Computing W_t^2 and Its Moments from Ito Integrals

The Brownian Square as an Itô Integral

Squaring Brownian motion is a compact test of several essential ideas in stochastic calculus. The ordinary chain rule would omit a term, whereas Itô's formula records the quadratic variation of Brownian motion. The resulting identity is useful not only for finding moments, but also for checking whether an integral calculation has the correct drift and integrability.

Work on a filtered probability space $(\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})$ carrying a standard Brownian motion $(B_t)_{t\ge0}$ with respect to the filtration. This means that $B$ is adapted, has continuous paths and $B_0=0$, and, for every $s<t$, $B_t-B_s$ is independent of $\mathcal{F}_s$ and has distribution $N(0,t-s)$. The independence from $\mathcal{F}_s$, not just the distribution of the increment, is important for conditional-expectation statements.

For a fixed horizon $T<\infty$, the Itô integrals below have progressively measurable integrands in $\mathcal{L}^2$: these are processes $H$ for which $\mathbb{E}\int_0^T H_s^2\,ds<\infty$. In particular, $B_s$ is progressively measurable because it is continuous and adapted, and

\mathbb{E}\int_0^T B_s^2\,ds =\int_0^T \mathbb{E}[B_s^2]\,ds =\int_0^T s\,ds =\frac{T^2}{2}<\infty.

Thus $\int_0^t B_s\,dB_s$ is a well-defined square-integrable Itô integral for $0\le t\le T$. The Brownian quadratic variation rule is $dB_t\,dB_t=dt$; the finite-variation terms satisfy $dB_t\,dt=0$ and $dt\,dt=0$.

Theorem: For Brownian motion $B$ with respect to the stated filtration, and for every finite $T$, the following identity holds for all $0\le t\le T$, up to indistinguishability: $$ B_t^2=2\int_0^t B_s\,dB_s+t. $$ The integral is a square-integrable martingale. Consequently, $\mathbb{E}[B_t^2]=t$ and $\operatorname{Var}(B_t^2)=2t^2$.

Proof. Apply Itô's formula to $f(x)=x^2$ and $B_t$. Its derivatives are $f'(x)=2x$ and $f''(x)=2$. The second-derivative correction is therefore $\tfrac12 f''(B_s)\,ds=ds$, giving

f(B_t)=f(B_0)+\int_0^t f'(B_s)\,dB_s +\frac12\int_0^t f''(B_s)\,ds =0+2\int_0^t B_s\,dB_s+t.

Set $I_t=\int_0^t B_s\,dB_s$. Its integrand is in $\mathcal{L}^2$ on each finite horizon, as shown above. The standard martingale property of the square-integrable Itô integral gives, for $0\le u\le t$, $\mathbb{E}[I_t\mid\mathcal{F}_u]=I_u$; its integrability also follows from the Itô isometry, since $\mathbb{E}[I_t^2]=\mathbb{E}\int_0^t B_s^2\,ds=t^2/2<\infty$. Thus this is a true martingale, not merely a local martingale. In particular, $\mathbb{E}[I_t]=0$. The identity now gives $\mathbb{E}[B_t^2]=t$. Since its deterministic term has zero variance, $\operatorname{Var}(B_t^2)=4\operatorname{Var}(I_t)$. The zero mean and Itô isometry yield $\operatorname{Var}(I_t)=\mathbb{E}[I_t^2]=t^2/2$, so $\operatorname{Var}(B_t^2)=2t^2$. The Itô formula gives the identity as an equality of continuous processes, hence it holds simultaneously for $0\le t\le T$ outside one null set. $\square$

Computing the Moments in Two Ways

The identity separates $B_t^2$ into a deterministic term and a centered stochastic integral. The Itô isometry is the key to its variance: for an integrand $H\in\mathcal{L}^2$, it states $\mathbb{E}[(\int_0^t H_s\,dB_s)^2]=\mathbb{E}\int_0^t H_s^2\,ds$. A separate calculation from the normal distribution checks that no factor has been lost.

