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 ?
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 ?
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 !