# Expection of Brownian Squared conditional on the end of the path

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? – Did Nov 6 '12 at 16:42
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. – BlueTrin Nov 6 '12 at 16:45
The identity you cite is true but not much related to your problem. Do you know the variance-covariance matrix of the couple $(W_t,W_T)$? – Did Nov 6 '12 at 16:48
I do not know this at all but I will have a look at it. Thank you Mr. Did – BlueTrin Nov 6 '12 at 16:51
@did: will it be $\bigl(\begin{smallmatrix} t&t\\ t&T \end{smallmatrix} \bigr)$ ? I am not too sure how it will help me ? – BlueTrin Nov 6 '12 at 17:19

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 !

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