Mean and Variance of "white noise" integral I have to find the mean and variance of "white noise" - Brownian motion of
$$ X = \int^1_0 t dW(t)$$
So we start the mean:
$$ E[X] = E[\int^1_0 tdW(t)] $$
I'm thinking this way:
$$ E[X] = tW(1) - tW(0) - \int^1_0 1W(t)dt $$
Then I'm not sure where to go next as the mean of $W(t)$ is $0.$ 
 A: Sketch:
$$E[\int_0^1 t dW(t)] \sim E[\sum_{i=0}^{N - 1} \frac{i}{N} [W(\frac{i+1}{N}) - W(\frac{i}{N})]$$
$$E[\sum_{i=0}^{N - 1} \frac{i}{N} [W(\frac{i+1}{N}) - W(\frac{i}{N})] = \sum_{i=0}^{N - 1} \frac{i}{N}E[W(\frac{i+1}{N}) - W(\frac{i}{N})]$$
$$\sum_{i=0}^{N - 1} \frac{i}{N}E[W(\frac{i+1}{N}) - W(\frac{i}{N})] = \sum_{i=0}^{N - 1} \frac{i}{N}\cdot0 = 0$$
Assuming we can pass the expectation through the limit, we get:
$E[\int_0^1 t dW(t)] = 0$
For Variance, 
$$Var(\int_0^1 t dW(t)) \sim \sum_{i=0}^{N - 1} \frac{i^2}{N^2}Var[W(\frac{i+1}{N}) - W(\frac{i}{N})]$$
By independence.
$$\sum_{i=0}^{N - 1} \frac{i^2}{N^2}Var[W(\frac{i+1}{N}) - W(\frac{i}{N})] = \sum_{i=0}^{N - 1} \frac{i^2}{N^2}\frac{1}{N}$$
The above looks a lot like the expression: 
$$\int_0^1 t^2 dt$$
Which is equivalent in the limit.
Therefore: $$Var(\int_0^1 t dW(t)) = \int_0^1 t^2 dt = \frac{1}{3}$$
This result is based Ito Isometry FYI, which is the fundamental theorem of stochastic calculus.
$$E[(\int_0^T f(t)dW(t))^2] = \int_0^T f(t)^2dt$$ 
