How does one find $E\Bigl(\frac{(\sum_{i=1}^nX_i)^2}{\sum_{i=1}^n X_i^2}\Bigr)$ Suppose $X_1,\ldots,X_n$ are a random sample of $N(0,\sigma^2)$. How can find 
$$
E\left(\frac{\bigl(\sum_{i=1}^nX_i\bigr)^2}{\sum_{i=1}^n X_i^2}\right)
$$
 A: Assuming $X_i$ are independent identically distributed normals with zero mean.
Using $\frac{1}{\lambda} = \int_0^\infty \mathrm{e}^{-\lambda t} \mathrm{d} t$:
$$\begin{eqnarray}
   \mathbb{E}\left( \frac{\left(\sum_{i=1}^n X_i\right)^2}{\sum_{i=1}^n X_i^2} \right) &=& \int_0^\infty \mathbb{E}\left( \sum_{i=1}^n \sum_{j=1}^n X_i X_j \mathrm{e}^{-t \sum_{k=1}^n X_k^2} \right) \mathrm{d} t \\ &\stackrel{\text{symmetry}}{=}&  \int_0^\infty \sum_{i=1}^n \mathbb{E}\left( X_i^2 \mathrm{e}^{-t \sum_{k=1}^n X_k^2}\right) \mathrm{d} t \\ &\stackrel{\text{indep.}}{=}& \int_0^\infty \sum_{i=1}^n \mathbb{E}\left(X_i^2 \mathrm{e}^{-t X_i^2} \right) \prod_{k=1,k\not= i}^n \mathbb{E}\left(\mathrm{e}^{-t X_k^2} \right) 
\mathrm{d} t \\ &=& \int_0^\infty \left(-\frac{\mathrm{d}}{\mathrm{d}t} \left(\mathbb{E}\left(\mathrm{e}^{-t X^2} \right)\right)^n  \right)\mathrm{d} t \\&=&
   \left.-\left(\mathbb{E}\left(\mathrm{e}^{-t X^2} \right)\right)^n\right|_{t \to 0^+}^{t \to \infty} = 1
\end{eqnarray}
$$
A: Expand the numerator. The expectation value for the mixed terms vanishes by symmetry. The sum of the remaining terms is the denominator, so the remaining fraction is $1$, and so is its expectation value.
