Let $Z$ be a random variable of the standarized normal distribution. True or false? $E(\sin Z)=\sin (E(Z))$ ($E$ stands for the mean value).
Comments. Since $\displaystyle \sin Z=\sum_{n=0}^{\infty}(-1)^n\frac{Z^{2n+1}}{(2n+1)!}$, we would just like to interchange $E$ (i.e. the integral) with the series - unfortunatelly though we are not dealing with positive rv's, in order to work with the standard Beppo Levi theorem. I am aware of the alternative version for rv's with values on $[-\infty,+\infty]$, according to which I have to check that the series $\displaystyle \sum_{n=1}^{\infty}E\bigg(\frac{|Z|^{2n+1}}{(2n+1)!}\bigg)$ converges. But I am stuck at the estimations of $E(|Z|^{2n+1})$ - at this point I am thinking of evaluating the integrals $\displaystyle \int\limits_{-\infty}^{+\infty}|z|^{2n+1}f_Z(z)dz=2\int\limits_{0}^{+\infty}z^{2n+1}f_Z(z)dz$, which seems to be a bit troubleshooting though. Am I on the right path or not?
Thanks in advance!
You're thinking much too hard.
$E[Z]=0$ so $\sin E[Z]=0$. On the other hand, $Z$ has a symmetric distribution, so for any measurable function $f$, if $E[f(Z)]$ exists then so does $E[f(-Z)]$ and they are equal. $f(x) = \sin(x)$ is bounded so $E[\sin(Z)]$ certainly exists. Thus $E[\sin(Z)] = E[\sin(-Z)] = E[-\sin(Z)] = -E[\sin(Z)]$ because $\sin(x)$ is an odd function and $E$ is linear. So $E[\sin(Z)]=0$.
Even more simply, we have $$E[\sin(Z)] = \int_{-\infty}^{\infty} \sin(x) \frac{1}{\sqrt{2\pi}} e^{-x^2/2}\,dx.$$ The integrand is bounded in absolute value by $\frac{1}{\sqrt{2\pi}} e^{-x^2/2}\,dx$, whose integral converges, so this integral converges too. And the integrand is odd, so the integral equals 0.