# Show that $M_t = W^3_{t} - 3 \int_0^{t}W_sds$ is a martingale

G'day, I am able to derive the answer, but unsure about the given argumentation of the prof.

Problem: Show that $$M_t = W^3_{t} - 3 \int_0^{t}W_sds$$ is a martingale. Let $$W_t$$ be Brownian motion.

$$W^3_{t} = 3 \int_0^{t}W^2_{s}dW_s + 3 \int_0^{t}W_sds$$

1) Substituting $$W_t^{3}$$ into the equation $$M_t$$

$$M_t = 3 \int_0^{t}W^2_{s}dW_s + 3 \int_0^{t}W_sds - 3 \int_0^{t}W_sds$$

2) Terms of $$+ 3 \int_0^{t}W_sds - 3 \int_0^{t}W_sds$$ cancels out leaving us with the answer

$$M_t = 3 \int_0^{t}W^2_{s}dW_s$$ and hence a martingale. However am I allowed to do step 2) just like that?

The prof gives the following reasoning which I don't understand: $$E(\int_0^{t}(W_s^{2})^2ds) = E(\int_0^{t}W_s^{4}ds) = \int_0^{t}3s^{2}ds < ∞$$ for all t>= 0.

Can someone explain the reasoning?

• Which book are you reading?
– user9464
Nov 24, 2019 at 14:45
• Stochastic Calculus for Finance and Arbitrage in Continuous Time Nov 24, 2019 at 14:46
• Can't find this book. Who's the author of the book?
– user9464
Nov 24, 2019 at 14:48
• First one is Shreve, the second one Thomas Bjork.. Nov 24, 2019 at 14:49
• In the step 1), we used the Ito formula for the function $f(x) = x^3$ with the process $W$. Then in the step 2), we showed that the process $M$ is actually a local martingale. In fact, we can even show that is a real (square) martingale by showing that the $L^2$-norm is finite. This is done by applying the Ito isometry. Nov 25, 2019 at 10:43

In order to make sense of a stochastic integral of the form $$\int_0^t f(B_s) \, dB_s$$ you need to verify two properties:
• $$f$$ is suitably measurable, e.g. progressively measurable
• $$f$$ is suitably integrable, e.g. $$\mathbb{E}(\int_0^t f(s)^2 \, ds)<\infty$$ or $$\mathbb{E}(\int_0^{t \wedge \tau_n} f(s)^2 \, ds)<\infty$$ for a sequence of stopping times $$\tau_n \uparrow \infty$$
Under the above conditions, the stochastic integral $$M_t := \int_0^t f(B_s) \, dB_s$$ is a local martingale but, in general, it might fail to be a martingale. In order to ensure that $$(M_t)_{t \geq 0}$$ is a martingale (not only a local one), $$f$$ has to satisfy the stronger integrability condition of the above-mentioned conditions, that is, $$\mathbb{E}(\int_0^t f(s)^2 \, ds)<\infty$$ for all $$t\geq 0$$.
This is exactly the condition which the authors of the solution verified for the particular case $$f(s):=B_s^2$$.