What is $E(W_s|W_t)$ , $0<s<t$

Question

Any thoughts on what is $ X_{s,t} = E(Ws|Wt)$ when $0<s<t $.

i.e conditional expectation of a Brownian motion $W_s$ given its future state $W_t$.

E.g.

Expectation ( is =2 zero ? )

Variance ( for which $s$ does it reach its max in $[0,t]$)

Analytical formula ( $X_{s,t} = a W_s + b W_t$ ? Brownian Bridge ?)

Answer 1

Edit: About $W_s|W_t$ and $\mathbb{E}\left(W_s|W_t\right)$

Thanks to OP's clarification, the question is concerning about $W_s|W_t$, a Brownian bridge. Here are the respective updates.

Still make use of these two facts:

• $W_s-\left(s/t\right)W_t$ is a sum of two independent normal random variables with zero means, i.e., $$ W_s-\frac{s}{t}W_t=\left(1-\frac{s}{t}\right)W_s-\frac{s}{t}\left(W_t-W_s\right), $$ for which it is also a normal random variable with a zero mean, and
• $W_s-\left(s/t\right)W_t$ and $W_t$ are independent normal random variables with zero means, with their independence able to be checked via their zero correlation coefficient.

Thus we may write in the form of $$ W_s|W_t=\left(W_s-\frac{s}{t}W_t\right)\bigg|W_t+\frac{s}{t}W_t|W_t=\left(W_s-\frac{s}{t}W_t\right)+\frac{s}{t}W_t|W_t, $$ where $W_s-\left(s/t\right)W_t$ in the last equation shall be understood as a random variable independent from $W_t$, even if $W_t$ appears as part of its component.

With this in mind, we have $$ \mathbb{E}\left(W_s|W_t\right)=\mathbb{E}\left(W_s-\frac{s}{t}W_t\right)+\mathbb{E}\left(\frac{s}{t}W_t|W_t\right)=\frac{s}{t}W_t. $$

Likewise, \begin{align} \text{Var}\left(W_s|W_t\right)&=\mathbb{E}\left[\left(W_s-\mathbb{E}\left(W_s|W_t\right)\right)^2|W_t\right]\\ &=\mathbb{E}\left[\left(W_s-\frac{s}{t}W_t\right)^2\bigg|W_t\right]\\ &=\mathbb{E}\left(W_s-\frac{s}{t}W_t\right)^2. \end{align} Recall the first fact in above that $$ W_s-\frac{s}{t}W_t=\left(1-\frac{s}{t}\right)W_s-\frac{s}{t}\left(W_t-W_s\right) $$ is a sum of two independent normal random variables. Hence $$ \mathbb{E}\left(W_s-\frac{s}{t}W_t\right)^2=\left(1-\frac{s}{t}\right)^2s+\left(\frac{s}{t}\right)^2\left(t-s\right)=\frac{s}{t}\left(t-s\right). $$

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Original Post: Mistaken $W_s|W_t$ (a conditioned process) as $W_s/W_t$ (a ratio)

It could be hopeless to evaluate any moment of $W_s/W_t$, because it is not defined for Cauchy distributions.

In fact, if $X$ and $Y$ are independent standard normal random variables, then $X/Y$ follows a standard Cauchy distribution. For the non-standard-normal cases, you may refer to here for more details.

Now get back to the Brownian motion case. As per $0<s<t$, we have

• $W_s-\left(s/t\right)W_t$ is a sum of two independent normal random variables with zero means, i.e., $$ W_s-\frac{s}{t}W_t=\left(1-\frac{s}{t}\right)W_s-\frac{s}{t}\left(W_t-W_s\right), $$ for which it is also a normal random variable with a zero mean, and
• $W_s-\left(s/t\right)W_t$ and $W_t$ are independent normal random variables with zero means, with their independence able to be checked via their zero correlation coefficient.

Therefore, $$ \frac{W_s}{W_t}=\frac{W_s-\left(s/t\right)W_t}{W_t}+\frac{s}{t}, $$ as a shift of the ratio of two independent normal random variables with zero means, follows a Cauchy distribution.