Let $y_i$, $i=1,\dots,n$, be independent Gaussian rv's of mean $\theta_i$ and variance $\sigma^2$ and let $\mathbf{y}:=[y_1,\dots,y_n]^\top$. Consider the function $f\colon \mathbb{R}^n\to \mathbb{R}$, $$ f(\mathbf{y})\mapsto \frac{1}{\|\mathbf{y}\|^2}-2\frac{y_1^2}{\|\mathbf{y}\|^4}. $$
I'm looking for a (possibly simple) argument which shows $$ \mathbb{E}\left[\left|f(\mathbf{y})\right|\right]<\infty,\quad \text{for } n\geq 3, $$ where $\mathbb{E}[\cdot]$ denotes expectation. Any suggestions?
Thanks in advance.
Not yet a solution:
For $f(\mathbf{y})=a(\mathbf{y})-b(\mathbf{y})$ one can take another random variable $f^{'}(\mathbf{y})=a(\mathbf{y})+b(\mathbf{y})$ Since expectation is a monotone operation $E[|f(\mathbf{y})|]\leq E[f^{'}(\mathbf{y})]$, as $|f|\leq f^{'}$. Now it suffices to show that $E[f^{'}(\mathbf{y})]$ has a finite mean. As expectation is linear we have $E[f^{'}(\mathbf{y})]=E[a(\mathbf{y})]+E[b(\mathbf{y})]$.The rest is to find that the functions $a$ and $b$ seperately doesnt drive the mean of $\mathbf{y}$ to infinity.
Sum of squares of any random variable with finite mean has a finite mean. $||\cdot||^4$ is also similary. For the inverse function we have $E[1/\mathbf{y}]\geq 1/E[\mathbf{y}]$. For finite expectations this doesnt guarrantee that if $E[\mathbf{y}]$ is finite $E[1/\mathbf{y}]$ too.