# Variance of time integral of squared Brownian motion

I want to calculate the variance of

$$I = \int_0^t W_s^2 ds$$

I was thinking I could define the function $f(t,W_t) = tW_t^2$ and then apply Ito's lemma so I get

$$f(t,W_t)-f(0,0) = \int_0^t \frac{\partial f}{\partial t}(s,W_s)ds + \int_0^t \frac{\partial f}{\partial x}(s,W_s)dW_s+ \frac{1}{2}\int_0^t \frac{\partial^2 f}{\partial x^2}(s,W_s)ds \\= I + \int_0^t 2sW_sdW_s + \frac{t^2}{2}$$

By rearranging I get

$$I = tW_t^2 - \int_0^t 2sW_sdW_s - \frac{t^2}{2}$$

We then get that (I'm not sure here but i think the expectation is zero of any integral w.r.t BM?)

$$\mathbf{E}[I]=\frac{t^2}{2}$$

And variance

$$\mathbf{V}[I] = \mathbf{V}[tW_t^2 - \int_0^t 2sW_sdW_s - \frac{t^2}{2}] = t^2\mathbf{V}[W_t^2]+\mathbf{E}[(\int_0^t 2sW_sdW_s)^2] \\= 2t^4 + \mathbf{E}[\int_0^t 4s^2W_s^2ds]\quad\text{(Isometry property)}$$

Not sure if it is OK to change order of integration and expectation here, but if I do that, I get

$\mathbf{V}[I]= 2t^4 + \int_0^t 4s^2\mathbf{E}[W_s^2]ds = 2t^4 + \int_0^t 4s^2\mathbf{E}[W_s^2]ds = 2t^4 + \int_0^t 4s^3ds=3t^4$

However, the answer says the variance should be $\frac{t^4}{3}$, so I guess I do something wrong?

• $$2\text{Cov}\left(tW_t^2,\,-2\int_{0}^{t}2sW_sdW_s\right)=???$$ – user16651 Oct 25 '16 at 22:13

Other Way

By application of Ito's lemma , we have $$W^4_t=4\int_{0}^{t}W^3_tdW_s+6\int_{0}^{t}W^2_tds\tag 1$$ We know

\left\{ \begin{align} &\mathbb{E}\left[ {{W}^{2n+1}}(t) \right]=0\,\,\,\,\,\,\,\,\,\,\,\,\,\, \\ & \quad \mathbb{E}\left[ {{W}^{2n}}(t) \right]=\frac{(2n)!}{{{2}^{n}}n\,!}\,{{t}^{n}} \\ \end{align} \right.

therefore $$\text{Var}(W^4_t)=\mathbb{E}[W^8_t]-\mathbb{E}[W^4_t]^2=105t^4-(3t^2)^2=96t^4\tag 2$$ By application of Ito's Isometry, we have $$\text{Var}\left(4\int_{0}^{t}W^3_tdW_s\right)=16\int_{0}^{t}\mathbb{E}[W^6_s]ds=240\int_{0}^{t}s^3ds=60t^4\tag 3$$ on the other hand $$2\text{Cov}\left(4\int_{0}^{t}W^3_tdW_s\,,\,6\int_{0}^{t}W^2_tds\right)=24t^4\quad\text{(Why?)}\tag 4$$ Moreover $$\text{Var}(W^4_t)=\text{Var}\left(4\int_{0}^{t}W^3_tdW_s+6\int_{0}^{t}W^2_tds\right)\tag 5$$ thus $$96t^4=60t^4+36\text{Var}\left(\int_{0}^{t}W^2_tds\right)+24t^4$$ i.e $$\text{Var}\left(\int_{0}^{t}W^2_tds\right)=\frac{1}{3}t^4$$

• When trying to calculate to covariance you provided I eventually get to $\mathbf{E}[W_t^4 \int_0^t W_s^2ds]$ Then I don't know how to proceed. I tried to replace the $ds$-integral with the other terms in your equation $(1)$, but that did not help. – termachine Oct 26 '16 at 10:26

Here's another take on the question:

\begin{align} \int_0^t W_s^2 ds &= \int_0^t \int_0^s d(W_u^2) ds \\ &= 2 \int_0^t \int_0^s W_u dW_u ds + \int^t_0 \int^s_0 du ds \tag{Itô's lemma}\\ &= 2 \int_0^t \int_u^t W_u ds dW_u + \frac{t^2}{2}\tag{Stochastic Fubini}\\ &= 2 \int_0^t W_s (t-s) dW_s + \frac{t^2}{2} \end{align}

Now you can use Itô's isometry to conclude: \begin{align} \Bbb{V}\left[ 2 \int_0^t W_s (t-s) dW_s \right] &= 4 \int_0^t \Bbb{E}[W_s]^2 (t-s)^2 d\langle W, W \rangle_s \\ &= 4 \int_0^t s(t^2-2st+s^2) ds \\ &= 4 \left( \frac{t^4}{2} - 2\frac{t^4}{3} + \frac{t^4}{4} \right) = \frac{t^4}{3} \end{align}

• Nicely spotted! – Quantuple Nov 13 '18 at 13:34

A few hints I would like to suggest:

• How is $Var(W_t^2)$ computed? Note that \begin{align*} W_t^2 = 2\int_0^t W_s dW_s + t. \end{align*} Then \begin{align*} Var(W_t^2) &=E\left(W_t^2-t)^2\right) =2t^2. \end{align*}

• Generally, the variance of a sum is not the sum of variances, which only holds for uncorrelated random variables. That is, you also need to compute the expectation \begin{align*} E\left(W_t^2 \int_0^t 2s W_s dW_s \right) &=4\int_0^ts^2 ds = \frac{4}{3}t^3. \end{align*}

• Finally, \begin{align*} Var(I) &= Var\left(tW_t^2\right) + Var\left(\int_0^t 2s W_s dW_s \right) - 2tE\left(W_t^2 \int_0^t 2s W_s dW_s \right) = \frac{1}{3}t^4. \end{align*}