# 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
Commented Oct 25, 2016 at 22:13

Other Way

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

• 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. Commented Oct 26, 2016 at 10:26
• I've completed the covariance calculation above in the answer to another post: quant.stackexchange.com/questions/57206/… (note that this is not the most direct method to solve the problem, Quantuple's answer below is very succinct, but the above post does demonstrate many nice properties of stochastic integration) Commented Aug 9, 2020 at 3:01

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! Commented Nov 13, 2018 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*}