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EDIT: Showing this using Ito's lemma is easy, that's NOT what I want to do. I also realised that $2\mathbb{E}[S\xi]\neq 2\xi\mathbb{E}[S]$ since $\xi$ is also a random variable. Nontheless, if this is the case I have no idea how to calculate the expectation of $S\xi$ anyway.


Given a Brownian motion $W(t)$ I want to show that

\begin{align} \lim_{n\rightarrow\infty}\mathbb{E}\left[\left|\sum_{j=0}^{n-1}\frac{jT}{n}\left(W\left(\frac{(j+1)T}{n}\right)-W\left(\frac{jT}{n}\right)\right) - TW(T)+\int\limits_0^TW(t)dt\right|^2\right]=0 \tag1. \end{align}

For simplicity we denote the sum by $S$ and set $\xi=-TW(t)+\int_{0}^TW(t) \ dt$, now we have that \begin{align} \mathbb{E}[(S+\xi)^2]=\mathbb{E}[S^2]+2\xi\mathbb{E}[S]+\xi^2. \end{align}

However I have trouble calculating $\mathbb{E}[S^2]$. I know that $\mathbb{E}[S]=0$ since

\begin{equation} \mathbb{E}[S]=\sum_{j=0}^{n-1}\frac{jT}{n}\mathbb{E}[W_{j+1}-W_j]=0 \end{equation} since the increments are $\sim\mathcal{N}(0,T/n)$ so the sum above is just a sum of zeroes. So I'm kind of left to show that

\begin{equation} \lim_{n\rightarrow\infty}\mathbb{E}[S^2]=-\xi^2. \end{equation}

I tried the following: For Brownian motion I know that $\mathbb{E}[S^2]=\text{Var}[S]$ so

\begin{align} \mathbb{E}[S^2]&=\sum_{j=0}^{n-1}\frac{j^2T^2}{n^2}\text{Var}[W_{j+1}-W_j]=\sum_{j=0}^{n-1}\frac{j^2T^2}{n^2}\frac{T}{n}=\frac{T^3}{n^3}\sum_{j=0}^{n-1}j^2\\&={\frac {{T}^{3} \left( 2\,{n}^{2}-3\,n+1 \right) }{6{n}^{2}}} \end{align}

which gives $T^3/3$ when $n\rightarrow\infty$. But this is not the result I want.

So you might ask how did I end up in (1)? Well, I wanted to show, using the definition of the Ito integral that

$$\int_0^TW(t)dt+\int_0^TtdW(t) = TW(T).$$

I rewrote it as

$$\int_0^TtdW(t) = TW(T) - \int_0^TW(t)dt$$

and used the definition: If there exists a stochastic process $I(T)$ such that $||I_n(T)-I(T)||_2=\lim\limits_{n\rightarrow\infty}\mathbb{E}[|I_n(T)-I(T)|^2]=0$ then $I(T)$ is an Ito integral. Plugging in $I_n(T)$ and $I(T)$ I arrived to (1).

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    $\begingroup$ What you can do is to discretize $TW(T)$ and $\int_0^TW(t)dt$ using the same partition, and then combine all them together to reach the term $-\frac{T}{n}W(T)$. $\endgroup$ – Gordon Feb 1 at 22:05
  • $\begingroup$ Thanks a lot, I will try this tomorrow! $\endgroup$ – Parseval Feb 1 at 22:59
  • $\begingroup$ @Gordon Is it correct partitioning the integral as $$\int_0^{T}W(t)dt\approx\sum_{0}^{n-1}\frac{jT}{n}W\left(\frac{jT}{n}\right) \ ?$$ For $TW(T)$, does the partition simply become the same as above? $\endgroup$ – Parseval Feb 2 at 8:47
  • $\begingroup$ If I only partition $\int_0^TW(t)dt$ I obtain 0 in the expectation. I leave $-TW(T)$ as is then I get the combined sum in the argument of the expectation to be only $TW(T)$. Can this be correct? $\endgroup$ – Parseval Feb 2 at 9:27
  • $\begingroup$ $W(T)=\sum_{j=0}^{n-1} \big(W(\frac{(j+1)T}{n})-W(\frac{jT}{n})\big)$ and $\int_0^TW(t)dt\approx \sum_{j=0}^{n-1}\frac{T}{n}W(\frac{jT}{n})$. $\endgroup$ – Gordon Feb 2 at 12:36
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Note that \begin{align*} \int_0^T W(t)dt \approx \sum_{j=0}^{n-1}\frac{T}{n}W\Big(\frac{jT}{n}\Big). \end{align*} Then, \begin{align*} &\ \sum_{j=0}^{n-1}\frac{jT}{n}\bigg(W\Big(\frac{(j+1)T}{n}\Big)-W\Big(\frac{jT}{n}\Big)\bigg) - TW(T) + \int_0^T W(t)dt\\ \approx &\ \sum_{j=0}^{n-1}\frac{jT}{n}\bigg(W\Big(\frac{(j+1)T}{n}\Big)-W\Big(\frac{jT}{n}\Big)\bigg) - TW(T) + \sum_{j=0}^{n-1}\frac{T}{n}W\Big(\frac{jT}{n}\Big)\\ =&\ \sum_{j=0}^{n-1}\frac{jT}{n}W\Big(\frac{(j+1)T}{n}\Big) - \sum_{j=0}^{n-1}\frac{(j-1)T}{n}W\Big(\frac{jT}{n}\Big) - TW(T)\\ =&\ \sum_{j=0}^{n-1}\frac{jT}{n}W\Big(\frac{(j+1)T}{n}\Big) - \sum_{j=-1}^{n-2}\frac{jT}{n}W\Big(\frac{(j+1)T}{n}\Big) - TW(T)\\ =&\ \frac{(n-1)T}{n}W(T) - TW(T)\\ =&\ -\frac{T}{n} W(T). \end{align*}

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  • $\begingroup$ Ah, alright, so you did not actually discretize $TW(T)$ but only approximated the integral with the Riemann sum. But how did you combine the last two sums, when they differ in indexes? I assume that what you've done is this: $$\frac{T}{n}\left(\sum_{j=0}^{n-1}W\left(\frac{(j+1)T}{n}\right)-\sum_{j=-1}^{n-2}W\left(\frac{jT}{n}\right)\right)=\frac{T}{n}\sum_{j=0}^{n-1}\left(W\left(\frac{(j+1)T}{n}\right)-W\left(\frac{jT}{n}\right)\right)=\frac{T}{n}W(T)$$ $\endgroup$ – Parseval Feb 2 at 17:07
  • $\begingroup$ I missed the $j$ factor above that multiplies the $W'$s. What happens to them? Things are clear till the 4th line, I don't see how you go from 4th to 5th. $\endgroup$ – Parseval Feb 2 at 17:22
  • $\begingroup$ I assume he did the following: Remove the last term on the left hand sum, the first term on the right hand sum. Then you get two sums from $j =0$ to $n-2$. Combine those sums, and then add back in the terms you popped off separately. $\endgroup$ – rubikscube09 Feb 2 at 17:34
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    $\begingroup$ @rubikscube09- what do you mean by "remove the last term"? I can't just remove it because it will change the value of the expression. EDIT: Ok I think I understand what you mean now. $\endgroup$ – Parseval Feb 2 at 17:42

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