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I have to show that the following two processes are martingales

  1. $M_t=g(t)B_t-\int_0^tg'(s)B_sds$
  2. $X_t=\exp\left(e^tB_t-\int^t_0e^sB_sds-\frac{e^{2t}}{4}+\frac{1}{4}\right)$

Where $g(t)$ is a real valued continuously differentiable function and $B_t$ is a brownian motion.

I tried the first part straight forward but without success. I guess I can not simply use Fubini here to switch expectation and integration. So I believe it has something to do with Ito. But i really don't know how to apply it here to show that 1. is a martingale.

For the second process: I know that the first two terms in the exponent are 1. with $g(t)=e^t$ so this is a martingale. I assume I also have to apply Ito here aswell, but like in 1., I'm not sure how.

Larry
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Toxxiqq
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    For the first one you can use Fubini's theorem for conditional expectations. Alternatively, you can apply Itô's formula to $f(t,x) := g(t) \cdot x$. – saz Dec 12 '18 at 09:53
  • I tried using Fubini and I got for $u<t$:$\mathbb{E}[M_t|\mathcal{F}_u]=g(t)B_u-\int^u_0g'(s)B_sds$. I don't see where the first $g(t)$ gets to a $g(u)$ – Toxxiqq Dec 12 '18 at 10:34
  • You went wrong when you calculated the conditional expectation of the integral. Compute $\mathbb{E}(B_s \mid \mathcal{F}_u)$ separately for $s \leq u$ and $s>u$. What do you get? – saz Dec 12 '18 at 10:42
  • I did separate it and I got $\int_0^ug'(s)\mathbb{E}[B_s|\mathcal{F}_u]ds+\int^t_ug'(s)\mathbb{E}[B_s|\mathcal{F}_u]$. In the first integral we have $s\leq u$ and therefore, all $B_s$ are $\mathcal{F}_u$ measurable, i.e. we just get $\int^u_0g'(s)B_sds$. In the second integral we just get $\int_u^tg'(s)\mathbb{E}[B_s]ds$ which is zero since $\mathbb{E}[B_s]=0$. Or am I wrong with that? – Toxxiqq Dec 12 '18 at 11:21
  • Yes, the second part is wrong. Note that $\mathbb{E}(B_s \mid \mathcal{F}_u) = B_u$ for $s \geq u$.... that's the martingale property of Brownian motion. – saz Dec 12 '18 at 11:26
  • Ah of course... That was dumb... Then, for the second part, I get $B_u(g(t)-g(u))$ and alltogether I get $M_u$. Thanks a lot. – Toxxiqq Dec 12 '18 at 11:33
  • Do you have an idea for the second process? Reformulating yields that I have to verify that $\mathbb{E}[e^{M_t}]=e^{M_u}e^{\frac{2(t-u)+1}{4}}$ for $M_t$ like in 1. with $g(t):=e^t$. – Toxxiqq Dec 12 '18 at 12:51
  • Could you please check whether there is a typo in the definition of $X_t$? Do you really mean $\exp(\ldots - \frac{e^{2t}}{4}\ldots)$ or rather $\exp(\ldots- \frac{2t}{4} \ldots)$? (Your previous comment suggests the later). If you feel somewhat familiar with Gaussian random variables (and their exponential moments), then you can calculate the conditional expectation by hands. Pull out the deterministic terms and the $\exp(-\int_0^u B_s , ds)$ and then use the independence of the increments (write $B_t = (B_t-B_u)+B_u$ and B_s = (B_s-B_u)+B_u$)... alternatively you can apply Itô's formula... – saz Dec 12 '18 at 13:20
  • My comment was not right, but the definition of $X_t$ is correct. I assumed $-\frac{2t}{4}$ which is not right. Sorry. I would like to use Itô but I sadly have no idea how. I was looking for some examples but I'm still stuck. In the lecture we had no examples. This makes it hard for me to understand what I really have to do. – Toxxiqq Dec 12 '18 at 13:33
  • Which version of Itô's formula do you know? The one for Itô processes (see e.g. wikipedia? – saz Dec 12 '18 at 13:38
  • For a function $f\in C^2(\mathbb{R}^n)$ and an Itô process $X_t$ we have $df(X_t)=f'(X_t)dX_t+\frac{1}{2}f''(X_t)dt$ in terms of differential notation. – Toxxiqq Dec 12 '18 at 13:48
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    There is no "dt" in the 2nd term on the right-hand side.. it should read $\langle X \rangle_t$ (or, perhaps, $(dX_t)^2$ ... depending on which notation you are using). – saz Dec 12 '18 at 13:56
  • @saz In an attempt to revive this question a bit; how come we can immediately say that we are able to apply Fubini's theorem for the first one? Don't we need that the function for which we wish to interchange expectation and integral, has to be both measurable and integrable? Why can we conclude that this is indeed the case? Thanks in advance! – Charlie Shuffler Apr 10 '19 at 10:19
  • @saz I looked into applying Tonelli's theorem, but for that we would need that $g'(s)B_s\geq 0$ for all $s$, which can also not say up front. So I am no further to figuring out why we can immediately say that we can apply Fubini's theorem. – Charlie Shuffler Apr 10 '19 at 11:31
  • @S.Crim Yes, we do. Note that $g$ is bounded on compact sets (since it is continuous) and $B_t$ is integrable (since it is Gaussian), so there are no integrability issues here. Fubini's theorem does not require non-negativity (perhaps you looked up Tonellli's theorem?) – saz Apr 10 '19 at 11:44
  • @saz Fair enough, thanks alot once again! – Charlie Shuffler Apr 10 '19 at 11:52

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