Conditional Expectations for an Exponential Brownian Motion Sum
Summary
The document solves for the expectation and conditional expectation of an exponential built from Brownian motion at two different times. It first combines the correlated Brownian values by expressing the later value as the earlier value plus an independent increment, giving the variance needed to compute the unconditional expectation.
For the conditional expectation, the answer treats the intervals before and after the midpoint separately. Before the midpoint, both Brownian terms retain future uncertainty, so conditioning is applied through the midpoint and uses the martingale property of an exponential Brownian motion. At or after the midpoint, the earlier Brownian value is known and only the later value requires conditioning. The response also shows that a scaled process formed from the sum of Brownian values at t and t/2 is not itself Brownian, because its increments are not independent. The derivation depends on the natural Brownian filtration specified in the question.
Key ideas
- The exponential expectation follows from the normal distribution and variance of the sum of Brownian values.
- The conditional expectation has different forms before and after the midpoint because different terms are then observable.
- The tower property reduces conditioning before the midpoint to conditioning through the midpoint.
- A scaled sum of Brownian motion evaluated at t and t/2 does not have independent increments.
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# Questions about exponential Brownian motion
# Questions about exponential Brownian motion
Let $(\Omega,\mathcal{F},P)$ be a probability space, equipped with a filtration $(\mathcal{F})_{0 \leq t \leq T}$ that is the natural filtration of a standard Brownian motion $(W_{t})_{0 \leq t \leq T}$.
Let $X=\exp(W_{T/2}+W_{T})$. Find the expectation $E[X]$;
Let $X_{t}=E[X|\mathcal{F}_{t}]$ for $0 \leq t \leq T$. Find $X_{t}$.
The first question is easy for me: $W_{T/2}+W_{T}=2W_{T/2}+W_{T}-W_{T/2}$, by independence of increments and the property of Brownian motion, $W_{T/2}+W_{T} \sim N(0,5T/2)$,therefore, $E[X]=\exp(5T/4)$.
I have tried to solve the second question as:
Since $W_{t/2}+W_{t}\sim N(0,5t/2)$, $B_{}t:=\sqrt{2/5}(W_{t/2}+W_{t})\sim N(0,t)$ Can I say that B_{t} is a Brownian motion? If not, Is there any rigorous way to prove this?
If B_{t} is a Brownian motion, then, $E[e^{\sqrt{\frac{5}{2}}B_{T}}|\mathcal{F}_{t}]=E[e^{\sqrt{\frac{5}{2}}(B_{T}-B_{t}+B_{t})}|\mathcal{F}_{t}]=e^{\sqrt{\frac{5}{2}}B_{t}}e^{5(T-t)/4}$.
i.e.$X_{t}=e^{W_{t}+W_{t/2}}e^{5(T-t)/4}$.
By the way, how can we solve by discuss the cases $t<T/2$ and $T/2 \leq t < T$ seperately?
Thanks!
## Answer by Gordon (score 4, accepted)
https://quant.stackexchange.com/a/17980
For $T/2 \leq t \leq T$, \begin{align*} E(X\mid \mathcal{F}_t) &= \exp\big(W_{\frac{T}{2}}+\frac{1}{2}T\big) E\big(\exp\big(W_{T}-\frac{1}{2}T\big)\mid \mathcal{F}_t\big)\\ &= \exp\big(W_{\frac{T}{2}}+\frac{1}{2}T\big) \exp\big(W_{t}-\frac{1}{2}t\big)\\ &=\exp\big(W_{\frac{T}{2}}+W_{t} + \frac{1}{2}T-\frac{1}{2}t\big). \end{align*} For $0 \leq t \leq T/2$, \begin{align*} E(X\mid \mathcal{F}_t) &= E\big( E(X\mid \mathcal{F}_{T/2})\mid\mathcal{F}_t)\big)\\ &=E\big(\exp(2W_{\frac{T}{2}}+ T/4)\mid\mathcal{F}_t\big)\\ &=\exp\big(\frac{5}{4}T\big)E\big(\exp\big(2W_{\frac{T}{2}} - \frac{1}{2}\times 2^2 \times T/2\big)\mid\mathcal{F}_t\big)\\ &= \exp\big(\frac{5}{4}T\big)\exp\big(2W_{\frac{t}{2}} - \frac{1}{2}\times 2^2 \times t/2\big)\\ &=\exp\big(2W_{\frac{t}{2}} +\frac{5}{4}T - t\big). \end{align*} We can then also have that \begin{align*} E(X) = \exp\big(\frac{5}{4}T\big). \end{align*}
Additionally, to show that $B_t = \sqrt{2/5}(W_t+W_{t/2})$ is not a Brownian motion, we need only note that \begin{align*} B_t - B_{\frac{t}{2}} &= \sqrt{2/5}\big(W_t - W_{\frac{t}{4}}\big) \\ &=\sqrt{2/5}\big(W_t - W_{\frac{t}{2}} + W_{\frac{t}{2}} - W_{\frac{t}{4}}\big) \end{align*} and \begin{align*} B_{\frac{t}{2}} - B_{\frac{t}{4}} &= \sqrt{2/5}\big(W_{\frac{t}{2}} - W_{\frac{t}{8}}\big) \\ &=\sqrt{2/5}\big(W_{\frac{t}{2}} - W_{\frac{t}{4}} + W_{\frac{t}{4}} - W_{\frac{t}{8}}\big) \end{align*} are not independent. That is, $(B_t)_{t\geq 0}$ does not have independent increments.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.