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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.

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  • $\begingroup$ Probably not. Do you have explicit forms for $\alpha,\beta$? $\endgroup$ – Alex R. Nov 28 '16 at 18:37
  • $\begingroup$ Unfortunately I don't. I just have means and upper bounds on their variances. Any thoughts on an analytic lower bound on the expectation? It's always between 0 and 1. I thought of doing something with Chebyshev's inequality but wondered whether there was a better way. $\endgroup$ – Jeff Nov 28 '16 at 20:31
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    $\begingroup$ Do you know the joint distribution of $\alpha$ and $\beta$? Eg. Multiivariate normal? $\endgroup$ – Matthew Gunn Nov 29 '16 at 6:08
  • $\begingroup$ No, I can 't assume they're multivariate normal. I just have that they are independent. I expect them each to be roughly normal, but I can't rely on that. I need to a true lower bound. Thank for asking! $\endgroup$ – Jeff Nov 30 '16 at 5:14
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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?

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    $\begingroup$ I don't believe this lower bound of $0.28$. As a counterexample, let $\alpha$ take on the value $(1+p)/(1-p)$ with probability $1-p$ and $-1$ with probability $p$, so its mean is $1$. Let $\beta$ be essentially zero (compared to $|\alpha|$). Then $\alpha/\sqrt{\alpha^2+\beta^2}$ takes on the value $-1$ with probability $p$ and $1$ with probability $1-p$, making its expectation $1-2p$. Choosing $p\approx 1$ shows the expectation is bounded only by $-1$, and this is the best possible lower bound. $\endgroup$ – whuber Nov 30 '16 at 6:41
  • $\begingroup$ @whuber -- As $p$ goes to 1, doesn't the variance of $\alpha$ in your counterexample go to infinity? But in the question, the variance of both $\alpha$ and $\beta$ are bounded by $0.3$. Sorry for not writing that more clearly in the question. $\endgroup$ – Jeff Dec 1 '16 at 8:56
  • $\begingroup$ I did notice a defect in my answer: I assumed $\alpha / \sqrt{\alpha ^2 + \beta^2} \ge 0$ but that's wrong. Rather $\alpha / \sqrt{\alpha ^2 + \beta^2} \ge -1$, as you note. I wonder if the answer can be patched up. $\endgroup$ – Jeff Dec 1 '16 at 8:57
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    $\begingroup$ You can achieve an infimum of $-\sigma^2(2+\sigma^2)$ when the variance of $\alpha$ is $\sigma^2$. Do this by making $\beta$ identically zero and letting $\alpha$ take on two values: one is infinitesimal but negative, with probability $(1+\sigma^2)/(2+\sigma^2)$; the other value is $2+\sigma^2$. $\endgroup$ – whuber Dec 1 '16 at 17:18
  • $\begingroup$ I think that solves my problem. Thabks much. Would you post it as an answer so I can accept it? $\endgroup$ – Jeff Dec 13 '16 at 14:57

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