Conditional Expectation of an Exponential Brownian Motion
Summary
The document checks the conditional expectation of an exponential function of Brownian motion given the information available at an earlier time. Brownian increments after that time are independent of the earlier information and normally distributed, so the conditional expectation separates into the known exponential of the earlier Brownian value and the moment-generating function of the future increment. Completing the square in the normal density gives the increment’s exponential moment.
The resulting conditional expectation includes both the earlier Brownian value and a variance adjustment proportional to the elapsed time. The question’s proposed calculation omits the exponential around that adjustment and also mixes the coefficient on Brownian motion in its final line; the answer supplies the corrected form for an exponential with volatility coefficient. A second explanation uses the stochastic exponential martingale. These derivations rely on standard Brownian motion and the stated filtration, and the stochastic-exponential route assumes familiarity with stochastic calculus.
Key ideas
- A future Brownian increment is independent of information available at the conditioning time.
- The exponential moment of a centered normal variable includes half its variance in the exponent.
- The conditional expectation is the exponential of the current Brownian value plus the variance adjustment.
- The same result can be expressed through the stochastic exponential martingale.
Tags
Full text
# Conditional expectation of a geometric brownian motion
# Conditional expectation of a geometric brownian motion
I'm reviewing stuff from the past and I'm very confused all of a sudden. Some verification would help about the following.
$$ \mathbb{E}[e^{\sigma W(t)}|{\cal F}_s] = \mathbb{E}[e^{\sigma (W(t) - W(s) + W(s))}|{\cal F}_s] = \mathbb{E}[e^{\sigma (W(t) - W(s))}|{\cal F}_s]e^{W(s)} $$ $$ =\mathbb{E}[e^{\sigma (W(t) - W(s))}]e^{W(s)} =\frac{1}{2}\sigma^2 (t-s) e^{W(s)} $$
Is this true?
## Answer by LocalVolatility (score 2, accepted)
https://quant.stackexchange.com/a/30707
We have
\begin{equation} \sigma \left( W_t - W_s \right) \sim \mathcal{N} \left( 0, \sigma^2 (t - s) \right). \end{equation}
Let $X \sim \mathcal{N} \left( 0, \xi^2 \right)$, then
\begin{eqnarray} \mathbb{E} \left[ e^{X} \right] & = & \frac{1}{\sqrt{2 \pi} \xi} \int_\mathbb{R} \exp \left\{ x -\frac{x^2}{2 \xi^2} \right\} \mathrm{d}x\\ & = & \frac{1}{\sqrt{2 \pi} \xi} \int_\mathbb{R} \exp \left\{ -\frac{x^2 - 2 x \xi^2 \pm \xi^4}{2 \xi^2} \right\} \mathrm{d}x\\ & = & \frac{1}{\sqrt{2 \pi} \xi} e^{\xi^2 / 2} \int_\mathbb{R} \exp \left\{ -\frac{\left( x - \xi^2 \right)^2}{2 \xi^2} \right\} \mathrm{d}x\\ & = & e^{\xi^2 / 2}, \end{eqnarray}
where we recognize the integrand in the second last line as the density of of a $\mathcal{N} \left( \xi^2, \xi^2 \right)$ normal random variable which integrates to one. Thus
\begin{equation} \mathbb{E} \left[ e^{\sigma \left( W_t - W_s \right)} \right] = \exp \left\{ \frac{1}{2} \sigma^2 (t - s) \right\}. \end{equation}
Your other steps are correct, i.e.
\begin{equation} \mathbb{E} \left[ \left. e^{W_t} \right| \mathcal{F}_s \right] = \exp \left\{ W_s + \frac{1}{2} \sigma^2 (t - s) \right\} \end{equation}
## Answer by Quantuple (score 3)
https://quant.stackexchange.com/a/30711
Depending on how well you are familiar with stochastic calculus, another way of getting to this result is to recognise that: $$ e^{\sigma W_t} = \mathcal{E}[\sigma W_t] e^{\frac{1}{2}\sigma^2 t} $$ where $$ \mathcal{E}(X_t) = \exp\left(X_t - \frac{1}{2}\langle X \rangle_t\right) $$ denotes the Doléans-Dade exponential (stochastic exponential) of an Itô process $X_t$, which is well-known to verify the martingale property i.e. $$ \Bbb{E}_0\left[\mathcal{E}(X_t)\right] = \exp(X_0)$$ (of course the proof relies on @LocalVolatility's answer)
Using this result along with the fact that $W_0=0$ by definition for a standard Brownian motion, one gets: $$ \Bbb{E}_0 \left[ e^{\sigma W_t} \right] = \underbrace{\Bbb{E}_0 \left[ \mathcal{E}(\sigma W_t) \right]}_{=\exp(\sigma W_0)=1} \Bbb{E}_0 \left[ e^{\frac{1}{2}\sigma^2 t} \right] = e^{\frac{1}{2}\sigma^2 t} $$Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)
This summary was written by Stratmill's research agent from the original; it is not a copy of the source.