Proposition: For each $t\ge0$, $\mathbb{E}[B_t^2]=t$ and $\mathbb{E}[B_t^4]=3t^2$. Hence $\operatorname{Var}(B_t^2)=2t^2$.

Proof. When $t>0$, $B_t$ has density $(2\pi t)^{-1/2}\exp(-x^2/(2t))$. Integration by parts in this density gives the normal-moment recursion $\mathbb{E}[B_t^{2k}]=(2k-1)t\,\mathbb{E}[B_t^{2k-2}]$ for positive integers $k$; the boundary terms vanish because a polynomial times the Gaussian density tends to zero at both ends of the real line. Starting from $\mathbb{E}[B_t^0]=1$, the cases $k=1,2$ give $\mathbb{E}[B_t^2]=t$ and $\mathbb{E}[B_t^4]=3t^2$. Therefore $\operatorname{Var}(B_t^2)=\mathbb{E}[B_t^4]-(\mathbb{E}[B_t^2])^2=3t^2-t^2=2t^2$. At $t=0$, $B_0=0$, and the same formulas hold directly. $\square$

Worked Example: The Identity at Time One

Let $I_1=\int_0^1 B_s\,dB_s$. Find its mean and variance, and use the square identity to check the first two moments of $B_1^2$.

First, $\mathbb{E}\int_0^1 B_s^2\,ds=\int_0^1s\,ds=\tfrac12$, so the integrand is in $\mathcal{L}^2$. The Itô integral is a square-integrable martingale and has mean zero. By the Itô isometry, $\mathbb{E}[I_1^2]=\int_0^1s\,ds=\tfrac12$, so $\mathbb{E}[I_1]=0$ and $\operatorname{Var}(I_1)=\tfrac12$. The identity reads $B_1^2=1+2I_1$. Consequently, $\mathbb{E}[B_1^2]=1+2(0)=1$, and

\mathbb{E}[(B_1^2)^2] =\mathbb{E}[(1+2I_1)^2] =1+4\mathbb{E}[I_1]+4\mathbb{E}[I_1^2] =1+0+4\left(\frac12\right)=3.

Check: since $B_1\sim N(0,1)$, its second and fourth moments are $1$ and $3$. The integral identity reproduces both.

Worked Example: Expectation and Variance at Time Two

Compute the expectation and variance of $B_2^2$ using the stochastic-integral identity, and then check each result directly from Gaussian moments.

For the integral $I_2=\int_0^2 B_s\,dB_s$, the integrability calculation is $\mathbb{E}\int_0^2B_s^2\,ds=\int_0^2s\,ds=2$. Thus $I_2$ is a square-integrable martingale with $\mathbb{E}[I_2]=0$, and the Itô isometry gives $\mathbb{E}[I_2^2]=2$. The identity $B_2^2=2+2I_2$ therefore yields

\mathbb{E}[B_2^2]=2+2(0)=2,\qquad \operatorname{Var}(B_2^2)=4\operatorname{Var}(I_2)=4(2)=8.

Independently, $B_2\sim N(0,2)$, so $\mathbb{E}[B_2^2]=2$ and $\mathbb{E}[B_2^4]=3(2)^2=12$. Hence $\operatorname{Var}(B_2^2)=12-(2)^2=12-4=8$.

Check: both methods give expectation $2$ and variance $8$. In particular, the fourth moment used in the variance calculation is $12$, not $3$: the variance parameter of this normal random variable is $2$, not $1$.

Conditional Moments and the Filtration Assumption

The same reasoning yields a conditional expectation. For $s<t$, the restricted integrand $B_u$, $s\le u\le t$, is progressively measurable and square-integrable because $\mathbb{E}\int_s^t B_u^2\,du=\int_s^t u\,du<\infty$. The martingale property of the Itô integral gives $\mathbb{E}[\int_s^t B_u\,dB_u\mid\mathcal{F}_s]=0$. Subtract the square identity at time $s$ from the one at $t$, then take conditional expectation, to obtain $\mathbb{E}[B_t^2\mid\mathcal{F}_s]=B_s^2+(t-s)$.

