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I am trying to calculate $E(\int_0^T {W_s ds})$, where $W_s$ is a standard Brownian motion.

Now two approaches I can think of:

1) Take a partition of $[0,T]$. Calculate $E(\sum {W_{t_i}(t_{i+1} - t_i)})$ and take the limit as you shrink the size of the partition.

2) Calculate $\int_0^T { E(W_s)ds }$.

However, for approach 1), its not clear what function would dominate the absolute value of the terms inside the E() for all possible partitions, and that it would have a finite expectation. So, interchanging limit and expectation is dicey.

For approach 2), Fubini's theorem would require me to know a-priori that $E(\int_0^T {|W_s|ds})$ is finite, and I don't see how I could show that.

How can any of these approaches be fixed, if at all? Or is there another way to solve the problem?

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    $\begingroup$ Brownian motion is a martingale (en.wikipedia.org/wiki/Martingale_%28probability_theory%29); the expectation you want is always zero. Also voting to close as this would be better suited to another site mentioned in the FAQ. $\endgroup$
    – Steve Huntsman
    Commented Oct 14, 2010 at 3:28
  • $\begingroup$ If BM is a martingale, why should its time integral have zero mean ? (Although, yes, both approaches will give me an answer of 0). Also, the problems with the approaches I mentioned are valid for questions like calculating E(\Int_0_T {W_s^2 ds}) $\endgroup$
    – Cosmonut
    Commented Oct 14, 2010 at 3:56
  • $\begingroup$ If I have the distribution of Max{|W_s|:0<s<T}, I could try showing that its expectation if finite. This would let me validate both the approaches. If someone has a reference where this is calculated, please let me know. $\endgroup$
    – Cosmonut
    Commented Oct 14, 2010 at 4:04
  • $\begingroup$ Martingales are in $L^1$. $\endgroup$
    – Steve Huntsman
    Commented Oct 14, 2010 at 5:34
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    $\begingroup$ More appropriate for math.stackexchange.com $\endgroup$
    – zhoraster
    Commented Oct 14, 2010 at 6:34

2 Answers 2

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For approach 2, Fubini's theorem works just as well if you show $$ \int_0^T E|W_s|ds < \infty $$ which is easy. Indeed, perhaps even easier is to note $$ \int_0^T E(|W_s|^2)ds = \int_0^T s ds = \frac{1}{2}T^2 < \infty$$ and use Jensen/Hölder/Cauchy-Schwarz.

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    $\begingroup$ This is the way to do it. Any invocation of martingales, running maximum, etc., is overkill for this problem. $\endgroup$
    – Tom LaGatta
    Commented Oct 15, 2010 at 1:03
  • $\begingroup$ Thanks, I like this. Looks like this approach can be generalized to work out the expectation of integrals of various functions of a Weiner process as well, such as exp(W_s). $\endgroup$
    – Cosmonut
    Commented Oct 15, 2010 at 4:19
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It is possible to integrate by parts in $\int_0^T B(t) dt$ and obtain

$-B(t) (T-t)|_{t=0}^{t=T} + \int_0^T (T-t) dB(t) \overset{d}{=} \int_0^T (T-t) dB(t)$

The Wiener integral on the right has a normal distribution with mean $0$ and variance $\int_0^T (T-t)^2 dt = T^3/3$.

Edit: Sorry, I used $B$ instead of $W$ to denote Brownian motion.

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  • $\begingroup$ This is really cool! I had worked out the variance of \Int_0_T {W_s ds} the hard way by partitioning [0, T] into n parts, calculating the variance of the sum and then letting n go to infinity. I got the correct answer, (thanks for verifying that :)) but was once again worried about limit interchange issues. Although, I'm guessing there must be some technical isses involved in justifying the integration by parts formula and equating the time integral to a stochastic integral. (I guess you are implicitly using Ito's lemma ?) $\endgroup$
    – Cosmonut
    Commented Oct 15, 2010 at 4:29
  • $\begingroup$ Slightly technical, but not bad: If h is a function with bounded variation on [a,b] and g is continuous on [a,b], then the Riemann-stieltjes integral \int_a^b g(x) dh(x) can be defined by approximating with step functions. Then, applying integration by parts, you can define the integral \int h(x) dg(x) as h(x)g(x)|_a^b - \int g(x) dh(x). Since W is almost surely continuous, and (T-t) has bounded variation, you apply this fact omega by omega to g(t) = W(t,\omega). See pages 12 and 13 in Kuo's book Introduction to Stochastic Integration for more detail. $\endgroup$
    – MarkV
    Commented Oct 18, 2010 at 15:22
  • $\begingroup$ This is the canonical answer according to: math.nyu.edu/faculty/goodman/teaching/StochCalc2004/notes/… $\endgroup$
    – David Addison
    Commented Sep 8, 2017 at 1:29
  • $\begingroup$ Unclear why you're able to conclude that the Wiener integral on the right is normally distributed with these properties $\endgroup$
    – user20672
    Commented May 1, 2022 at 8:20

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