Let $X_t= e^{\left(\mu-\sigma^2/2 \right)t+\sigma W_t}$ be a geometric Brownian motion with drift $\mu$ and volatility $\sigma$. I am trying to find an analytical solution to

$$\mathbb{E}\left[ \max(a X_T + b X_S -K,0)\right],$$ where $a$, $b$ and $K$ are constants and $0<S<T$.

My objective is to find the critical point as from which $aX_T + bX_S$ will be greater than $K$ so that I can disregard the maximum function and evaluate the expectation.

Am I right to say that if I had only $Y= X_T + X_S$, I could use the relation $X_T + X_S=2X_S + X_T - X_S$ to find its mean and variance and subsequently find the critical point?

Is there any way to proceed in the same way for my original problem?


You can write

$$\mathbb{E}\left[ \max(a X_T + b X_S -K,0)\right] = \mathbb{E}\left[ \max(a X_S Y_{S,T} + b X_S -K,0)\right],$$

with $Y_{S,T} = X_T/X_S.$

For a given value of $X_S$ we can write $$\mathbb{E}\left[ \max(a X_S Y_{S,T} + b X_S -K,0)\right] = X_S \mathbb{E}\left[ \max(a Y_{S,T} + b -K/X_s,0)\right],$$ since $Y_{S,T}$ is log-normal this can be evaluated by a BS type formula.

We then integrate numerically over the value of $X_S$ with a log-normal density.

  • $\begingroup$ I am slightly confused. Are we using the law of iterated expectations? That is, first condition on $X_S$ to use the BS formula and then use numerical integration to find the value at some time $t<S$? $\endgroup$ – Bhavish Suarez Aug 10 '15 at 5:55
  • $\begingroup$ yes if you like $\endgroup$ – Mark Joshi Aug 10 '15 at 6:45
  1. let $X$ be a log-normal random with mean $\mu$ and variance $\sigma^2$ then $aX$ is said to have a scaled log-normal distribution with mean $a\mu$ and and variance $a^2\sigma^2$.
  2. Let $X_j$ be a independent log-normally distributed variables with mean $\mu_j$ and variance $\sigma_j^2$ and $Y=\sum_{j=1}^{n}X_j$.The distribution of $Y$ has no closed-form expression, but can be reasonably approximated by another log-normal distribution $Z$ at the right tail. Its probability density function at the neighborhood of $0$ has been characterized and it does not resemble any log-normal distribution.A commonly used approximation due is obtained by matching the mean and variance of another log-normal distribution: \begin{align} &\mu_Z=\ln\left(\sum_{j=1}^{n}e^{2\mu_j+\sigma_j^2/2}\right)-\frac{1}{2}\sigma^2_Z\\ &\sigma^2_Z=\ln\left(\frac{\sum_{j=1}^{n}{e^{2\mu_j+\sigma_j^2}}(e^{\sigma_j^2}-1)}{\left[\sum_{j=1}^{n}e^{2\mu_j+\sigma_j^2}\right]^2}+1\right) \end{align}
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