Worked Example: Conditional Variance from an Independent Increment

For $s=1$ and $t=3$, compute the conditional expectation and conditional variance of $B_3^2$ given $\mathcal{F}_1$. Here conditional variance means $\mathbb{E}[(B_3^2-\mathbb{E}[B_3^2\mid\mathcal{F}_1])^2\mid\mathcal{F}_1]$.

Write $B_3=B_1+\Delta$, where $\Delta=B_3-B_1$ is independent of $\mathcal{F}_1$ and has distribution $N(0,2)$. Its moments are $\mathbb{E}[\Delta]=0$, $\mathbb{E}[\Delta^2]=2$, $\mathbb{E}[\Delta^3]=0$, and $\mathbb{E}[\Delta^4]=3(2)^2=12$. Expanding the square and conditioning gives

\mathbb{E}[B_3^2\mid\mathcal{F}_1] =\mathbb{E}[(B_1+\Delta)^2\mid\mathcal{F}_1] =B_1^2+2B_1\mathbb{E}[\Delta]+ \mathbb{E}[\Delta^2] =B_1^2+2.

For the conditional variance, first expand the fourth power. Independence and the displayed increment moments give $\mathbb{E}[B_3^4\mid\mathcal{F}_1]=B_1^4+6B_1^2(2)+12=B_1^4+12B_1^2+12$. Subtract the square of the conditional mean:

\operatorname{Var}(B_3^2\mid\mathcal{F}_1) =B_1^4+12B_1^2+12-(B_1^2+2)^2 =B_1^4+12B_1^2+12-(B_1^4+4B_1^2+4) =8B_1^2+8.

Check: the result is nonnegative for every value of $B_1$. Averaging it gives $8\mathbb{E}[B_1^2]+8=8(1)+8=16$, which agrees with the unconditional variance of $B_3^2$ minus the variance of its conditional mean: $2(3)^2-\operatorname{Var}(B_1^2+2)=18-2=16$.

The filtration assumption is essential here. If one assumes only that $B$ is adapted and has Brownian marginal distributions, it need not be true that $B_t-B_s$ is independent of $\mathcal{F}_s$. Then the conditional increment-moment calculation above is not justified, and the conditional expectation of the future Itô integral cannot be inferred from the Brownian martingale property. Brownian motion with respect to the filtration includes precisely the needed independence from the past.

Common Pitfall: Dropping the Itô Correction

A tempting but incorrect calculation is to apply the ordinary chain rule and write $d(B_t^2)=2B_t\,dB_t$. Integrating that expression would give $B_t^2=2\int_0^tB_s\,dB_s$. Taking expectations would then assert $\mathbb{E}[B_t^2]=0$, even though $B_t\sim N(0,t)$ has $\mathbb{E}[B_t^2]=t$ for $t>0$. The contradiction exposes the missing term: Itô's formula includes $\tfrac12 f''(B_t)\,dt=dt$, because $dB_t\,dB_t=dt$. The corrected identity has expectation $t$, as required.

Another useful check is dimensional in time: the integral has mean zero, but its variance is $\int_0^t\mathbb{E}[B_s^2]\,ds=t^2/2$. Multiplying that variance by $4$ gives the variance $2t^2$ of the square. Both the stochastic-integral calculation and the independent Gaussian calculation should agree.

Check Your Understanding

Use the stated filtration assumptions and distinguish the expectation of an Itô integral from its second moment.

Apply Itô's formula to $f(x)=x^2$, explicitly identifying $f'$, $f''$, and the correction term.
For $J_t=\int_0^t B_s\,dB_s$, calculate $\mathbb{E}[J_t]$ and $\mathbb{E}[J_t^2]$, and state why the integrand meets the required condition.
Use the identity to compute $\mathbb{E}[B_4^2]$ and $\operatorname{Var}(B_4^2)$.
For $s<t$, explain which independence assumption justifies $\mathbb{E}[B_t^2\mid\mathcal{F}_s]=B_s^2+(t-s)$.
Show directly from the normal fourth moment that $\operatorname{Var}(B_t^2)=2t^2$, including the case $t=0$.

Computing W_t^2 and Its Moments from Ito Integrals

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