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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 $(X, Y)$.

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

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

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

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