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 $(X, Y)$.



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.


$E(Y) = E(X/X>0) = \mu + \sigma \lambda(\alpha)$

$Var(Y) = \sigma^2(1-\delta(\alpha))$

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 $(X,Y) = \sqrt{Var(X)Var(Y)}$

$COV(X,Y) = COV(Y,X)= \sigma.\sigma(\sqrt{1-\frac{2}{\pi}}) = \sigma^2(\sqrt{1-\frac{2}{\pi}})$

$$D = \begin{bmatrix}\sigma^2 & \sigma^2(\sqrt{1-\frac{2}{\pi}})\\\ \sigma^2(\sqrt{1-\frac{2}{\pi}}) & \sigma^2(1-\frac{2}{\pi})\end{bmatrix}$$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.