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How do I compute the following 6 expectations given the various conditions:

(i) $E\left[W_s \Big| \mathcal{F}_t \right]$, where t > s

(ii) $E\left[W_s \Big| \mathcal{F}_t \right]$, where t < s

(iii) $E\left[W_s \Big| W_t \right]$, where t > s

(iv) $E\left[W_s \Big| W_t \right]$, where t < s

(v) $E\left[\int_{0}^{s} W_udu\Big|W_t\right]$, where t > s

(vi) $E\left[\int_{0}^{s} W_udu\Big|W_t\right]$, where t < s

Given that $W_t$ is a standard Brownian motion and $\{\mathcal{F}_t, t\ge 0\}$ denotes its standard filtration.


EDIT:

From @SpettroDiA's hints:

(i) $=W_s$

(ii) $=W_t$

(iii) See @SpettroDiA answer

(iv) $=E\left[W_s-W_t \Big| \mathcal{F}_{W_t}\right]+ E\left[W_t \Big| \mathcal{F}_{W_t}\right] = 0 + W_t = W_t$

(v)

(vi)

Any hints on how to approach (v) and (vi)?

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  • $\begingroup$ Two remarks. First, the (valid) hints in the accepted answer will not allow you to treat (iii). Second, the conditions "where t > s" and "where t < s" in your items (v) and (vi) are absurd since the conditional expectations in these items involve no s. $\endgroup$ – Did Mar 21 '17 at 7:54
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I'll give you some hints:

(i) Remember that if $X$ is $\Sigma$-measurable, then $E(X | \Sigma)=X$.

(ii) The Wiener process is a martingale with respect to its standard filtration.

(iii) / (iv) Recall the definition of conditional expectation with respect to a random variable, $E(X|Y)= E(X| \sigma(Y))$, hence $E(W_t | W_s)= E(W_t | \mathcal{F}_{W_s})$ for every $t,s$. Furthermore, if $X$ is independent of $\Sigma$ then $E(X| \Sigma)=E(X)$. With this in mind, exploit the role of the increments (on which we have a lot of information): for instance, if $t < s$

$$ E(W_s | W_t )=E(W_s | \mathcal{F}_{W_t}) = E(W_s-W_t | \mathcal{F}_{W_t})+ E(W_t | \mathcal{F}_{W_t}) = 0 + W_t=W_t $$

(v)/(vi) On which set are these integrals calculated? Maybe you want to consider the conditional expectation of the process $Y_t= \int_{0}^{t} W_r \, dr $.

Hope this helps, let me know if you need a deeper discussion.

As pointed out in a comment, (iii) requires a different approach. Let $\phi(x)=E(W_t | W_s = x) $, we want an explicit form for $\phi$ that will be eventually evaluated at $W_s$. $$ \phi(x)=\int y \mathbb{P}(W_t \in dy | W_s=x ) $$

Now recall that, by definition of Brownian motion:

$$ \mathbb{P}(W_s \in I, W_t \in J )=\int_{I} \int_{J} g_t(x) g_{s-t}(y-x) \, dx dy $$

where $I,J$ are any Borel set of $\mathbb{R}$ and $g_t(x)=(2 \pi t)^{-1/2} \exp\left( {\frac{x^2}{2t}} \right)$ is the classic Gaussian density. Furthermore, the definition of conditional probability leads (with some boring but easy algebraic passages) to the following density equation: \begin{align*} \mathbb{P}(W_t \in dy | W_s=x) & =\frac{\mathbb{P}(W_t \in dy, W_s \in dx)}{\mathbb{P}(W_s \in dx)}=\frac{g_t(y)g_{s-t}(x-y)dx dy}{g_s(x) dx} \\ & = g_{\frac{t}{s}(s-t)}\left( y-\frac{tx}{s} \right) dy \end{align*} Hence we recognize that $X_x= (W_t | W_s=x) \sim \mathcal{N}\left(\frac{tx}{s},\frac{t}{s}(s-t) \right)$. In conclusion:

$$ E(W_t | W_s )=\phi(x) \bigg\rvert_{x=W_s} = \int y g_{\frac{t}{s}(s-t)}\left(y-\frac{tW_s}{s}\right) \, dy = \frac{t}{s} W_s $$

Interesting remark

If we set $s=1 $ and $x=0$ we obtain:

$$ X_0=(W_t | W_1=0) \sim \mathcal{N}(0,t(1-t)) \ \ \text{with} \ \ t \in [0,1] $$

which is a Brownian motion starting at zero and conditioned to have $W_1=0$, also known as the famous "Brownian Bridge". See https://en.wikipedia.org/wiki/Brownian_bridge for more details and more connections with part (iii) of this exercise.

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  • $\begingroup$ thanks for your reply, will give it a go myself :) $\endgroup$ – chengcj Mar 16 '17 at 15:22
  • $\begingroup$ Hi @SpettroDia, thanks for helping. But I think I still need a bit more hints to continue. Do you know what keywords should I search in google to find out how to solve these problems? Thanks. $\endgroup$ – chengcj Mar 21 '17 at 7:52
  • $\begingroup$ No problem, editing the answer as soon as I have time. Well, first of all I recommend you to carefully study basic definition and properties of conditional expectation, classic Brownian motion, martingales and integration of Brownian paths. It is not so easy to explicitly find papers about "how to solve these problems": these questions are often part of stochastic processes books (usually with no solution) or proposed and maybe solved in a stochastic processes class. Once you have good knowledge of the above concepts, it shouldn't be hard to solve them alone. $\endgroup$ – GaC Mar 21 '17 at 9:38
  • $\begingroup$ A note on your edited question: both summands in (iv) can be expanded a bit more, for instance, what can you say about the measurability of $W_t$ wrt to $\mathcal{F}_{W_t}$? $\endgroup$ – GaC Mar 21 '17 at 9:53
  • $\begingroup$ Thanks again for editting your answer :) I'm currently self-studying stochastic calculus. Could you recommend some books / website which I can read and learn how to integrate Brownian paths please. $\endgroup$ – chengcj Mar 28 '17 at 16:36

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