Wiener process identities

Question

Let $W(t)$ be the Wiener process. Prove the following identities by taking limits of the Forward Euler discretization\begin{gather*} \int_0^Tt\,dW(t)=TW(T)-\int_0^T W(t)\,dt\\ \int_0^T W(t)\,dW(t)=\frac{W(T)^2}{2}-\frac{T}{2}. \end{gather*} My idea is to use Abel’s summation formula on the Forward Euler discretization \begin{equation}\label{1.1} \sum_{n=0}^{N-1}t_n(W(t_{n+1})-W(t_n)) \end{equation} and take the limit to get strong convergence to Itô's integral, but I can't get things to work. Any ideas?

Answer 1

In the meantime, I believe I solved it. Here's my solution.

Consider the first identity. Given a $N$-point time discretization we may use the Forward Euler method to write \begin{equation} \sum_{n=0}^{N-1}t_n(W(t_{n+1})-W(t_n)). \end{equation} Recall Abel's summation formula (summation by parts)\begin{align*} \sum_{k=m}^nf_k(g_{k+1}-g_k)=(f_{n+1}g_{n+1}-f_mg_m)-\sum_{k=m}^ng_{k+1}(f_{k+1}-f_k). \end{align*} Then, we get \begin{align*} \sum_{n=0}^{N-1}t_n(W(t_{n+1})-W(t_n))=(t_NW(t_N)-t_0W(t_0))-\sum_{n=0}^{N-1}W(t_{n+1})(t_{n+1}-t_n). \end{align*} Since $t_N=T$, $t_0=0$ and $W(0)=0$ we get\begin{align*} \sum_{n=0}^{N-1}t_n(W(t_{n+1})-W(t_n))=TW(T)-\sum_{n=0}^{N-1}W(t_{n+1})(t_{n+1}-t_n). \end{align*} By taking the limit $N\to\infty$, which naturally implies $t_{n+1}-t_n\to 0,\forall n$, we get that the left-hand side converges strongly to Ito's integral (this is shown by constructing a Cauchy sequence in a Hilbert space), while the summation on the right-hand side gives the Riemann integral of $W(t)$ in $L^2$. Indeed, using the Cauchy Schwartz inequality we obtain \begin{align*} E\left(\left|\sum_{n=0}^{N-1} W(t_{n+1})(t_{n+1}-t_n)-\int_0^TW(t)dt\right|^2\right) &=E\left(\left|\sum_{n=0}^{N-1} \int_{t_n}^{t_{n+1}}\!\!(W(t_{n+1})-W(t))dt\right|^2\right)\\ &\hspace{-25mm}\leq N \sum_{n=0}^{N-1} E\left(\left|\int_{t_n}^{t_{n+1}}\!\!(W(t_{n+1})-W(t))dt\right|^2\right)\\ &\hspace{-25mm}\leq N \sum_{n=0}^{N-1} (t_{n+1}-t_n) E\left(\int_{t_n}^{t_{n+1}}\!\!|W(t_{n+1})-W(t)|^2dt\right)\\ &\hspace{-25mm}= N \sum_{n=0}^{N-1}\frac{(t_{n+1}-t_n)^3}{2} = N \sum_{n=0}^{N-1}\frac{T^3}{2N^3} =\frac{T^3}{2N}\to 0 \end{align*} as $N\to\infty,$ and thus the following holds \begin{align*} \int_0^T W(t)dW(t)=TW(T)-\int_0^T W(t)dt. \end{align*} For the second identity, the binomial expansion yields\begin{align*} \sum_{n=0}^{N-1}W(t_n)(W(t_{n+1})-W(t_n))&=\frac12\left( W(t_{n+1})^2-W(t_{n})^2-(W(t_{n+1})-W(t_{n}))^2 \right)\\ &=\frac12 \sum_{n=0}^{N-1}(W(t_{n+1})^2-W(t_{n})^2)-\frac12\sum_{n=0}^{N-1}(W(t_{n+1})-W(t_{n}))^2. \end{align*} For the first summation, since it's a telescoping sum, we get\begin{align*} \sum_{n=0}^{N-1}W(t_{n+1})^2-W(t_{n})^2=W(t_N)^2-W(t_0)^2=W(T)^2. \end{align*} For the second summation we claim that, as $N\to\infty$,\begin{align*} \Sigma_W \equiv \sum_{n=0}^{N-1}(W(t_{n+1})-W(t_{n}))^2\to T \end{align*} in the $L^2$-norm sense, with $\|I\|_{L^2}\equiv \sqrt{E[I^2]}$. To prove this we need to show that\begin{align*} E\left[\left(\Sigma_W-T\right)^2\right]\to 0 \end{align*} as $N\to\infty$. Let us first find the expected value of the random variable $\Sigma_W$. By definition and since the terms of the summation are independent, we have\begin{align*} E\left[\sum_{n=0}^{N-1}(W(t_{n+1})-W(t_{n}))^2 \right]=\sum_{n=0}^{N-1}E[(W(t_{n+1})-W(t_{n}))^2]=\sum_{n=0}^{N-1} t_{n+1}-t_n=T. \end{align*} Hence, the expected value $E\left[\left(\Sigma_W-T\right)^2\right]$ is the variance of $\Sigma_W$. The variance of the sum of independent variables is the sum of their variances, i.e.,\begin{align*} \text{Var}\left[(\Sigma_W-T)^2 \right]=\sum_{n=0}^{N-1}\text{Var}\left[(W(t_{n+1})-W(t_{n}))^2\right]. \end{align*} Then, for each term we have\begin{align*} \text{Var}\left[(W(t_{n+1})-W(t_{n}))^2\right]&=E\left[(W(t_{n+1})-W(t_{n}))^4\right]-\left(E\left[(W(t_{n+1})-W(t_{n}))^2\right] \right)^2. \end{align*} A quick computation with integrals can show that if $X\sim N(0,b)$ then $E[X^4]=3b^2$. Hence,\begin{align*} \text{Var}\left[(W(t_{n+1})-W(t_{n}))^2\right]=3(t_{n+1}-t_n)^2-(t_{n+1}-t_n)^2=2(t_{n+1}-t_n)^2 \end{align*} and so\begin{align*} E\left[\left(\Sigma_W-T\right)^2\right]=\sum_{n=0}^{N-1}2(t_{n+1}-t_n)^2. \end{align*} Since$$(t_{n+1}-t_n)^2\leq \max_{0\leq n'\leq N-1}(t_{n'+1}-t_{n'})\cdot (t_{n+1}-t_{n}),$$we get\begin{align*} E\left[\left(\Sigma_W-T\right)^2\right]\leq 2\max_{0\leq n'\leq N-1}(t_{n'+1}-t_{n'})\sum_{n=0}^{N-1}(t_{n+1}-t_n)=2\max_{0\leq n'\leq N-1}(t_{n'+1}-t_{n'}) \cdot T\to 0 \end{align*} as $N\to\infty$. Hence, in the limit we get\begin{align*} \int_0^T W(t)dW(t)=\frac{W(T)^2}{2}-\frac{T}{2}. \end{align*}