Expected value of Ito integral

Question

I am trying to understand why $$E\bigg[\exp\big(a^\frac{1}{2}\int f(t)dW(t)\big)\bigg]=\exp\left(\frac{1}{2}a\int f(t)^2dt\right),$$ but I'm having trouble with Ito integration in general. I was able to show that if $Z(t)=\exp(a\cdot W(t))$, then $E[Z(t)]=\exp\left(\frac{1}{2}a^2t\right),$ but in my current problem, I have the integral $\int f(t)dW(t)$ rather than just $W(t)$, so I'm starting to doubt whether this is the right approach.

Answer 1

Hint:

1. The random variable $X := \int f(t) \, dW_t$ is a Gaussian random variable. Determine by its mean $\mu$ and variance $\sigma^2$.
2. Recall that for any Gaussian random variable $X$ with mean $\mu$ and variance $\sigma^2$ it holds that $$\mathbb{E}(e^{\lambda X} )= \exp \left(-\lambda \mu+ \frac{1}{2} \lambda^2 \sigma^2 \right)$$ for all $\lambda \in \mathbb{R}$.
3. Conclude.

Remark: I take it that $f$ is deterministic, i.e. $f=f(t)$ does not depend on $\omega$. Otherwise the claim does obviously hold not true since the right-hand side is a (non-trivial) random variable and the left-hand side a constant.