Why is the integral $\int_0^1t\,dW_t$ a normal random variable? Consider the random variable $X=\int_0^1t\,dW_t$, where $W_t$ is a Wiener process. The expectation and variance of $X$ are
$$E[X]=E\left[\int_0^1t\,dW_t\right]=0,$$
and $$
Var[X]=E\left[\left(\int_0^1t\,dW_t\right)^2\right]=\int_0^1t^2\,dt=\left.\frac{1}{3}t^3\right|_0^1=\frac{1}{3}.$$
A typical textbook, like Klebaner's one, concludes that the random variable $X$ is a normal random variable with mean $0$ and standard variation $1/\sqrt{3}$. However, two moments cannot lead to the conclusion that $X$ is a normal random variable.
So, here is my question: How to prove that $X$ is normal? Is is necessary to calculate all moments of $X$ and to compare them with those of a normal distribution? Or is there a more elegant way?
 A: Another approach (which I outline) is via Ito's formula. Define $I_t=\int_0^t s\,dW_s$ for $t\ge 0$, and compute the characteristic function of $I_t$. By Ito:
$$
\exp(i\lambda I_t)=1+i\lambda\int_0^t \exp(i\lambda I_s)s\,dW_s-{\lambda^2\over 2}\int_0^t \exp(i\lambda I_s)s^2\,ds.\tag1
$$
Write $g(s):=\Bbb E[\exp(i\lambda I_s)]$. Take expectations in (1):
$$
g(t)=1-{\lambda^2\over 2}\int_0^t g(s)s^2\,ds,\qquad t\ge 0. 
$$
Now differentiate with respect to $t$ to see that $g$ solves the initial value problem
$$
g(0)=1,\qquad g'(t)=-{\lambda^2t^2\over 2}g(t),\quad t\ge 0.
$$
Therefore $g(t) = \exp(-t^3\lambda^2/6)$ for $t\ge 0$. In particular, for $t=1$, the random variable $I_1$ has characteristic function
$\Bbb E[\exp(i\lambda\int_0^1 s\,dW_s)]=\exp(-\lambda^2/6)$, which means that $\int_0^1 s\,dW_s\sim\mathscr N(0,1/3)$.
A: You can use the definition of the stochastic integral as a limit of integrals of step functions. The integral of a step function is a linear combination of independent normal variables.
