Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 ?

share|cite|improve this question
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 !

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.