Characteristic function of Bernoulli distribution

Question

How do I show that if $X$ is a Bernoulli-r.v. with parameter $p$, then $$ \varphi_X(u)=E[e^{iuX}]=e^{iu0}(1-p)+e^{iu}p $$ Without using the characteristic function of a binomial-r.v. with parameters n,p.

Answer 1

$Ee^{iuX }=Ee^{iuX }I_{\{X=0\}}+Ee^{iuX }I_{\{X=1\}}=Ee^{0 }I_{\{X=0\}}+Ee^{iu }I_{\{X=1\}}=P(X=0)+e^{iu}P(X=1)=(1-p)+e^{iu}p$.

Answer 2

We can evaluate the expected value $E[e^{iuX}]$ using the LOTUS. We have that $$ E[e^{iuX}]=(1-p)e^{iu0}+pe^{iu}=(1-p)+pe^{iu}. $$