# Why is $\int_0^t t \, dW_s$ not a martingale?

Assuming $W_t$ is the standard Brownian motion, then $\int_0^t t \, dW_s=tW_t$ is not a martingale since $\mathbb{E}(tW_t \mid \mathcal{F}_s)=tW_s$. However, by martingale property of Ito's integral it seems that $\int_0^t t \, dW_s$ should be a martingale...

The only explanation I can think of is that the "stochastic process" $t$ looks like not adapted to $\mathcal{F}_s$. But this sounds very strange because $t$ is actually a constant in that integral. And if that is the case, the answer here that uses Ito's isometry to compute a variance is also problematic because there $(t-s)$ is also not adapted to $\mathcal{F}_s$ but Ito's isometry requires such adaptation.

## 2 Answers

The process $(t W_t)_{t \geq 0}$ is indeed not a martingale. The reason is simply that you cannot use the martingale property of stochastic integrals to deduce the martingale property of the process. Correctly stated the martingale property of stochastic integrals driven by Brownian motion reads as follows:

Theorem Let $f:\Omega \times [0,\infty) \to \mathbb{R}$ be a progressively measurable function such that $\mathbb{E}\left( \int_0^t f(s)^2 \,ds \right)<\infty$. Then the process $$M_t := \int_0^t f(s) \, dW_s$$ is a martingale.

Note that $f$ is not allowed to depend on $t$, i.e. the statement does not claim that

$$N_t :=\int_0^t f(s,t) \, dW_s$$

is a martingale for any progressively measurable and square integrable function $f$.

If we choose $f(s) := s$, then the above statement yields that

$$M_t = \int_0^t s \, dW_s$$

is a martingale - which is true.

We can also choose $f(s)=t_0$ for some fixed $t_0$; then we find that

$$M_t = t_0 W_t$$

is a martingale - which is also true (because $t_0$ is a fixed constant).

I am not sure whether you are referring to $\int_0^t s d B_s$ or to $(t B_t)$. The beauty is that I don't actually have to know because I can discuss the martingale properties of the two processes simultaneously.

Note that

\begin{align} \int\limits_0^t s d B_s =t B_t - \int\limits_0^t B_s ds, \end{align}

as you can see easily with Ito's formula. According to the general theory cited by saz, we know that $\int_0^t s d B_s$ must be a martingale. While saz's remarks were focused on why the general theorems do not apply, this gives you a nice argument why $(t M_t)$ is not a martingale - because $\int_0^t B_s d s$ is not a martingale:

\begin{align*} \mathbb{E}\left[ \int\limits_0^t W_r dr \middle| \left(\int\limits_0^z W_r dr\right)_{z\in [0,s]} \right] &= \int\limits_0^s W_r dr + \underbrace{\mathbb{E}\left[ \int\limits_s^t W_r dr \right]}_{=0 \text{ by Fubini}} +(t-s)\underbrace{\mathbb{E}\left[ W_s \middle| \left(\int\limits_0^z W_r dr\right)_{z\in [0,s]} \right]}_{\neq 0 \text{ by continuity of } W_r}. \end{align*}

• To see that $(tB_t)$ is not a martingale you don't need any computations. It holds for any non-trivial martingale $(M_t)_t$ that $(t M_t)_t$ fails to be a martingale. – saz Jan 21 '18 at 9:25
• I never realized! Like this? Suppose $(M_t)_t$ is a non-trivial martingale such that $(tM_t)_t$ is a martingale, too. Wlog $M_0=0$. Because it is non-trivial, there exists a $t_0$ s.t. $\mathbb{P}[M_{t_0}>0]>0$. Wlog, $t_0=1$. Then for all $t>t_0$, $\mathbb{E}[tM_t|t_0M_{t_0}]=tM_0=M_0$ almost surely which contradicts $\mathbb{P}[M_{t_0}>0]>0$. – Ecthelion Jan 21 '18 at 10:09
• There are some typos in your calculation, and I'm not sure whether you were thinking in the right direction. Since $(M_t)_t$ is a martingale we have $$\mathbb{E}(tM_t \mid \mathcal{F}_s)= t M_s$$ for any $s < t$; now if $(t M_t)_t$ was also a martingale, then $$\mathbb{E}(t M_t \mid \mathcal{F}_s) = s M_s,$$ and so $$s M_s =t M_s \quad \text{almost surely}$$ which is only possible if $M_s=0$ almost surely; as $0 \leq s < t < \infty$ was arbitrary, we get $M=0$. – saz Jan 21 '18 at 10:35
• Thank you! In fact, the same proof works to show the following statement: Fix any $f:\mathbb{R}_+ \rightarrow \mathbb{R}$. Suppose $(M_t)_t$ is a martingale (and something like for all $t$, $M_t | M_{t-\varepsilon}$ does not follow a Dirac distribution for all $\varepsilon>0$). Then $(f(t)M_t)_{t}$ is a martingale if and only if $f\equiv c$ for some constant $c>0$. – Ecthelion Jan 21 '18 at 18:17