I want to calculate $E(W_t | W_1)$, $E(W_t^2 | W_1)$ and $E(W_t^2 | W_1, W_2)$, where $(W_t)_{t\geq0}$ is a Wiener process.

For the first one I used the conditional distribution formula for the multivariate normal distribution and I got:

$E(W_t | W_1)=EW_t+\sum_{12}\sum_{22}^{-1}(W_1-EW_1)=0+ (1\wedge t)\cdot 1\cdot(W_1-0)=(1 \wedge t) W_1$.

For the second I did:

$E(W_t^2 | W_1)=VAR(W_t |W_1)+E^2(W_t|W_1)=t-(t\wedge 1)^2+ ((1\wedge t )W_1)^2$

How can I do the third one?

  • $\begingroup$ What about $t$...? $t \geq 1$ or $t \leq 1$ or...? $\endgroup$ – saz Mar 13 '15 at 10:46
  • $\begingroup$ @saz $t$ can be either greater or less than $1$ $\endgroup$ – klmgg1 Mar 13 '15 at 11:16

You can extend your first two answers readily to the third. Consider that for $t\in[0,1]$ we know that $\mathbb{E}(W_t|W_1) = t\, W_1$ but this is really $\mathbb{E}(W_t|W_1,W_0) = t\,W_1 + (1-t)W_0$ just that $W_0=0$. In general we have

$$\mathbb{E}(W_t|W_r,W_s) = \frac{t-r}{s-r} W_s + \frac{s-t}{s-r} W_r \qquad\text{for }\,\, r\leq t\leq s$$

This is the general result for a Brownian bridge. We can combine the $\min$ and $\max$ functions ($\wedge$ and $\vee$ respectively) to produce an arbitrary continuous piecewise linear function. Specifically in our case we get

$$\mathbb{E}(W_t|W_2,W_1,W_0=0) = (t\wedge1)W_1 + (0\vee(1\wedge t))(W_2-W_1)$$

For the Brownian bridge we also have that

$$\mathrm{Var}(W_t|W_r,W_s) = \frac{(s-t)(t-r)}{s-r} \qquad\text{for }\,\, r\leq t\leq s$$

Thus we have

$$\begin{align}\mathrm{Var}(W_t|W_1,W_0) &= t(1-t) \qquad&\text{for }\,\, 0 \leq t < 1\\ \mathrm{Var}(W_t|W_2,W_1)&= (2-t)(t-1) \qquad&\text{for }\,\, 1 \leq t < 2 \end{align}$$

These are parabolas, which are negative outside the intervals $[0,1]$ and $[1,2]$ respectively thus we can combine them easily with the $\max$ function as well as accounting for the linearly increasing variance for $t>2$ giving

$$\mathrm{Var}(W_t|W_2,W_1,W_0=0) = \big(t(1-t)\big)\vee\big((2-t)(t-1)\big)\vee(t-2)$$

Now, exactly as you did for your second answer, we combine the variance and square of the expectation through $$\mathbb{E}(W_t^2|W_2,W_1,W_0=0) = \mathrm{Var}(W_t|W_2,W_1,W_0=0) + \mathbb{E}(W_t|W_2,W_1,W_0=0)^2,$$ to get the final answer. (I haven't actually written out in full but you should be able to see what it looks like and if it simplifies.)


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