# Prove $X_t = {W_t}^3 - 3\int^{t}_{0} W_s \, ds$ is a martingale

Any help/ guidance in showing that $$X_t$$ above is integrable would be much appreciated! I am able to show, with respect to information set $$\mathcal F_t = \sigma(W_s : s \leq t)$$, $$X_t = E[X_{t'}|\mathcal F_t], \quad t' > t$$

• note: $$W_t$$ is Brownian motion

However, I am stuck in showing $$E[|X_t|] < \infty$$ My approach is: \begin{align} E[|X_t|]& = E\left[|W_t^3 - 3\int_{0}^{t}W_s \,ds|\right] \\& \leq E\left[|W_t^3| + 3|\int_{0}^{t}W_s \,ds|\right] \\&= E[|W_t^3|] + 3E\left[ |\int_{0}^{t}W_s\,ds|\right] \end{align}

I am unsure what next. Would this equality $$E\left[\int_{s}^{t}W_u \, du|\mathcal F_s\right] = (t-s)W_s$$ be helpful?

Thank you for reading.

## 1 Answer

$$E|X_t| \leq E|W_t^{3}| +3\int_0^{t} E|W_s| ds$$. The first term is finite because normal distributions have finite moments of all orders. Also $$Y=\frac 1 {\sqrt s} W_s$$ has standard normal distribution so $$E|W_s|=\sqrt s E|Y|$$. Hence $$\int_0^{t} E|W_s| ds=E|Y| \int_0^{t} \sqrt s ds <\infty$$.

• thank you. I realized showing $X_t$ was easy and should not have posted the question here. However, may I verify showing $E[X_t|\mathcal F_s] = X_s$ for all $s < t$ with you? I start by considering $E[X_t|\mathcal F_s] = E[{W_t}^3 - 3\int_{0}^{t} W_s \, ds|\mathcal F_s] = E[{W_t}^3|\mathcal F_s] - 3E\left[\int_{0}^{t} W_s \, ds| \mathcal F_s\right] = 3(t-s)W_s + {W_s}^3 - 3tW_s = {W_s}^3 - 3sW_s = {W_s}^3 - 3E \left[ \int_{0}^{s} W_u\,du|\mathcal F_s \right] = {W_s}^3 - 3\int_{0}^{s} W_u \, du$ – Galvin Ng Jul 1 at 8:32
• Please note that in the comment above, I used 2 results $$E[{W_t}^3|\mathcal F_s] = 3(t-s)W_s + {W_s}^3$$ and $$E \left[ \int_s^{t} W_u \, du |\mathcal F_s \right] = (t-s)W_s$$ – Galvin Ng Jul 1 at 8:39