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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!

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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.

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    $\begingroup$ Note that, for $\sigma>0$, $\{\exp(-\frac{1}{2}\sigma^2 t + \sigma W_t)\}_{t \geq 0}$ is a martingale. $\endgroup$
    – Gordon
    May 25, 2015 at 23:55
  • $\begingroup$ I added the proof to show that $B_t = \sqrt{2/5}(W_t+W_{t/2})$ is not a Brownian motion. $\endgroup$
    – Gordon
    May 26, 2015 at 13:19
  • $\begingroup$ Thank you for your careful solution. I encounter the difficulty when I obtain the $X_{t}$ by discussing the two situations separately when I try to find the martingale representation of X, that is, find an adapted process g such that $E[\int^{T}_{0}g^{2}_{s}d_{s}]<\infty$ and $X=E[X]+\int^{T}_{0}g_{s}dW_{s}$. $\endgroup$
    – oamc
    May 27, 2015 at 23:31

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