I am not sure if I got this exercise right... I have 2 questions:

  1. Have I obtained the final result correctly?
  2. If so, I used Wolfram Alpha to obtain the value of the series, but how else can I obtain $\sum_{k=0}^{n-1}\frac{k}{n^2}=\frac{n-1}{2n}$ using my own calculations?

Thanks a lot for your help!


Let $W(t)$, $t\in\mathbb{R}_+$ be a Brownian motion with its natural filtration $\mathcal{F}_t, t\in\mathbb{R}_+$. Let $$\mathcal{H}:=\{h(t):h(t)\text{ is an adapted process, }\mathbb{E}\left[\int_0^{\infty}h^2(t)dt\right]<\infty\}$$ denote the set of general integrands with respect to $W(t)$.

Consider the stochastic processes $$X_n(t):=\sum_{k=0}^{n-1}W\left(\frac{k}{n}\right)\mathbb{1}_{\left(\frac{k}{n},\frac{k+1}{n}\right]}(t), t\ge 0,$$ for $n\ge 1$ and define $X(t):=W(t)\mathbb{1}_{[0,1]}(t),t\ge 0$.

Q) Verify that $X_n\in\mathcal{H}$ for all $n\ge 1$. (You may use Fubini's theorem without its proof)


We have, using Fubini's theorem, $$\mathbb{E}\int_0^{\infty}X_n^2(t)dt=\sum_{k=0}^{n-1}\int_{\frac{k}{n}}^{\frac{k+1}{n}}\mathbb{E}\left(W\left(\frac{k}{n}\right)\right)^2dt=\sum_{k=0}^{n-1}\frac{k}{n}\int_{\frac{k}{n}}^{\frac{k+1}{n}}1 dt=\sum_{k=0}^{n-1}\frac{k}{n}\left[t\right]_{\frac{k}{n}}^{\frac{k+1}{n}}=\sum_{k=0}^{n-1}\frac{k}{n}\left(\frac{1}{n}\right)=\sum_{k=0}^{n-1}\frac{k}{n^2}=\frac{n-1}{2n}<\infty$$

  • $\begingroup$ What exactly do you mean by the notation $W1_{(r,s]}(t)$ for $r,s \geq 0$? $\endgroup$ – saz Aug 9 '15 at 17:16
  • $\begingroup$ $\mathbb{1}_{[r,s]}$ is just the indicator function, so it is $1$ for $\{t:t\in[r,s]\}$ and $0$ otherwise. Also, I re-edited the question... Just in case you read the previous version. Thanks in advance! $\endgroup$ – s1047857 Aug 9 '15 at 17:22
  • 1
    $\begingroup$ Yes, I did read the previous version. Concerning your edited version: It is a well-known fact that $$\sum_{k=0}^{n-1} k = (n-1) n \frac{1}{2};$$ it can be proved e.g. by induction (a standard calculus1-exercise, I would say; see also this question math.stackexchange.com/q/122546). $\endgroup$ – saz Aug 9 '15 at 17:27

The question doesn't ask you to compute $E\int_0^\infty X_n^2(t)dt$, it asks you to show that it is finite. Do you need to know $\sum_{k=0}^{n-1}\frac{k}{n^2} = \frac{n-1}{2n}$ in order to realize this is a finite number? Nope!, finite sum of finite numbers is finite.

Also you should mention something about why $X_n$ is adapted (trivial but since the question asks to show $X_n \in \mathcal{H}$ you should do it.

  • $\begingroup$ But that is how my professor wants it (showing $\mathbb{E}\int_0^{\infty}X_n^2(t)dt<\infty$). The same question has appeared in all the past papers, I have his answers for the last 2 and he did it like that. Thanks, I forgot to mention that $X_n$ is adapted, they don't ask us to prove it though. $\endgroup$ – s1047857 Aug 9 '15 at 17:34
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    $\begingroup$ While it is not unreasonable to expect undergrads to know how to compute $\sum_{k=0}^{n-1} k$ explicitly, it is beyond absurd to require that in this problem. It has nothing to do with the problem at all, and as such I guarantee you that no professor would mark you wrong for not explicitly calculating the sum. $\endgroup$ – nullUser Aug 9 '15 at 20:46
  • $\begingroup$ By analogy, that would be like requiring them to be able to show that $\sum \frac{1}{n^2} = \frac{\pi^2}{6}$ if the question asked "does the sum converge?" The particular value of the sum does not matter, only its finiteness. $\endgroup$ – nullUser Aug 9 '15 at 20:48
  • $\begingroup$ OK, thanks! So I guess I will not panic during the exam if I cannot calculate such sums since it is enough to show that it is finite. $\endgroup$ – s1047857 Aug 9 '15 at 21:15

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