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In Appendix 12A, p. 262 of this book, the author Hull derives a handy, tractable formula for the expression $E[\max(V-K, 0)]$, where $V$ is a lognormally distributed random variable and $K$ is a constant.

I would like to see if a similarly handy formula can be derived for $Var[\max(V-K, 0)]$.

I know that if I can just somehow use Hull's approach to find $E[\max(V-K, 0)^2]$ then I could find $Var[\max(V-K, 0)]$, but I am stuck.

Actually, what I am really interested in is $E[]$ and $Var[]$ of $\max(V, K)$, but I can probably figure this out if you can help me with the $\max(V-K, 0)$ case.

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  • $\begingroup$ Express $\max(V-K,0)$ as a mixture of an atom at $0$ and a lognormal distribution left-truncated at $K.$ The moments and weights of both components are straightforward to compute. The general problem of finding the moment of a mixture is also straightforward: see, for instance, stats.stackexchange.com/a/16609/919. $\endgroup$ – whuber Jun 6 '18 at 18:24
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I know that if I can just somehow use Hull's approach to find $E[\max(V−K,0)^2]$ then I could find $\text{Var}[\max(V−K,0)]$, but I am stuck.

If you apply the tricks in your reference, you’d get \begin{align*} E[\max(V−K,0)^2] &= \int_K^{\infty}(v-k)^2f_V(v)dv \\ &= \int_K^{\infty} (v-k)^2 \frac{1}{v}(2\pi\sigma^2)^{-1/2}\exp\left[ -\frac{(\log v - \mu)^2}{2 \sigma^2}\right] dv\\ &= \int_{\log K}^{\infty}\left(\exp[s] - k\right)^2 (2\pi \sigma^2)^{-1/2}\exp\left[- \frac{(s-\mu)^2}{2\sigma^2} \right]ds\\ &= \int_{\frac{\log k - \mu}{\sigma} }^{\infty}(e^{2(\mu + r \sigma)} - 2ke^{\mu + r \sigma} + k^2)(2\pi)^{-1/2}e^{-r^2/2}dr. \end{align*}

This breaks up into three integrals. Looking at the first: \begin{align*} \int_{\frac{\log k - \mu}{\sigma} }^{\infty}e^{2(\mu + r \sigma)}(2\pi)^{-1/2}e^{-r^2/2}dr &= e^{2\mu}\int_{\frac{\log k - \mu}{\sigma} }^{\infty}(2\pi)^{-1/2}e^{-\frac{(r^2 -4 r \sigma)}{2}}dr \\ &= e^{2\mu+\sigma^2}\int_{\frac{\log k - \mu}{\sigma} }^{\infty}(2\pi)^{-1/2}e^{-\frac{(r -2 \sigma)^2}{2}}dr\\ &= e^{2\mu+\sigma^2} P\left(R > \frac{\log k - \mu}{\sigma}\right) \end{align*} where $R \sim \text{Normal}(2\sigma,1)$. So by "completing the square" like this, we can turn expectations of censored normal variates into probabilities.

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    $\begingroup$ I'm simply not seeing how this answers the question. It doesn't provide a "handy, tractable" formula for the desired expression or, alternatively, explain why one doesn't exist. $\endgroup$ – jbowman Jun 6 '18 at 16:43
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The calculation of the expectation $E[max(V-k,0) ]$ is the value of a call option, prior to discounting for time.

To answer your question, it is given that $V \sim \text{Lognormal}(a,b)$ with pdf $f(v)$:

enter image description here

Then, you seek:

enter image description here

and

enter image description here

where:

  • Expect and Var are the expectation and variance functions from the mathStatica package for Mathematica, which I am using to automate the calculation;
  • Erf[z] denotes the error function $\text{erf}(z)=\frac{2}{\sqrt{\pi }}\int _0^z e^{-t^2} d t$, and where the cdf of a standard Normal variable is given by $\frac{1}{2} \left(1+\text{erf}\left(\frac{z}{\sqrt{2}}\right)\right)$;
  • and making the substitution $c = \frac{a-\log (k)}{\sqrt{2} b}$ will make the result look neater.
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  • $\begingroup$ This was very helpful. Wish I could give the check to both answers. $\endgroup$ – ben Jun 7 '18 at 18:49

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