I want to compute the Wiener Integral $\int_{0}^{t}W_sdW_s$, where $W_s$ is the Brownian Motion. I am given with the following answer $$\int_{0}^{t}W_sdW_s = -\frac{1}{2}t + \frac{1}{2}W_t^2.$$ However from my computation I got that the integral is $0.$
This is what I've done.
Since the mean and the variance of $W_s$ are continuous, we use the formula $$\int_{0}^{t}W_sdW_s = \lim_{n\rightarrow\infty}\sum_{i=1}^{n}W_{t_{i-1}}(W_{t_{i}} - W_{t_{i-1}}).$$ We have that $$2\sum_{i=1}^{n}W_{t_{i-1}}(W_{t_{i}} - W_{t_{i-1}}) = -\sum_{i=1}^{n}(W_{t_{i}} - W_{t_{i-1}})^2 + \sum_{i=1}^{n}(W_{t_{i}}^2 - W_{t_{i-1}}^2). $$ Taking the partition $t_i = \frac{ti}{n}$, I compute that $$\sum_{i=1}^{n}(W_{t_{i}}- W_{t_{i-1}})^2 = \sum_{i=1}^{n}\left(N\left(0,\frac{t}{n}\right)\right)^2 = \frac{t}{n}\sum_{i=1}^{n}\chi^2(1)=t\chi^2(1)$$ For the same partition $t_i = \frac{ti}{n}$, I compute that $$\sum_{i=1}^{n}(W_{t_{i}}^2 - W_{t_{i-1}}^2) = \sum_{i=1}^{n}\left(\left(N\left(0,\frac{ti}{n}\right)\right)^2 - \left(N\left(0,\frac{t(i-1)}{n}\right)\right)^2\right)=\\ \sum_{i=1}^{n}\left(\frac{ti}{n}\chi^2(1) - \frac{t(i-1)}{n}\chi^2(1)\right)=\frac{t}{n}\sum_{i=1}^{n}\chi^2(1)=t\chi^2(1).$$ Thus $\int_{0}^{t}W_sdW_s=0.$
Where is the mistake in my computations?
The problem with your manipulations is that you are treating probability distributions as objects that can be added and subtracted, which doesn't really make sense. For instance, you cannot conclude that the difference of two random variables with the same distribution is identically 0 without knowing something about their joint distribution.
Instead, I recommend working with the two sums in your first line directly. Note that the sum $$ \sum_{i=1}^n(W_{t_i}^2-W_{t_{i-1}}^2)$$ telescopes, while the sum $$ \sum_{i=1}^n(W_{t_i}-W_{t_{i-1}})^2$$ will converge to the quadratic variation of $W$ on $[0,t]$ as the mesh of the partition tends to $0$. Putting these two pieces together should allow you to compute the integral.