,y) is the joint PDF of X and Y
Cov(X,Y) = E[XY] - E[X] * E[Y]
Now, let's calculate E[Y|X=x], E[Y], and Cov(X,Y) step-by-step.
1. E[Y|X=x]:
Using the given conditional PDF, we plug it into the formula for E[Y|X=x] and integrate:
E[Y|X=x] = ∫ Y * (1/√(2π)) * exp(-(y+2x)^2/2) dy
This integral can be simplified using techniques such as completing the square.