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Question: Let: $$ Y_t=\int_0^t\alpha_s \, dW_s $$ where $\alpha_t$ is a deterministic, continuous integrand and $W_t$ is a P Brownian motion. Calculate the moment generating function of $Y$.

I can solve this question using the fact that $Y$ is a stochastic integral and is therefore normal, then just finding the mgf as you normally would for any normal rv. But how would you go about proving that $Y$ is normal by finding the mgf? I'm getting confused with how to manipulate the integral in the exponential term.

Cheers

saz
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WeakLearner
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    Hi dimebucker91, it seems I always find your posts. Use the fact that $Y_t=\lim\sum_{k=1}^n\alpha_{t_k}(B_{t_{k+1}}-B_{t_k})$ in probability, and where the limit is as the mesh of the sequence of partitions goes to zero. By the way, a small comment: the stochastic integral is known to be a normal r.v. only if the integrand is deterministic (which is indeed the case here). – Ian May 26 '14 at 14:20
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    Hi Ian, Yes it seems I am lucky like that, haha.

    Here's what I get, does this make sense:

    $$ B_{t_{i+1}}-B_{t_i}=\alpha\sqrt{({t_{i+1}}-{t_i})}Z $$ Where Z is standard normal

    This eventually gives me:

    $$ m_Y(\theta)=E[exp(Z\theta{\int_0^t}\alpha_sZ\sqrt{ds})] $$

    Which gives me the right result, is this mathematically correct though?

    Also, one minor technical question, is the square of any term with an integral in it, just the integral of the integrand sqared?

     – WeakLearner May 26 '14 at 14:42
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    What is $\sqrt{ds}$? – Did May 26 '14 at 14:46
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    I assumed that $\sqrt{t_{i+1}-t{i}}$ would translate into $\sqrt{ds}$ I guess this is what I was asking if it was a mathematically correct thing to do.. – WeakLearner May 26 '14 at 14:51

2 Answers2

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If we apply Itô's formula to the function

$$f(x) := \exp(\lambda x)$$

and the Itô process $(Y_t)_{t \geq 0}$, then we find

$$e^{\lambda Y_t}-1 = \lambda \int_0^t e^{\lambda Y_s} \alpha_s \, dW_s + \frac{\lambda^2}{2} \int_0^t e^{\lambda Y_s} \alpha^2(s) \, ds.$$

Since the first term on the right-hand side is a martingale, we get for $\phi_{\lambda}(t):= \mathbb{E}e^{\lambda Y_t}$

$$\phi_{\lambda}(t) -1 = \frac{\lambda^2}{2} \int_0^t \phi_{\lambda}(s) \alpha^2(s) \, ds.$$

This ordinary differential equation can be solved explicitely,

$$\mathbb{E}e^{\lambda Y_t} = \phi_{\lambda}(t) = \exp \left( \frac{\lambda^2}{2} \int_0^t \alpha(s)^2 \, ds \right).$$

This proves that $Y_t$ is normal with mean $0$ and variance $\int_0^t \alpha(s)^2 \, ds$.

Edit As @NateEldredge pointed out, we have to ensure that $(e^{\lambda Y_s})_{s \geq 0}$ is suitable integrable; for a proof that this is indeed the case see e.g. René L. Schilling/Lothar Partzsch: Brownian motion - An introduction to stochastic processes, Chapter 18 (2nd edition) or my other answer.

saz
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  • There's a bit more work to do to check that the first term is a martingale. A priori it is only a local martingale; we need to know something about the integrability of $e^{\lambda Y_s}$. – Nate Eldredge May 26 '14 at 16:44
  • @NateEldredge Yes, of course - but in fact $e^{\lambda Y_s}$ is suitable integrable under the given assumptions; see e.g. Schilling/Partzsch: Brownian motion, Chapter 18. Alternatively, one may apply the standard stopping argumentation. – saz May 26 '14 at 16:50
  • @saz I was looking for a proof of the martingale Property of the first term but I couldn’t find it. At which page or Theorem can I find it ? Thank you – ez43eg May 16 '23 at 13:45
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It follows easily from Itô's formula that

$$M_t := \exp \left( \lambda Y_t - \frac{\lambda^2}{2} \int_0^t \alpha(s)^2 \, ds \right)$$

satisfies

$$M_t -1 = \lambda \int_0^t M_s dY_s, \tag{1}$$

i.e. $(M_t)_{t \geq 0}$ is a non-negative local martingale with continuous sample paths. This implies that $(M_t)_{t \geq 0}$ is a supermartingale (see e.g. this question); in particular, $\mathbb{E}M_t\leq 1 $. As $\alpha$ is a deterministic function, we get $$\mathbb{E}e^{\lambda Y_s} \leq \exp \left( \frac{\lambda^2}{2} \int_0^t \alpha(s)^2 \, ds \right)<\infty.$$ This means that the moment-generating function of $Y_t$ is well-defined. Moreover, using this estimate, we see that $(M_t)_{t \geq 0}$ is a true martingale, i.e. $\mathbb{E}M_t = 1$. By definition of $M_t$, this is equivalent to

$$\mathbb{E}\exp(\lambda Y_t) = \exp \left( \frac{\lambda^2}{2} \int_0^t \alpha^2(s) \, ds \right).$$

saz
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    Shouldn't it be that (according to https://en.wikipedia.org/wiki/Dol%C3%A9ans-Dade_exponential) $M_t - 1 = \int_0^t e^{\lambda Y_s - \frac{\lambda^2}{2} \int_0^s \alpha(r)^2 , dr} dY_s$ since the stochastic exponential is a solution to $M_t -1 = \int_0^t M_s dY_s$ – Slade Oct 31 '19 at 17:17
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    @Slade Thank you; fixed it. – saz Oct 31 '19 at 22:54