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I have been asked as a brainteaser to compute the value of:

$\mathbb{E}[W_t^2|W_T]$ with $t < T$ ?

Does anyone know how to proceed ?

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    Yes. What are your thoughts? Which similar problems can you solve? Which related results do you know?2012-11-06
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    I do not know much about processes. I thought I could say that $\mathbb{E}[W_t^2] = \mathbb{V}ar[W_t] + \mathbb{E}[W_t]^2 = t$ but it seems that I was wrong.2012-11-06
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    The identity you cite is true but not much related to your problem. Do you know the variance-covariance matrix of the couple $(W_t,W_T)$?2012-11-06
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    I do not know this at all but I will have a look at it. Thank you Mr. Did2012-11-06
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    @did: will it be $\bigl(\begin{smallmatrix} t&t\\ t&T \end{smallmatrix} \bigr)$ ? I am not too sure how it will help me ?2012-11-06
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    Here is how: based on this matrix, try to write $W_t=aW_T+bZ$ for some scalar $a$ and $b$ and some $Z$ standard gaussian independent of $W_T$. Then come back to your question...2012-11-06
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    $W_T - W_t$ is independent of $W_t$ but not from $W_T$2012-11-06
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    Sure. This does not prevent $(a,b,Z)$ to exist as in my last comment...2012-11-06
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    $a = \sqrt{-\frac{b}{T}}$2012-11-06
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    No. Hint: compute $E(W_tW_T)$.2012-11-06
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    $Cov(W_{T}W_{t})=E(W_{T}W_{t}) = E[(W_{T}-W_{t})W_{t}]+E[W_t^2]=t$2012-11-06
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    Indeed. Now, what must be a and b if one wants this covariance?2012-11-06
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    I am slow: I do not see the link between the expressions of covariance and the equation with a and b2012-11-06
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    Come! Surely you can compute $(E(W_tW_T)$ and $E(W_t^2)$ if $W_t=aW_T+bZ$.2012-11-07
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    let us [continue this discussion in chat](http://chat.stackexchange.com/rooms/6361/discussion-between-bluetrin-and-did)2012-11-07

1 Answers 1

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First let's assume that we can find:

$$X = W_t + aW_T$$

, such as $\mathbb{E}\left[X\cdot W_T\right] = 0$

This can be rewritten as:

$$ \mathbb{E}\left[X \cdot W_T\right] = \mathbb{E}\left[\left(W_t + a \cdot W_T\right) \cdot W_T\right] $$

$$ \mathrm{E}\left[X \cdot W_T\right] = t + a \cdot T $$

Therefore we want $a = - \frac{t}{T}$.

The mean of a Brownian bridge is the interpolated value between the two extremities and we know that $W_0 = 0$: $$ \mathbb{E}\left[W_t|W_T\right]=\frac{t}{T}W_T$$

The variance of a Brownian bridge is: $$ \mathbb{Var}\left[W_t|W_T\right]=\frac{(T-t)t}{T} $$

We can compute the quantity we are interested in: $$ \mathbb{E}\left[W_t^2|W_T\right]=\mathbb{Var}\left[W_t^2|W_T\right] - \mathbb{E}\left[W_t|W_T\right]^2 $$

$$ \mathbb{E}\left[W_t^2|W_T\right]=\frac{(T-t)t}{T} - \left(\frac{t}{T} \cdot W_T \right)^2 $$

Thank you Did for being so patient !