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$B_t$ is a Brownian motion and $Y_t:=e^{aB_t+bt}$. For which $a$ and $b$ is $Y_t\in M^2$?

I found a theorem that says that sufficient for $Y_t\in M^2$ would be $E[\int_0^\infty Y_t^2 dt]<\infty $

But how can I integrate over a function of a Brownian motion?

I don't think that the Ito isometry is helpful, because I would need $Y_t\in M^2$ before I can use it.

Or is there a simpler way to prove $Y_t\in M^2$ or $E[\int_0^\infty Y_t^2 dt]<\infty $ without having to calculate the integral?

milfor
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  • What is $M^2$...? By definition, $B_t$ is Gaussian with mean zero and variance $t$. Gaussian random variables have exponential moments and they can be calculated explicitly (see e.g. wikipedia). This will allow you to compute $\mathbb{E}(Y_t^2)$ explicitly. – saz Sep 18 '19 at 11:13
  • @saz: I think that $M^2$ is the vector space of square integrable martingales. – Surb Sep 18 '19 at 11:36
  • We have defined $M^2_{step}$ as the set of the random step functions $\sum \eta_j*1_{[t_j,t_{j+1})}$ where $\eta_j$ are random variables and $M^2$ as set of functions that can be approximated by them in the sense of $\lim E[\int_0^\infty (f_n(t)-f(t))^2 dt] =0$ with $f_n\in M^2_{step}$ and $f\in M^2$. – milfor Sep 18 '19 at 11:51

2 Answers2

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  • If $$\mathbb E\int_0^\infty Y_t^2dt<\infty $$ then $\left(\int_0^t Y_sdB_s\right)_{t\geq 0}$ is a martingale, but in general, not $(Y_t)_{t\geq 0}$.

  • Let $s<t$. $$\mathbb E[e^{aB_t+bt}\mid \mathcal F_s]=e^{bt+aB_s}\mathbb E[e^{a(B_t-B_s)}\mid \mathcal F_s]=e^{bt+aB_s}\mathbb E[e^{a(B_t-B_s)}].$$ Using the fact that $B_t-B_s\sim \mathcal N(0,t-s)$ allow you to find $a,b$ s.t. $(e^{aB_t+bt})_{t\geq 0}$ is a Martingale. Now, $\mathbb E[Y_t^2]$ shouldn't be to complicate to calculate, and will allow you to conclude.

An other way is using Itô formula, $$e^{aB_t+bt}=1+\int_0^t\left(\frac{a^2}{2}+b\right)e^{aB_t+bt}\,\mathrm d t+\int_0^tae^{aB_s+bs}\,\mathrm d B_s.$$

As far as you proved that $$\mathbb E\int_0^\infty e^{2aB_t+2bt}\,\mathrm d t<\infty ,$$ then, $(e^{aB_t+bt})_{t\geq 0}$ is a Martingale $\iff$ $\frac{a^2}{2}+b=0$.

Surb
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  • That this equation is equivalent to $Y_t$ being a martingale was an earlier example. But how does that help? Couldn't it be that this equation is violated and still $Y_t\in M^2$? – milfor Sep 18 '19 at 12:03
  • In this precise example no, but in general yes of course ! There is no reason that $Y_t=\int Y_t^2,\mathrm d B_t$. – Surb Sep 18 '19 at 12:05
  • Sorry, but i don't understand that at all. For which a and b is $Y_t\in M^2$? I have complemented our definition of $M^2$ above. (I tried to copy it into the original question too but that doesn't work properly.) – milfor Sep 18 '19 at 12:23
  • @Martin: Your definition of $M^2$ is not clear. What do you want ? $a,b$ s.t. $(e^{aB_t+bt})_{t\geq 0}$ is a Martingale or $a,b$ s.t. $\mathbb E\int_0^\infty e^{2aB_t+2bt}dt<\infty $ ? – Surb Sep 18 '19 at 12:26
  • Both. But the first is an earlier example that is already solved. The second is the problem. – milfor Sep 18 '19 at 12:29
  • The second is just $\int_0^\infty e^{bt}\mathbb E[e^{aB_t}]dt$ which is not so difficult to calculate since $B_t\sim \mathcal N(0,t)$. – Surb Sep 18 '19 at 12:34
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Evaulte

$$E\left[\int_0^\infty Y_t^2 dt\right] = E\left[\int_0^\infty e^{2aB_t+2bt} dt\right] $$

$$= \int_0^\infty E\left[e^{2aB_t}\right] e^{2bt} dt=\int_0^\infty e^{2a^2t}e^{2bt} dt $$ $$=\frac{1}{2(a^2+b)}e^{2(a^2+b)t}|_0^\infty $$

Thus, the condition on $a$ and $b$ is

$$a^2+b < 0$$

Quanto
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