I am looking to solve this exercise.
Let $X_1,X_2,\ldots,X_n$ be i.i.d random variables with pdf $$\frac{\sqrt{\theta}}{\sqrt{2\pi}}e^{-\theta x^2/2}$$
Determine the mean of $\frac{1}{\sum_{i=1}^{n}X_i^2}$.
Question: Can I say that since $X_1,X_2,\ldots,X_n$ are independent and identically distributed with a pdf of $\frac{\sqrt{\theta}}{\sqrt{2\pi}}e^{-\theta x^2/2}$, we can write the denominator of the expression inside the expectation as follows:
\begin{align*} \sum_{i=1}^{n}X_i^2 &= X_1^2 + X_2^2 + \ldots + X_n^2 \\ &= \frac{1}{\theta}\left(\chi_{1}^2 + \chi_{2}^2 + \ldots + \chi_{n}^2\right)\\ &= \frac{1}{\theta}\chi_{n}^2. \end{align*}
Where $\chi_{i}^2$ is a chi-squared random variable with one degree of freedom.
Therefore, we have:
\begin{align*} \mathbb{E}[\frac{1}{\sum_{i=1}^{n}X_i^2}] &= \frac{1}{\theta} E\left[\frac{1}{\chi_{n}^2}\right]\\ &= \frac{1}{\theta} \int_{0}^{\infty} \frac{1}{x^2}\frac{1}{2^{1/2}\Gamma(1/2)}e^{-x^2/2} dx\\ &= \frac{1}{\theta} \int_{0}^{\infty} \frac{1}{x^2}\frac{1}{\sqrt{2\pi}}e^{-x^2/2} dx\\ &= \frac{1}{\theta} \cdot 1\\ &= \frac{1}{\theta} \end{align*}
Therefore, $ \mathbb{E}\left[\frac{1}{\sum_{i=1}^{n}X_i^2}\right] = \frac{1}{\theta}$.