Let $W_t$ be a Brownian motion, and let $F_t$ be its filtration then for $t > s$ we are asked to compute


We have $$W_t = W_s + (W_t - W_s)$$


$$W_t^{2} = W_s^{2} + 2W_s(W_t - W_s) + (W_t - W_s)^2$$


$$\mathbb{E}\left[W_t^{2}|F_s\right] = W_s^{2} + t - s$$

I don't see how

$$2\mathbb{E}\left[W_s(W_t - W_s)|F_s\right] = t - s$$

  • $\begingroup$ You could have provided the reference - this is exercise 6.22 from Joshi's book. $\endgroup$ Jan 21, 2018 at 6:25

1 Answer 1


\begin{equation} \mathbb{E} \left[ \left. W_s \left( W_t - W_s \right) \right| \mathfrak{F}_s \right] = W_s \mathbb{E} \left[ W_t - W_s \right] = 0 \end{equation}

The first step uses that $W_s$ is $\mathfrak{F}_s$-measureable and that the increment $W_t - W_s$ is independent of $\mathfrak{F}_s$. Next,

\begin{equation} \mathbb{E} \left[ \left. \left( W_t - W_s \right)^2 \right| \mathfrak{F}_s \right] = \mathbb{E} \left[ \left( W_t - W_s \right)^2 \right] = \mathbb{E} \left[ W_{t - s}^2 \right] = t - s. \end{equation}

Here we used again independence in the first step. In the second one we use that the unconditional distribution of $W_t - W_s$ is the same as that of $W_{t - s}$.

  • $\begingroup$ I see that $$\mathbb{E}\left[(W_t - W_s)^2\right] = \mathbb{E}\left[W_{t-s}^{2}\right]$$ but I do not see how the last part gives us $t-s$ $\endgroup$
    – Wolfy
    Jan 21, 2018 at 21:25
  • $\begingroup$ What is the distribution of $W_{t - s}$? $\endgroup$ Jan 21, 2018 at 21:29
  • $\begingroup$ $W_{t-s}\sim N(0,t-s)$? $\endgroup$
    – Wolfy
    Jan 21, 2018 at 21:35
  • $\begingroup$ Nearly there.. So you know that $\mathbb{E} \left[ W_{t - s} \right] = 0$ and $\text{Var} \left( W_{t - s} \right) = t - s$. Now use that for some random variable $X$, $\text{Var}(X) = \mathbb{E} \left[ X^2 \right] - \left( \mathbb{E}[X] \right)^2$. This is really so fundamental that you should review it if that is sth. you struggle with. $\endgroup$ Jan 21, 2018 at 21:38

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