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Let $W$ be a Wiener process, $r,\sigma \in \mathbb{R}_+$ and $S(T) = S(t)e^{(r-\frac12 \sigma^2)(T-t) + \sigma(W(T)-W(t))}$.

I want to evaluate $$A:=E[e^{- \frac12 \sigma^2 (T-t) - \sigma(W(T)-W(t)) } * \mathbf{1}_{\{S(T)>1.5\}}|\mathscr{F}_t].$$


You will see that the left-hand factor of the expectation is an exponential martingale of the form $\mathcal{E}(-\sigma W)_t$.

Question: What can I do to progress further? I feel like I can't go:

$$A = E[e^{- \frac12 \sigma^2 (T-t) - \sigma(W(T)-W(t)) }|\mathscr{F}_t] * E[\mathbf{1}_{\{S(T)>1.5\}}|\mathscr{F}_t]$$

because $W(T)$ appears both inside the $e^{...}$ and inside the $\mathbf{1}_{\{...\}}$, making them dependent.

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  • $\begingroup$ Have tried Girsanov Theorem ? $\endgroup$ – TheBridge Nov 24 '12 at 8:08
  • $\begingroup$ @TheBridge I am not allowed to change probability measure in this question (they specifically told us so). $\endgroup$ – Jase Nov 24 '12 at 8:31
  • $\begingroup$ But $[S(T)\gt0]=[S(t)\gt0]$, no? Which would imply that the event belongs to $\mathcal F_t$, and make the answer trivial. $\endgroup$ – Did Nov 24 '12 at 11:58
  • $\begingroup$ @did I apologize for the typo. $\endgroup$ – Jase Nov 24 '12 at 12:16
  • $\begingroup$ Jase: Just out of curiosity, who is asking you to solve this? $\endgroup$ – Did Nov 24 '12 at 19:44
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Since $W(T)-W(t)$ is independent of $\mathcal F_t$, $A=\mathbb E(X\cdot\mathbf 1_{S(t)Y\gt x}\mid\mathcal F_t)$ for $x=1.5$ and for some random variables $X$ and $Y$ which are independent of $\mathcal F_t$. Hence $A=\mathbb E(X)\cdot G(S(t))$ where $G(z)=\mathbb P(zY\gt x)$.

To go further, one can note that $X=\exp(- \frac12 \sigma^2 (T-t) - \sigma(W(T)-W(t)))$ hence $\mathbb E(X)=1$, and that $Y=\exp((r- \frac12\sigma^2) (T-t)-\sigma\sqrt{T-t}Z)$, where $Z$ is standard normal, hence $$ G(z)=\Phi\left(\frac{\log z-\log x+(\frac12\sigma^2-r) (T-t)}{\sigma\sqrt{T-t}}\right), $$ or some similarly ugly and mostly useless formula.

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  • $\begingroup$ Just saying, this formula is related to Black Schole's modell. $\endgroup$ – user519338 Jan 27 '18 at 1:01

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