conditional expected value of a brownian motion

Question

Professor gave us this homework: given $B_t$ a standard brownian motion and $0<s<t$ compute

• $\mathbb{E}[B_t|B_s]$
• $\mathbb{E}[B_s|B_t]$
• $\mathbb{E}[B_{(t+s)/2}|B_s,B_t]$

The first one is easy: $\mathbb{E}[B_t|B_s]=\mathbb{E}[B_t-B_s+B_s|B_s]=B_s$ because of independent increments.

The second one was a little more difficult: I define ${W_t=tB_{1/t}}$ so that $0<\frac 1 t<\frac 1 s$ and then $$\mathbb{E}[B_s|B_t]=s\mathbb{E}[W_{\frac 1 s}|tW_{\frac 1 t}]=s\mathbb{E}[W_{\frac 1 s}|W_{\frac 1 t}]$$ Then using the previous exercise $$s\mathbb{E}[W_{\frac 1 s}|W_{\frac 1 t}]=s\mathbb{E}[W_{\frac 1 s}-W_{\frac 1 t}+W_{\frac 1 t}|W_{\frac 1 t}]=sW_{\frac 1 t}=\frac s t B_t$$ I don't know if I'm right on this one.

For the last one I don't know where to start: have you any suggestions?

Answer 1

Your solution for the second point is OK. Here is another solution.

Let's find a constant $c$ such that $B_s - cB_t$ and $B_t$ are uncorrelated: $$\mathrm{cov}(B_s - cB_t, B_t) = \mathrm{cov}(B_s, B_t) - c \ \mathrm{cov}(B_t, B_t) = s - ct = 0 \Rightarrow c = \frac st.$$

Thus, $B_s - \frac st B_t$ and $B_t$ are uncorrelated and hence independent (because they are Gaussian).

Then subtract and add the term $\frac st B_t$ inside the conditional expectation: $$\mathbb E[B_s \ | \ B_t] = \mathbb E\left[B_s - \frac st B_t + \frac st B_t\ | \ B_t\right] = \mathbb E\left[B_s - \frac st B_t | \ B_t\right] - \mathbb E\left[\frac st B_t\ | \ B_t\right].$$ The first conditional expectaion turns into the usual expectaion due to independency: $$\mathbb E\left[B_s - \frac st B_t | \ B_t\right] = \mathbb E\left[B_s - \frac st B_t\right] = \mathbb E[B_s] - \frac st \mathbb E[B_t] = 0.$$ Since $B_t$ is $\sigma(B_t)$-measurable, $$\mathbb E\left[\frac st B_t\ | \ B_t\right] = \frac st B_t.$$ So finally, $\mathbb E[B_s, B_t] = \frac st B_t$.

Answer 2

Here is an approach to find $\mathbb{E}\left(B_{\frac{s+t}{2}}\big|B_s,B_t\right)$.

The condition density of $B_{\frac{s+t}{2}}|B_s,B_t$ is given by $$\begin{eqnarray*}f_{B_{\frac{s+t}{2}}\big|B_s,B_t}(y|x,z)&=&\frac{f_{B_s,B_{\frac{s+t}{2}},B_t}(x,y,z)}{f_{B_s,B_t}(x,z)} \\ &=& \frac{f_{B_t|B_s,B_{\frac{s+t}{2}}}(z|x,y)f_{B_{\frac{s+t}{2}}|B_{s}}(y|x)f_{B_s}(x)}{f_{B_t|B_s}(z|x)f_{B_s}(x)} \\ &=& \frac{f_{B_t|B_s,B_{\frac{s+t}{2}}}(z|x,y)f_{B_{\frac{s+t}{2}}|B_{s}}(y|x)}{f_{B_t|B_s}(z|x)}\end{eqnarray*}$$ We know that $$B_t|B_s,B_{\frac{s+t}{2}}\sim \mathcal{N}\left(B_{\frac{s+t}{2}},\frac{t-s}{2}\right)$$ $$B_{\frac{t+s}{2}}|B_s\sim \mathcal{N}\left(B_s,\frac{t-s}{2}\right)$$ $$B_t|B_s\sim \mathcal{N}\left(B_s,t-s\right)$$ Plugging in these densities and simplifying yields $$f_{B_{\frac{s+t}{2}}\big|B_s,B_t}(y|x,z)=\frac{1}{\sqrt{2\pi\left(\frac{t-s}{4}\right)}}\exp\left(-\frac{1}{2}\cdot\frac{\left(y-\frac{x+z}{2}\right)^{2}}{(t-s)/4}\right)$$ This suggests $B_{\frac{s+t}{2}}|B_s,B_t\sim \mathcal{N}\left(\frac{B_s+B_t}{2},\frac{t-s}{4}\right)$ and so $$\mathbb{E}\left(B_{\frac{s+t}{2}}\Big|B_s,B_t\right)=\frac{B_s+B_t}{2}$$