Does this hold in every case, and if only this one, why? Expectation, mean of random variable.

Question

Characteristic function of random variable $X$ let us denote as $f_X(t)$ and $EX$ it's mean or expectation. Does the following hold in all cases, because it keeps coming up and I don't know why it is correct:$$f_{X_k}(t)=E(e^{itX_k})=\color{#f00}{e^{itEX_k}} $$

This is driving me nuts: It's given that the sequence of random variables $X_n$ are independent and of equal distribution with mean $\color{#f00}{EX_n=a}$. The it says $$f_{\frac{1}{n}\sum_{k=1}^{n}X_k}(t)=\prod_{k=1}^{n} f_{X_k}(\frac{t}{n})=\color{#f00}{(1+ai\frac{t}{n}+o(\frac{t}{n}))^n}$$.

But looking at this I can only see that $f_{X_k}(\frac{t}{n})=E(e^{i\frac{t}{n}X_k})\color{#f00}{=e^{ai\frac{t}{n}}}?$ How in the world?

Answer 1

It is not true in general that $\mathbb{E}[e^{itX}]=e^{it\mathbb{E}[X]}$.

For instance, suppose that $X=1$ with probability $\frac{1}{2}$ and $X=-1$ with probability $\frac{1}{2}$. Then $\mathbb{E}[X]=0$, hence $e^{it\mathbb{E}[X]}=1$. On the other hand, $$ \mathbb{E}[e^{itX}]=\frac{1}{2}\Big(e^{it}+e^{-it}\Big)=\cos(t) $$

Answer 2

No, it does not. Consider for example a Gaussian variable $X\sim N(0, 1)$

$f_X(x) = e^{-1/2\ x^2} \ne e^{it EX} = e^{0} = 1$

BS