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!