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I'm currently working about Brownian Bridge and I have to compute the following expectation

\begin{equation} E[W_tW_s\vert W_T]\end{equation}

We consider here that the Brownian bridge is defined as a Brownian motion conditioned to hit 0 at time T, and for $(W_t)_{t≥0}$ a Brownian motion, we give that for $s≤ t ≤ 1$, \begin{equation} E[W_s \vert W_t = x] = \frac{s}{t}x \text{ and } Var[W_s \vert W_t = x] = \displaystyle\frac{s(t-s)}{t}\end{equation}

I found 2 different ways to do it that seems good, but I find a different result. Can somebody explain me why there is this difference, and if there's a mistake where it is ?

Method 1 (this is the result I'm supposed to find)

\begin{equation}E[W_tW_s\vert W_T = 0] = E[E(W_s W_t \vert W_t) \vert W_T] = E[W_t E[W_s \vert W_t] \vert W_T] = E[W_t^2 \frac{s}{t} \vert W_T] = \frac{s}{t} Var(W_t \vert W_T) = \frac{s}{t} t \frac{(T-t)}{T} = \displaystyle\frac{s(T-t)}{T}\end{equation}

Method 2 (not the same result but (I think) mathematically true) \begin{equation}E[W_tW_s\vert W_T = 0] = E[(W_t - W_s)W_s + W_s^2 \vert W_T] = E[W_t - W_s \vert W_T] E[W_s \vert W_T] + E[W_s W_s\vert W_T] = Var(W_s\vert W_T) = \displaystyle\frac{s(T-s)}{T}\end{equation}

Thanking you in advance !

Arthur Boivert
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1 Answers1

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I am not convinced by the use (?) of a tower property in method 1, nor of the independence in method 2. Suppose $s<t<T$. Note $$\begin{bmatrix} W_T\\ W_t\\ W_s \end{bmatrix}\sim \mathcal{N}(0,\Sigma),\,\quad \Sigma=\begin{bmatrix} T&t&s\\ t&t&s\\ s&s&s \end{bmatrix}$$ So we use the conditional density $$(W_t,W_s)|W_T=w\sim \mathcal{N}(\mu(w),\Gamma),\,\quad \mu(w)=\begin{bmatrix} wt/T\\ ws/T\\ \end{bmatrix},\,\quad \Gamma=\begin{bmatrix} t&s\\ s&s \end{bmatrix}-\frac{1}{T}\begin{bmatrix} t^2&ts\\ ts&s^2 \end{bmatrix}$$ Now recall $$\textrm{Cov}[W_t,W_s|W_T]=E[W_tW_s|W_T]-E[W_t|W_T]E[W_s|W_T]$$ So $$E[W_tW_s|W_T=w]=s-\frac{st}{T}+\frac{st w^2}{T^2}$$

Snoop
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