In an example I was working through it was shown that $W_{t}^{2} - t$ was a martingale with respect to the Brownian motion filtration $\mathcal{F}_{s}^{W}$ with $t>s$. Everything was fine except a part in the proof where the author used the fact \begin{equation} E(t|\mathcal{F}_{s}^{W}) = s \end{equation}

I can't quite see the rationale for this. For example if we take a process $X(t,\omega) = t$, then it seems that $X$ is not stochastic, and in fact is independent of $\omega$ for all $\omega$ in the sample space -- so why does the conditional expectation in the equation above make sense?


2 Answers 2


The above question was a typo due to the author -- the expression should be evaluated as \begin{equation} E(t|\mathcal{F}_{s}^{W}) = t \end{equation}

due to the reasoning in the question. Sorry for the noise.


You might want to give us the exact statement of the author.

Let the Wiener process $W_{s}$ be a r.v. from $\left(\mathcal{F}_{s},\Omega\right)\to\left(\mathcal{B}\left(\mathbb{R}\right),\mathbb{R}\right)$. The Borel-$\sigma$-algebra $\mathcal{B}\left(\mathbb{R}\right)$ contains all intervals of the form $\left[x,y\right]$ for $x\neq y\in\mathbb{R}$, because you have to be able to tell at time $s\geq 0$ if the Wiener process $W_{s}$ has its value in this interval or not. In order for $W_{s}$ to be measurable all the pre-images of this intervals have to be in the $\sigma$-algebra $\mathcal{F}_{s}^{W}$. So the (deterministic) random variable $X\left(t,\omega\right)=t$ is also measurable at time $s\geq 0$ because we can say in which interval its value is. But the deterministic r.v. $X\left(t,\omega\right)=t$ does not depend on $\omega$, so the pre-image of every obtainable resp. not obtainable interval is $\Omega$ resp. $\emptyset$.

Every deterministic r.v. is measurable to the trivial $\sigma$-algebra $\mathcal{F}_{0}:=\left\{\emptyset,\Omega\right\}$, which is contained in every other $\sigma$-algebra $\mathcal{F}_{s}^{W}$. So even if we condition on the coarser (smaller) $\sigma$-algebra $\mathcal{F}_{s}^{W}$ a deterministic r.v. is measurable and we only need the trivial $\sigma$-algebra $\mathcal{F}_{0}$. But that is \begin{equation} \mathbb{E}\left[t\mid\mathcal{F}_{0}\right] = \mathbb{E}\left[t\right] = t \mathrm{.} \end{equation}


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