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Suppose we have a Wiener stochastic integral $$\int_{t-h}^t f(r_v) d W_v, \tag{1}$$ It is well known that, by the Ito isometry property, $$\mathbb E \left [\left(\int_{t-h}^t f(r_v) d W_v\right)^2 \bigg | \mathcal F_{t-h}\right] = \int_{t-h}^t \mathbb E[f(r_v)^2|\mathcal F_{t-h}] d v.$$

I am wondering how to extend this result to fourth conditional moments in general $$\mathbb E \left[\left(\int_{t-h}^t f(r_v) d W_v\right)^4 \bigg | \mathcal F_{t-h}\right].$$

There is an answer to a similar question here that implies: $$ \mathbb E \left[\left(\int_{t-h}^t f(r_v) d W_v\right)^4 \bigg| \mathcal F_{t-h}\right] = 3 \left[\int_{t-h}^t \mathbb E[f(r_v)^2 |\mathcal F_{t-h}] d v\right]^2 \tag{2} $$

However, it seems that the answer relies on the fact that $\mathbb E (Y^4) = 3 (\sigma^2)^2$ for any $Y \sim N(0,\sigma^2).$

So my question goes, how can I compute the conditional fourth moment of (1) when $\int_{t-h}^t f(r_v) d W_v$ is not normally distributed? Should I expect result (2)?

An example with $f(x) = x$ would be computing $$\mathbb E \left[ \left(\int_{t-h}^t r_v d W_v\right)^4 \bigg | \mathcal F_{t-h} \right]$$ given, e.g. $$d r_{t} = \mu(r_t) d t + \eta \sqrt{r_t} d W_t.$$

Thank you.

lcu
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  • I think that in general, you need to do do a case-by-case analysis. So for instance with your process $r_t$, start by applying Itô's lemma to $\left(\int_{t-h}^t r_v d W_v\right)^4$ and see if you can compute the expectations of the terms you get. – Stratos supports the strike May 15 '23 at 10:00
  • @StratosFair so you are not awear of any general approach to end up with something like: $h\left(\int_{t-h}^t \mathbb E[g(r_v) |\mathcal F_{t-h}] d v\right)$, where $h(\bullet)$ $g(\bullet)$ are some functions. – lcu May 15 '23 at 10:12
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    Indeed, I am not aware of such a result, sorry. In the specific example you give though I think the "brute force" approach will work, although it will probably be quite tedious. – Stratos supports the strike May 15 '23 at 10:15
  • The product of two integrals is the integral over the square with two independent variables. with Itos rule, that identical dW^2 -> dt, so one simply has to identify in a product of four dW .. dW all identical pairs. – Roland F May 15 '23 at 10:33

1 Answers1

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This falls into studying multiple iterated Wiener integrals (see Wiener chaos). Here one can at least get an upper bound. As mentioned in "Notes on the Itô Calculus" by Steven P. Lalley

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and so for $m=2$ we get

$$E\left(\int f dW\right)^{4}\leq c_{2}E\left[\left(\int f^{2} dW\right)^{2}\right]^{2}=c_{2}\left(\int Ef^{2} ds\right)^{2}.$$

In case the link dies, the proof simply uses that Hermite polynomials $H_{2m}(x,t)$ of Itô integrals are martingales and that the leading coefficient is $x^{2m}$.

Thomas Kojar
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