3
$\begingroup$

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

$\endgroup$
  • 1
    $\begingroup$ 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
2
$\begingroup$

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*}

$\endgroup$
  • $\begingroup$ 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
1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.