How to compute the conditional expected value of a geometric brownian motion?

Question

I'm working on a project, and I have to use the cumulative and conditional expected value of the variations of a stock following a Geometric Brownian Motion.

I know that the cumulative is as follows : $$ \mathbb{E}\left[ \mathbb{1}_{ \frac{S_{i+1}}{S_{i}} < z}\right] = \mathbb{P} \left[ \frac{S_{i+1}}{S_{i}} < z \right] = \Phi\left(\frac{\log(z) - (r- \frac{\sigma^2}{2})(t_{i+1}-t_i)}{\sigma \sqrt{t_{i+1}-t_i}}\right) $$

$\Phi$ being the standard normal distribution cumulative function.

But I couldn't find the expression of the conditional expected value : $$ \mathbb{E}\left[\frac{S_{i+1}}{S_i} 1_{\frac{S_{i+1}}{S_i}<z}\right] $$

Answer 1

Note that \begin{align*} E\bigg(\frac{S_{i+1}}{S_i}\mathbb{I}_{\frac{S_{i+1}}{S_i} < z}\bigg) &=zE\bigg(\mathbb{I}_{\frac{S_{i+1}}{S_i} < z}\bigg)-E\bigg(\Big(z-\frac{S_{i+1}}{S_i}\Big)\mathbb{I}_{\frac{S_{i+1}}{S_i} < z}\bigg) \\ &=zP\bigg(\frac{S_{i+1}}{S_i}<z\bigg)-E\bigg(\Big(z-\frac{S_{i+1}}{S_i}\Big)^+\bigg). \end{align*} Then you can compute the expectation using the put option pricing formula.

Alternatively, note that \begin{align*} \frac{S_{i+1}}{S_i} &= e^{(r-\frac{\sigma^2}{2})(t_{i+1}-t_i) + \sigma (W_{t_{i+1}}-W_{t_i})}\\ &=e^{(r-\frac{\sigma^2}{2})(t_{i+1}-t_i) + \sigma \sqrt{t_{i+1}-t_i} \xi}, \end{align*} where $\xi$ is a standard normal random variable. Then $\frac{S_{i+1}}{S_i}<z$ is equivalent to \begin{align*} \xi <\frac{\ln z-(r-\frac{\sigma^2}{2})(t_{i+1}-t_i)}{\sigma \sqrt{t_{i+1}-t_i}}. \end{align*} Let \begin{align*} d_2 = -\frac{\ln z-(r-\frac{\sigma^2}{2})(t_{i+1}-t_i)}{\sigma \sqrt{t_{i+1}-t_i}}. \end{align*} We then have that \begin{align*} E\bigg(\frac{S_{i+1}}{S_i}\mathbb{I}_{\frac{S_{i+1}}{S_i} < z}\bigg) &= \int_{-\infty}^{-d_2}e^{(r-\frac{\sigma^2}{2})(t_{i+1}-t_i) + \sigma \sqrt{t_{i+1}-t_i} x}\frac{1}{\sqrt{2\pi}}e^{-\frac{x^2}{2}}dx\\ &=\int_{-\infty}^{-d_2}e^{r(t_{i+1}-t_i) }\frac{1}{\sqrt{2\pi}}e^{-\frac{\big(x - \sigma \sqrt{t_{i+1}-t_i}\big)^2}{2}}dx\\ &=\int_{-\infty}^{-d_2- \sigma \sqrt{t_{i+1}-t_i}}e^{r(t_{i+1}-t_i) }\frac{1}{\sqrt{2\pi}}e^{-\frac{x^2}{2}}dx\\ &=e^{r(t_{i+1}-t_i) }\Phi(-d_1), \end{align*} where \begin{align*} d_1 = d_2+ \sigma \sqrt{t_{i+1}-t_i}. \end{align*}

Answer 2

What you are looking for is the partial expectation of $\frac{S_{i+1}}{S_i}$. Since $\frac{S_{i+1}}{S_i}$ is lognormally distributed, you can use the following result:

For a lognormal random variable $X \sim LND(m,v^2)$, $$ E(X | X < z) = E[X] \Phi\left( \frac{\log(z)-m-v^2}{v} \right) $$ In your case, $m = (r-\frac{1}{2}\sigma^2) (t_{i+1}-t_{i})$, $v^2 = \sigma^2 (t_{i+1}-t_{i})$, and $E[X] = S_i e{(r+\frac{1}{2}\sigma^2) (t_{i+1}-t_{i})}$.

You can then use the fact that $\mathbb{E}[X|X<z] = \frac{\mathbb{E}[\mathbb{I}_{X<z} X]}{\mathbb{P}(X<z)}$ to get the desired expression.