conditional distribution of $W(t/2)$ given $W(t)=x$ How to get the conditional distribution of $W(t/2)$ given $W(t)=x$ where $W(t)$ represents wiener process.
This was a problem in my exam and i couldn't think how to start :(
Any help!!
Thanks in advance. 
 A: Joint distribution of $(W(t/2), W(t))$ is a binormal distribution with zero means, and covariance:
$$
    \mathbb{Var}(W(t/2)) = \frac{t}{2}, \quad \mathbb{Var}(W(t)) =t, \quad \mathbb{Cov}(W(t/2),W(t)) = t/2  
$$
translating into $\rho = \frac{\sqrt{2}}{2}$, $\sigma_1^2 = \frac{t}{2}$, $\sigma_2^2 = t$. Now, conditional distribution of binormal  is well known:
$$
    W(t/2)|W(t)=x \sim \mathcal{N}\left(\frac{x}{2}, \frac{\sqrt{t}}{2} \right)
$$
Alternatively, you could have used that the Wiener process conditioned upon $W(t)=x$ gives the Brownian bridge process.
A: Here is the trick. Find $\alpha$ such that $W_{t/2}+\alpha W_t$ and $W_t$ are independent i.e. -since $W$ is a Gaussian process with zero mean- such that
$$ \mathbb{E} \left[( W_{t/2}+ \alpha W_t) W_t\right] = 0 $$
You find $\alpha = -1/2$ using $\mathbb{E}\left[ W_a W_b \right] = a \wedge b$. Then
$$ \mathbb{E} \left[ W_{t/2} \,\big|\, W_t = x \right] = \mathbb{E} \left[ W_{t/2} - \frac{1}{2} W_t + \frac{1}{2} W_t \,\big|\, W_t = x \right] = \underbrace{\mathbb{E} \left[ W_{t/2} - \frac{1}{2} W_t \right] }_{=0}+ \frac{1}{2} x = \frac{1}{2} x$$
More generally we have (using the same trick) the famous interview question
$$ t \le s, \quad \mathbb{E} \left[ W_t \,\big|\, W_s \right] = \frac{t}{s} W_s $$
