I want to compute the fourth moment of a standard Wiener process: $E[W_t^4]$. My solution is not equal to the one in my textbook but I don't understand where I am wrong. I started by applying Ito's lemma to $g(W_t) = W_t^4$ and got

$dg(W_t) = 6W_t^2dt + 4W_t^3dW_t$,

implying that

$W_t^4 = \int_0^t 6W_t^2ds + 4 \int_0^t W_s^3 dW_s$.

Then, I took expectations and got for the first term

$ E[\int_0^t 6W_t^2ds] = 6\int_0^t V[W_t]ds = 6\int_0^t sds = 3 t^2$.

In order to calculate the second term, I applied Ito's lemma to $f(W_t) = W_t^4/4$ and got

$df(W_t) = \frac{3}{2}W_t^2dt + W_t^3dW_t$,

implying that

$\int_0^t W_s^3 dW_s = W_t^4/4 - \int_0^t \frac{3}{2} W_s^2ds $.

So, I took again expectations and got

$E[\int_0^t W_s^3 dW_s] = E[W_t^4]/4 - \frac{3}{2} \int_0^t V[W_t]ds \\ = 3 t^4 /4 - 3 t^2 /4$.

Hence, my result is $E[W_t^4]=3 t^2 + 4(3 t^4/4 - 3 t^2/4) = 3t^4 $.

However, in my textbook the second term was just dropped and the result was $3 t^2$. What's wrong with my way?


First of all, there are several typos in your calculations (e.g. it should read $\int_0^t W_s^2 \,ds$ instead of $\int_0^t W_t^2 \, ds$). Your calculation goes wrong when you write

$$\mathbb{E} \left( \int_0^t W_s^3 \, dW_s \right) = \frac{\mathbb{E}(W_t^4)}{4} - \frac{3}{2} \int_0^t V(W_s) \, ds = \frac{\color{red}{3t^4}}{4} - \frac{3t^2}{4}.$$

(I don't get what you did in this last step - you want to calculate $\mathbb{E}(W_t^4)$; so why replace it with $3t^4$?)

Note that applying Itô's lemma is overkill: Since $W(_t)_{t \geq 0}$ is a Wiener process, we know that $W_t \sim N(0,t)$ (i.e. $W_t$ is Gaussian with mean $0$ and variance $t$) and the moments of Gaussian random variables can be calculated explicitly. However, if you really want to invoke Itô's formula, then it goes like that: By Itô's formula, we have

$$W_t^4 = 4 \int_0^t W_s^3 \, dW_s + 6 \int_0^t W_s^2 \, ds. \tag{1}$$

Since $(W_s^3)_{s \geq 0}$ is properly integrable, we know that the stochastic integral

$$M_t := \int_0^t W_s^3 \, dW_s$$

is a martingale and therefore $\mathbb{E}M_t = \mathbb{E}M_0=0$. Taking expectation in $(1)$ yields

$$\mathbb{E}(W_t^4) = 6 \int_0^t \mathbb{E}(W_s^2) \, ds$$

by Fubini's theorem. Finally, since $\mathbb{E}(W_s^2)=s$, we get $\mathbb{E}(W_t^4) = 3t^2$.

  • $\begingroup$ Thanks for your answer. Why isn't it possible just to apply $\sigma^p(p-1)!!$ to this problem? This gives $3 t^4$. Moreover, does this mean if one integrates $W_t^p$ wrt to $W_t$ and takes expectation, it is always zero? $\endgroup$ – stats_guy Jan 23 '15 at 19:15
  • $\begingroup$ In addition, what is the solution for $M_t$ without taking integrals? $\endgroup$ – stats_guy Jan 23 '15 at 19:24
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    $\begingroup$ @stats_guy 1. Yes, $$\mathbb{E} \left( \int_0^t W_s^p \, dW_s \right)=0$$ holds for any $p \geq 1$. 2. The rule $\sigma^p (p-1)!!$ can be applied; note however that $\sigma^2$ is the variance, i.e. $\sigma^2 = t$; hence $\sigma = \sqrt{t}$. This means that $\sigma^4 (4-1)!! = 3t^2$. $\endgroup$ – saz Jan 23 '15 at 19:24
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    $\begingroup$ What do you mean by "solution"? The Itô formula shows that we can rewrite $M_t$ as $$\frac{W_t^4}{4} - \frac{6}{4} \int_0^t W_s^2 \, ds,$$ but that's all we can do. $\endgroup$ – saz Jan 23 '15 at 19:26
  • $\begingroup$ The fact that I didn't take $\sqrt{t}$ is really worth a facepalm. Regarding my second comment: thanks making it clear. I will need some time to get an intuition for this. $\endgroup$ – stats_guy Jan 23 '15 at 19:33

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