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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 : $$ \mathbb{E}\left[\frac{S_{i+1}}{S_i} 1_{\frac{S_{i+1}}{S_i}<z}\right] $$

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    $\begingroup$ I think conditional expection is a more usual way to describe this. $\endgroup$ – SRKX Mar 18 '15 at 9:16
  • $\begingroup$ @SRKX No sorry, I don't think that's what I'm looking for. $\endgroup$ – Naucle Mar 18 '15 at 9:23
  • $\begingroup$ Ok then I changed it back, but aren't you looking to compute the expectation of the return give the return is below $z$? $\endgroup$ – SRKX Mar 18 '15 at 9:34
  • $\begingroup$ @SRKX Yes, that's what I'm looking for $\endgroup$ – Naucle Mar 18 '15 at 9:35
  • $\begingroup$ Well then isn't that the definition of conditional expectation? $\endgroup$ – SRKX Mar 18 '15 at 9:37
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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*}

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  • $\begingroup$ I guess this is the right answer, thank you. $\endgroup$ – Naucle Mar 18 '15 at 14:56
  • $\begingroup$ Note that I have changed the definitions for $d_1$ and $d_2$ so that they are consistent with the Black-Scholes formula defintions $\endgroup$ – Gordon Mar 19 '15 at 19:57
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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.

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  • $\begingroup$ Correct me if I'm wrong, but $\mathbb{E}[X|X<z] \neq \mathbb{E}[X\mathbb{I}_{X<z}]$ right? The first being conditional expectation and the second being called partial expectation apparently? $\endgroup$ – SRKX Mar 19 '15 at 2:28
  • $\begingroup$ Yes, the two are different. The two are related by $\mathbb{E}[X|X<z] = \frac{\mathbb{E}[\mathbb{I}_{X<z} X]}{\mathbb{P}(X<z)}$. I'll add that part to my answer $\endgroup$ – pbr142 Mar 19 '15 at 11:17

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