Let $(W_{t})_{t\geq 0}$ be a Brownian motion. It is well known how to calculate $\mathbb{E}[W_{t}|W_{s}]$ when $t\geq s\geq 0$.

What about $\mathbb{E}[W_{t}|W_{s}]$, when $t\in [0,s)$?

Inspired by Behrouz Maleki I decided to sum up his solutions:

I Attempt

We know that:

  1. $\mu=\mathbb{E}[W_t]=0$
  2. $\sigma^{2}=Var[W_t]=t$
  3. $W_t\sim\mathcal{N}(0,t)$
  4. $Cov[W_t,W_s]=\min{\{t,s\}}$
  5. $\rho=corr[W_t,W_s]=\sqrt{\frac{\min{\{t,s\}}}{\max{\{t,s\}}}}$

My source: Wikipedia

We also know that, if

$$ \begin{pmatrix} X_1 \\ X_2 \end{pmatrix} \sim \mathcal{N} \left( \begin{pmatrix} \mu_1 \\ \mu_2 \end{pmatrix} , \begin{pmatrix} \sigma^2_1 & \rho \sigma_1 \sigma_2 \\ \rho \sigma_1 \sigma_2 & \sigma^2_2 \end{pmatrix} \right) $$ then

$$\mathbb{E}(X_1 \mid X_2=x_2) = \mu_1 + \rho \frac{\sigma_1}{\sigma_2}(x_2 - \mu_2)$$

My source: Wikipedia

It means that, for $t\in [0,s)$, we have:

  1. $X_1=W_t$ and $X_2=W_s$
  2. $\mu_1=0$ and $\mu_2=0$
  3. $\sigma_1^2=t$ and $\sigma_2^2=s$
  4. $\sigma_1=\sqrt{t}$ and $\sigma_2=\sqrt{s}$
  5. $Cov[W_t,W_s]=t$
  6. $\rho=\sqrt{\frac{t}{s}}$


$$ \begin{pmatrix} W_t \\ W_s \end{pmatrix} \sim \mathcal{N} \left( \begin{pmatrix} 0 \\ 0 \end{pmatrix} , \begin{pmatrix} t & t \\ t & s \end{pmatrix} \right) $$ and

$$\mathbb{E}(W_t \mid W_s=x_2) = \frac{t}{s}x_2$$

so finally

$$\mathbb{E}(W_t \mid W_s) = \frac{t}{s}W_s$$

II attempt:

$$\mathbb{E}[W_t|W_s]=\mathbb{E}[W_t-\frac{t}{s}W_s|W_s]+\frac{t}{s}W_s=$$ $$\mathbb{E}[W_t-\frac{t}{s}W_s]+\frac{t}{s}W_s=\mathbb{E}[W_t]-\frac{t}{s}\mathbb{E}[W_s]+\frac{t}{s}W_s=\frac{t}{s}W_s$$

Here we used the fact that the pair $(W_t-\frac{t}{s}W_s, W_s)$ has multivariate normal distribution and

$$Cov[W_t-\frac{t}{s}W_s, W_s]=Cov[W_t,W_s]-\frac{t}{s}Cov[W_s,W_s]=$$ $$\min{\{t,s\}}-\frac{t}{s}Var[W_s]=t-\frac{t}{s}*s=0$$

It means that $W_{t}-\frac{t}{s}W_s$ and $W_s$ are independent random variables, so $W_{t}-\frac{t}{s}W_s$ is independent of $\sigma(W_s)$ and therefore we can use appropriate property of conditional expectation.

My source: link 1 and link 2

  • 2
    $\begingroup$ Start by looking at he joint distribution of $(W_t,W_s)$ (or $(N_t, N_s)$). $\endgroup$ – John Dawkins Nov 11 '16 at 17:40

Hint $$\mathbb{E}\left[W_t\middle | W_s\right]=\mathbb{E}\left[W_t-\frac{t}{s}W_s\middle |\, W_s\right]+\frac{t}{s}W_s=\frac{t}{s}W_s$$ other approach $$f_{W(s)}(y)=\frac{1}{2\sqrt{\pi\,s}}\text{exp}\left(-\frac{y^2}{2s}\right)$$ and $$f_{W(t),W(s)}(x,y)=\frac{1}{2\pi\sqrt{t(s-t)}}\text{exp}\left(-\frac{x^2}{2t}-\frac{(y-x)^2}{2(s-t)}\right)$$ therefore $$f_{W(t)|W(s)}(y)=\frac{f_{W(t),W(s)}(x,y)}{f_{W(s)}(y)}=\frac{1}{\sqrt{2\pi}}\frac{\sqrt{s}}{\sqrt{t(s-t)}}\text{exp}\left(-\frac{x^2}{2t}-\frac{(y-x)^2}{2(s-t)}+\frac{y^2}{2s}\right)$$ In other words $$f_{W(t)|W(s)}(y)=\frac{1}{\sqrt{2\pi}}\frac{1}{\sqrt{t-\frac{s}{t}}}\text{exp}\left(-\frac{\left(x-\frac{t}{s}y\right)^2}{2\left(t-\frac{t}{s}\right)}\right)$$ thus $$\mathbb{E}[W_t|W_s=y]=\frac{t}{s}y$$

  • $\begingroup$ Could you clarify what happen with the conditional expectation in your first attempt? Why it disappeared? $\endgroup$ – Leon Nov 13 '16 at 20:49
  • $\begingroup$ Set $X(t,s)=W_t-\frac ts W_s$ and $Y(t,s)=\frac ts W_s$. We know $\mathbb{E}[W_s^2]=s$ and $\mathbb{E}[W_s\, W_t]=\min\{t,s\}=t$ thus $X(t,s)$ and $W_s$ are independent. On the other hand $W_t=X(t,s)+Y(t,s)$ . thus $X(t,s)$ and $Y(t,s)$ are independent. $\endgroup$ – Behrouz Maleki Nov 13 '16 at 21:16
  • $\begingroup$ If we want to use property of conditional expectation, then $W_t-\frac{t}{s}W_s$ must be independent with $\sigma(W_s)$. Is it true? I can see that only $W_t-\frac{t}{s}W_s$ and $W_s$ are independent. $\endgroup$ – Leon Nov 13 '16 at 21:31
  • $\begingroup$ This result is true and clear $\endgroup$ – Behrouz Maleki Nov 13 '16 at 21:37
  • $\begingroup$ Thanks! I edited post with my question. Could you look at it and tell me whether everything is correctly? $\endgroup$ – Leon Nov 14 '16 at 17:48

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