Let i.i.d. $t_i\sim N(0,1), i = 1,2,\dots,n$, and
\begin{equation} X = \frac{\sum_i t_i}{\sum_i t_i^2}. \end{equation}
How to calculate the first two moments of $X$, i.e., $\mathrm{E}(X)$ and $\mathrm{E}(X^2)$?
I did some simulation studies and am almost sure that $\mathrm{E}(X) = 0$ and $\mathrm{E}(X^2) = \frac{1}{n-2}$. However, I failed to get the derivation in detail.
Thanks in advance.
I'll assume that you forgot to state that the $t_i$ are independent.
The expectation of any function that's antisymmetric in one of the $t_i$ is zero. Thus, $E(X)=0$ without further calculation.
For $E\left(X^2\right)$, the mixed pairs in the sum have zero expectation because they're odd in both factors, so what remains is
$$ E\left(X^2\right)=\frac{\sum_it_i^2}{\left(\sum_it_i^2\right)^2}=\frac1{\sum_it_i^2}=r^{-2}\;, $$
where $r$ is the radial coordinate in $n$-dimensional spherical coordinates for the $t_i$. The density for $r$ is proportional to $r^{n-1}\mathrm e^{-\frac12r^2}$, so the expectation of $r^{-2}$ is
$$ E\left(X^2\right)=\frac{\int_0^\infty r^{n-3}\mathrm e^{-\frac12r^2}\mathrm dr}{\int_0^\infty r^{n-1}\mathrm e^{-\frac12r^2}\mathrm dr}=\frac{2^{\frac n2-2}\Gamma\left(\frac n2-1\right)}{2^{\frac n2-1}\Gamma\left(\frac n2\right)}=\frac1{n-2}\;. $$