calculate the characteristic function of a random variable X with E(X)=0 and variance = 1

Question

The question ask to calculate the chf of a random variable $X$ with $E(X)=0$ and variance $= 1$.

I get a hint to use the Taylor expansion of the exponential $e^{iux}=\sum_{j=0}^{\infty}\frac{(iuX)^j}{j!}$. And call anything of order $u^3$ higher $o(u^2)$.

Can someone tell me how to use this hint? Thanks

Answer 1

$$\begin{align}\Bbb{E}[e^{i t X}]&=\Bbb{E}\left[\sum_{j=0}^\infty \frac{(i t X)^j}{j!}\right]\\&=\sum_{j=0}^\infty \frac{(it)^j}{j!}\Bbb{E}\left[ X^j\right]\\&=\Bbb{E}[1]+i\Bbb{E}[X] - \frac{1}{2}\Bbb{E}[X^2] + \sum_{j=3}^\infty t^j \frac{i^j}{j!} \Bbb{E}\left[ X^j\right]\end{align}$$ Just using linearity of expectation. Can you finish it off yourself from there?