# Add a compensator to submartingale to create a martingale.

Let $W_t$ be a Brownian motion. By Jensen's inequality, $W_t^2$ is a submartingale. I was wondering if it were possible to add another process to $W_t^2$, say $X_t$, such that $W_t^2 + X_t$ is a martingale. How would I go about doing this?

I know I need to solve for $X_t$ which satisfies the equation: $$E[W_t^2+X_t|\mathcal{F_s}] = W_s^2+ X_s$$ $\forall s<t$. How do I go about doing this?

• You could have been more precise@kerry Jan 9, 2018 at 14:44

For many stochastic processes (but certainly not all) you can find such a function $X_t$. It turns out that it will be a finite variation process, called the bracket process, or quadratic variation process.

In this particular case, one can use the markov property in place of Ito's formula:

\begin{align} \Bbb E[ W_t^2 | \mathcal{F}_s] &= \Bbb E\left[ (W_t-W_s)^2 + \left. 2W_sW_t - W_s^2 \ \right|\ \mathcal{F}_s\right] \\ &= \Bbb E[(W_t-W_s)^2 | \mathcal{F}_s] + 2W_s\Bbb E[W_t|\mathcal{F}_s] - W^2_s && \text{"taking out what is known"}\\ &= \Bbb E[(W_t-W_s)^2 | \mathcal{F}_s] + 2W^2_s - W_s^2 && W\text{ is a martingale}\\ &= \Bbb E[(W_t-W_s)^2] + W_s^2 &&\text{by independence from the past} \\ &= \Bbb EX_{t-s}^2 + W_s^2 && (\star)\\ &= (t-s) + W_s^2 &&(t>s)\end{align} where in the line marked $(\star)$, $X_{t-s}$ is a Brownian Motion restarted at time $s$. This tells us that $W_t^2 - t$ is a martingale.

• In the second line, why does $E[-W_s^2]$ become $W_s^2$? Sep 27, 2015 at 23:49
• @Ryan You need the same term $X_t$ that appears in $Y_t:=W_t^2 + X_t$ to appear in both $Y_t$ and $Y_s$, so you only move the $t$ across! Sep 28, 2015 at 0:12
• @Ryan in full, \begin{align} \Bbb E [W_t^2 | F_s] &= W_s^2 - s + t \\ \Bbb E [W_t^2 | F_s] - t&= W_s^2 - s \\ \Bbb E [W_t^2 - t | F_s] &= W_s^2 - s \end{align} Sep 28, 2015 at 0:14
• Honestly, it is a learned trick to me. But I imagine that my lecturer would have said, 'well, the only terms we like to see are those that are $\mathcal{F}_s$ measurable, and differences $W_{t_1} - W_{t_0}$ because [in my definition], thats all we assume we know about brownian motion!' If I recall correctly, a similar identity can work for $W_t^3$ but I couldn't figure that one out myself :P Sep 28, 2015 at 0:18
• @Ryan To me the point is, how can I write $W_t^2$ as a combination of things which are $W_s$-measurable and things which are independent of $W_s$? Intuitively we want to work with $W_s$ and $W_t-W_s$. Once you know you want $W_t-W_s$, you square it to get a $W_t^2$ term; then you add in whatever you are now missing.
– Ian
Sep 28, 2015 at 0:25

By Ito's formula:

$$W_t^2=2 \int_0^t W_s dW_s + \int_0^t ds = 2 \int_0^t W_s dW_s + t.$$

Now the Ito integral (of a square integrable, independent-of-future function) is a martingale, so $W_t^2-t$ is a martingale.

This procedure works for a wide class of diffusion processes.

• This looks very nice. I have not been taught Ito Calculus yet. So I don't understand the theory or the formula:( Sep 27, 2015 at 23:52
• Could you explain in a better way the last step? I understand that we have $W_t^2=W_t^2 + t$ from the Ito formula, but how you can say that $W^2_t-t$ is a martingale? May 3, 2018 at 13:00
• @GiacomoTabarelli It is a rather technical theorem that $\int_0^t f(s,\omega) dW_s$ is a martingale under suitable assumptions on the function $f$. You should be able to find this in most books on stochastic processes. If you know that, the last step is just algebra.
– Ian
May 3, 2018 at 13:13
• Ok thanks. But can you please compute $2\int_0^t W_s dW_s$? I can't understand why its mean is always zero. May 3, 2018 at 13:23
• It is $W_t^2-t$ but that does not tell you it has mean zero. The proof of the result you want requires inspection of the definition of integration dW and it relies on the Ito convention specifically.
– Ian
May 3, 2018 at 13:31