the Taylor-expansion of $e^{itx}$ is
$$1+itx+(itx)^2 / 2! + \cdots.$$
My question:
Why can one write $1+itx+o(t)$ for the sum I sated above?
$o(t)$ would mean that $(itx)^2 / 2! + \cdots$ would be growing linear in $t$, but I do not see why this is so.
My second question: If the $x$ in $e^{itx}$ is a real random variable, why can we conclude that the expectation of $e^{itx}$ is $1 + it E(x) + o(t)$, or more specific (it is clear that we use the linearity of the expectation):
Why is $E[o(t)] = o(t)$?
Thank you
Here is a useful estimate $$ \Big|e^{ix}-\sum_{k=0}^n{(ix)^k\over k!}\Big|\le\min(|x|^{n+1}/(n+1)!,2|x|^n/n!). $$ (For a proof see section 9.3 of Resnick's A Probability Path.) In particular (take $n=1$, and replace $x$ by $tX$) $$ \Big|e^{itX}-1-itX\Big|\le\min(t^2|X|^{2}/2,2|t|\cdot|X|). $$ From this, upon taking expectations, it follows that $\Bbb E[e^{itX}] = 1+it\Bbb E[X]+O(t)$ (when $\Bbb E|X|<\infty$) and that the $O(t)$ term is $O(t^2)$ (which is of course smaller than $o(t)$) when $\Bbb E[X^2]<\infty$.