Expected Integrated Powers of Brownian Motion

Summary

The document derives the expectation of the time integral of an integer power of standard Brownian motion. Fubini’s theorem moves expectation inside the time integral, reducing the problem to the moments of the Brownian value at each time. Since that value is centered normal with variance equal to elapsed time, odd powers have zero expectation, while even powers use the normal moment formula involving the double factorial. Integrating those moments over time gives the stated expectation for each parity of the exponent.

Although the question asks for both mean and variance, the included answer calculates only the mean; it provides no variance formula or derivation. The result is a probability calculation rather than a trading strategy, and the discussion does not address extensions to other processes, assumptions beyond Brownian motion, or applications in quantitative finance. Readers seeking the variance would need to compute joint moments across pairs of times, which are not covered in this document.

Key ideas

• Fubini’s theorem expresses the expected integral as the integral of pointwise expectations.
• Odd powers of a centered Brownian value have zero expectation.
• Even Brownian moments are determined by the normal moment formula and scale with time.
• The answer supplies the mean of the integral but leaves the requested variance unresolved.

Tags

# Expectation and variance of $\int_0^t (W_s)^n ds$ for any positive integer $n$?


# Expectation and variance of $\int_0^t (W_s)^n ds$ for any positive integer $n$?












It is well known that the integral $$\int_0^t W_s ds,$$ where $(W_s)_s$ is a Brownian motion, can be derived using Ito's Lemma. More precisely, Ito's lemma on $d(tW_t)$ implies that $$d(tW_t) = tdW_t + W_t dt.$$ Therefore, $$\int_0^t W_s ds = tW_t - \int_0^t sdW_s.$$ Its mean and variance can be obtained from this expression. This leads to my question below.

> Question: Given a positive integer $n,$ what is the mean and variance $$\int_0^t (W_s)^n ds?$$

Calculation above is for $n=1.$

## Answer by user39119 (score 4, accepted)

https://quant.stackexchange.com/a/50169

For the mean you can use Fubini's Theorem to change the order of integration $$ E\int_0^t (W_s)^n ds = \int_0^t E(W_s)^n ds$$ Then we can use the fact that $W_s \sim N(0,\sqrt{s})$ to obtain \begin{align*} E (W_s)^{2k+1} &= 0, k=0,1,2,... \tag*{(odd n)} \\ E (W_s)^{2k} &= (2k-1)!!s^k, k=1,2,... \tag*{(even n)} \end{align*} Therefore, \begin{align*} E\int_0^t (W_s)^{2k+1} ds &= 0, k=0,1,2,... \tag*{(odd n)} \\ E\int_0^t (W_s)^{2k}ds &= (2k-1)!! \int_0^t s^k ds = (2k-1)!! \frac{t^{k+1}}{k+1}, k=1,2,... \tag*{(even n)} \end{align*}

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