Q:

Distribution of Integral involving wiener process

Given $W(t)$ as a standard Wiener process, i.e. $W(t) \sim \mathcal{N}(0,t)$.
Prove the following statement:
$$\int_{0}^{1}tW(t)dt \sim \mathcal{N}(0,\frac{2}{15})$$

A:

The following proof is very unelegant/non-clever/direct/bruteforce, but it works nonetheless:
Define $X:= \int_0^1 tW_t dt$. For each $n \in \mathbb{N}$, define $$X_n=\frac{1}{n} \sum_{k=1}^n \frac{k}{n}\; W_{\frac{k}{n}}$$ Notice that $X_n$ is just the Riemann sum for $\int_0^1 tW_t dt$ with respect to the partition $(\frac{0}{n},\frac{1}{n}, \frac{2}{n},...,\frac{n}{n})$. Since $(W_t)$ is a.s. continuous, we know that $X_n \to X$ a.s.
Now, there is a theorem which states that if $X_n$ is a collection of normally distributed random variables with mean $\mu_n$ and variance $\sigma_n^2$, and if $\mu_n \to \mu$ and $\sigma_n^2 \to \sigma^2$, and $X_n \to X$ a.s, then $X$ is normally distributed with mean $\mu$ and variance $\sigma^2$.
Now, it is clear that the $X_n$ are normally distributed with mean $0$, so all we need to do is to prove that their variances converge to $\frac{2}{15}$. For this, notice that 
$var(X_n) = var \bigg(\frac{1}{n} \sum_{k=1}^n \frac{k}{n}\; W_{\frac{k}{n}}\bigg)$
$=\frac{1}{n^4}var\bigg(\sum_{k=1}^n k\; W_{\frac{k}{n}}\bigg)$
$=\frac{1}{n^4} \bigg(\sum_{k=1}^nvar\big(k\;W_{\frac{k}{n}}\big)+2\sum_{ i<j} cov\big(i\;W_{\frac{i}{n}},j\;W_{\frac{j}{n}}\big)\bigg)$
$=\frac{1}{n^4} \bigg(\sum_{k=1}^n\frac{k^3}{n}+2\sum_{ i<j} \frac{i^2j}{n}\bigg)$
$=\frac{1}{n^5} \bigg(\sum_{k=1}^nk^3+2\sum_{ j=2}^n \sum_{i=1}^{j-1} i^2j\bigg)$
$=\frac{1}{n^5} \bigg(\frac{1}{4}n^2(n+1)^2+\frac{1}{60}n(n^2-1)(8n^2+5n-2)\bigg)$
Now, as $n$ gets large, this last expression approaches $\frac{1}{60}\cdot 8=\frac{2}{15}$, which completes the proof.
(Note: Another, less direct way to compute those complicated summations is to associate them to Riemann sums of functions we know about.)

