# Expected value of ratio of correlated random variables?

For independent random variables $\alpha$ and $\beta$, is there a closed form expression for

$\mathbb E \left[ \frac{\alpha}{\sqrt{\alpha^2 + \beta^2}} \right]$

in terms of the expected values and variances of $\alpha$ and $\beta$? If not, is there a good lower bound on that expectation?

Update: I may as well mention that $\mathbb E[\alpha] = 1$ and $\mathbb E[\beta] = 0$. I can control the variance on $\alpha$ and $\beta$, and I have in mind a setting where the variances of both $\alpha$ and $\beta$ are pretty small relative to $\mathbb E[\alpha]$. Maybe both of their standard deviations are less than 0.3.

I thought of one lower bound, though I don’t think it’s very tight. I just pick an arbitrary value less than the mean of $\alpha$ and another arbitrary value around the mean of $\beta^2$. Since the expectation is of a non-negative random variable, and because $\alpha$ and $\beta$ are independent,

$\mathbb E \left[ \frac{\alpha}{\sqrt{\alpha^2 + \beta^2}} \right] \ge \frac{1}{\sqrt{2}}\mathbb P(\alpha \ge \frac{1}{2}) \mathbb P(\beta^2 \le\frac{1}{4})$.

By Chebyshev’s inequality,

$\mathbb P(\alpha \ge \frac{1}{2}) = \mathbb P(\alpha - 1 \ge -\frac{1}{2}) \ge \mathbb P(|\alpha - 1| \le \frac{1}{2}) = 1 - \mathbb P(|\alpha - 1| \ge \frac{1}{2}) \ge 1 - 4\mathrm{var}(\alpha)$

By Markov’s inequality,

$\mathbb P(\beta^2 \le\frac{1}{4}) = 1 - \mathbb P(\beta^2 \ge\frac{1}{4}) \ge 1 - 4 \mathbb E[\beta^2] = 1 - 4\mathrm{var}(\beta)$

Therefore,

$\mathbb E \left[ \frac{\alpha}{\sqrt{\alpha^2 + \beta^2}} \right] \ge \frac{1}{\sqrt{2}} (1 - 4 * 0.3^2) (1 - 4 * 0.3^2) > 0.28$

Is a more standard/systematic way to do what I’m doing here, that gets a tighter bound?