I'm completely stuck on how to do this problem. How can you go about calculating the variance of $Y$ and the covariance between $X$ and $Y$? I'm not sure how to use the information given to solve this problem.
Let X be normal with mean zero and variance $\sigma^2$. Let $Y$ be a call option on $X$ struck at $K = 0$. Calculate the $2 \times 2$ covariance matrix of $(X, Y)$.
Answer:
If X has mean 0 and variance $\sigma^2$, then Y is defined as $Max(X-K,0)$ and that is
$Y = X , X>0$
$Y = 0 ,X<0$
This translates into a truncated normal distribution.
I have added a link to finding the variance and expected value of a truncated normal distribution. I will attempt to give you the full solution.
https://en.wikipedia.org/wiki/Truncated_normal_distribution
$E(Y) = E(X/X>0) = \mu + \sigma \lambda(\alpha)$
$Var(Y) = \sigma^2(1-\delta(\alpha))$
where $\alpha = \frac{0-\mu}{\sigma}$
$\lambda(\alpha) = \frac{f(\alpha)}{(1-F(0))}$
$\delta(\alpha) = \lambda^2(\alpha)$
Substituting the value of $\mu = 0$ gives $\alpha = 0$
$\lambda(0) = 2f(\alpha)$ where $f(\alpha) = \frac{1}{\sqrt{2\pi}}$
$\lambda(0) = \sqrt{\frac{2}{\pi}}$
$E(Y) = E(X/X>0) = \sigma.\sqrt{\frac{2}{\pi}}$
$Var(Y) = Var(X/X>0) = \sigma^2(1-\frac{2}{\pi})$
Truncated Normal density function $= f(x,0,\sigma,0,\infty) = \frac{1}{\sigma \sqrt{2\pi}} \dfrac{e^\left(-\frac{x^2}{2\sigma^2}\right)}{(1-\phi(0))} $ $= \frac{2}{\sigma \sqrt{2\pi}} e^\left(-\frac{x^2}{2\sigma^2}\right)$
The stock price X should be perfectly positively correlated and hence the Covariance of $(X,Y) = \sqrt{Var(X)Var(Y)}$
$COV(X,Y) = COV(Y,X)= \sigma.\sigma(\sqrt{1-\frac{2}{\pi}}) = \sigma^2(\sqrt{1-\frac{2}{\pi}})$
$$D = \begin{bmatrix}\sigma^2 & \sigma^2(\sqrt{1-\frac{2}{\pi}})\\\ \sigma^2(\sqrt{1-\frac{2}{\pi}}) & \sigma^2(1-\frac{2}{\pi})\end{bmatrix}$$