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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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    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 !