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!!

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    $\begingroup$ 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) $\endgroup$
    – Surb
    Commented Jan 4, 2021 at 9:49
  • $\begingroup$ Thanks for pointing me out that. I will double check the notes. Unfortunately I should not use moments results! $\endgroup$ Commented Jan 4, 2021 at 9:55
  • $\begingroup$ 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 ! $\endgroup$
    – Surb
    Commented Jan 4, 2021 at 10:11
  • $\begingroup$ You linked a wikipedia page in which moments of Gaussian r.v. are listed. However I cannot use the fact I know them! $\endgroup$ Commented Jan 4, 2021 at 10:18

1 Answer 1


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 $a>-1$.

  • $\begingroup$ 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? $\endgroup$ Commented Jan 4, 2021 at 10:13
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    $\begingroup$ @YodaAndFriends Use the same approach. You will get $\int_0^{t} s^{(a-1)}ds$ which is finite if $ a>0$. $\endgroup$ Commented Jan 4, 2021 at 10:17
  • $\begingroup$ 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? $\endgroup$ Commented Jan 4, 2021 at 10:27
  • $\begingroup$ 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!!! $\endgroup$ Commented Jan 4, 2021 at 10:42
  • $\begingroup$ @YodaAndFriends Yes., it is correct. $\endgroup$ Commented Jan 4, 2021 at 11:30

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