# Expectation of stochastic integral, martingale

I have to show that $$\mathbb{E}\left[ \int_0^t a B_s^{a-1} \ dB_s \right] = 0$$ I started with the fact that, if I show that the inner integral $$\int_0^t a B_s^{a-1} \ dB_s$$ is a martingale I get the wanted result.

One way to show that the stochastic integral is a martingale is to show that: $$\mathbb{E}\left[ \int_0^t (a B_s^{a-1})^2 \ ds \right] < \infty$$ which is basically saying $$aB_s^{a-1} \in \Lambda^2(t)$$ (to have this we also need the process to be progressive, but it follows by BM properties I guess).

Using Fubini I can interchange expectation and integral, bring outside constants and write: $$\mathbb{E}\left[ \int_0^t (a B_s^{a-1})^2 \ ds \right] = \int_0^t a^2 \ \ \mathbb{E}\left[ B_s^{\ 2(a-1)} \right] ds< \infty$$ any hint to show this? I should not use results on Brownian motion moments.

NOTE: edited after @surb comment!!

• What has to be proved is $\mathbb E[\int_0^t a^2(B_s^{a-1})^2ds]<\infty$, and not what you wrote. I guess that $a\in \mathbb N$. Notice that if $X\sim \mathcal N(0,1)$, then all moments are very well known (see for example the normale distribution on wikipedia)
– Surb
Commented Jan 4, 2021 at 9:49
• Thanks for pointing me out that. I will double check the notes. Unfortunately I should not use moments results! Commented Jan 4, 2021 at 9:55
• What do you mean by : "you shouldn't use moment result" ? if you need to compute $\mathbb E B^{2(a-1)}_s$, you have to compute the $a-1$-moment... so obviously, you must use it !
– Surb
Commented Jan 4, 2021 at 10:11
• You linked a wikipedia page in which moments of Gaussian r.v. are listed. However I cannot use the fact I know them! Commented Jan 4, 2021 at 10:18

Here is a proof for $$\int_0^{t} aEB_s^{a-1}ds <\infty$$ which you wanted . You need to modify this based on the comment of Surb.
$$B_s \sim \sqrt s X$$ where $$X$$ has standard normal distribution. Hence, your integral $$\int_0^{t} aEB_s^{a-1}ds <\infty$$ is a constant times $$\int_0^{t} s^{(a-1)/2}ds$$ and this last integral equals $$\frac {t^{(a+1)/2}} {(a+1)/2}$$ if $$a>-1$$.
• Based on what Surb correctly spotted, I now have to deal with: $$\int_0^t E[B_s^{2(a-1)}] ds$$ where I just disregarded the constant a. Based on what I understand form your previous reasoning, should I first compute the expectation inside using the fact that $B_s$ is standard normal? If not what are you doing in your reasoning? Commented Jan 4, 2021 at 10:13
• @YodaAndFriends Use the same approach. You will get $\int_0^{t} s^{(a-1)}ds$ which is finite if $a>0$. Commented Jan 4, 2021 at 10:17
• Sorry but I find hard to discover what you are doing. I have: $$E[B_s^{2(a-1)}] = E[s^{a-1} X^{2(a-1)}]$$ with this equals to $$s^{a-1} \cdot E[X^{2(a-1)}]$$ is this right? How can I conclude your statement? Commented Jan 4, 2021 at 10:27
• I may have got it. You are basically saying the following: $$\int_0^t E[B_s^{2(a-1)} ]ds = \int_0^t s^{a-1} E[X^{2(a-1)}] ds$$ now, you bring outside the integral $$E[X^{2(a-1)}]$$ because it is constant wrt to s. Since standard normal moments are finite so it is $E[X^{2(a-1)}]$ and I am just left with $$\int_0^t s^{a-1} ds$$ So, showing this integral is finite suffices!! Hope to be correct!!! Commented Jan 4, 2021 at 10:42