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Suppose we have two independent random variables $X$ and $Y$, with expected values and standard deviations of $(\mu_X,\sigma_X)$ and $(\mu_Y,\sigma_Y)$, respectively. Can we say anything about the expected value and standard deviation of $|X-Y|$?

If it had been $X-Y$, the answer would be $(\mu_X-\mu_Y, \sqrt{\sigma_X^2+\sigma_Y^2})$. However, I think such a straightforward approach is not possible because $P(X>Y)$ is not independent from $P(X,Y)$.

sodiumnitrate
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    I would use the notation $\mu_X, \mu_Y, \sigma_X, \sigma_Y$, consistently denoting each random variable in the same way. The in expressions like $\Pr(X\le x)$ the lower-case $x$ would not be a random variable; rather $\Pr(X\le x)$ would be a function of the variable $x$. ${}\qquad{}$ – Michael Hardy Aug 05 '15 at 18:17
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    If we do not know anything else on the distribution of $X$ and $Y$ we cannot hope in something more than an inequality. Cantelli's inequality should be the way to go: if $X$ is concentrated around $\mu_x$ and $Y$ is concentrated around $\mu_y$, it is reasonable to expect that $|X-Y|$ is concentrated around $|\mu_x-\mu_y|$. – Jack D'Aurizio Aug 05 '15 at 18:25

1 Answers1

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Suppose we know the mean $\mu$ of $|X-Y|$. This is enough to determine the variance, since it is $E(|X-Y|^2)-\mu^2$. For note that $$E(|X-Y|^2)=E(X^2)-2E(XY)+E(Y^2),$$ and each of the terms above can be calculated from the mean and variance of $X$ and $Y$.

Unfortunately, $\mu$ is not determined by the individual means and variances of $X$ and $Y$. For example, let $X$ and $Y$ take on values $-1$ and $1$ each with probability $1/2$. Then $X$ and $Y$ have mean $0$ and variance $1$, and $E(|X-Y|)=1$. But if we use standard normals for $X$ and $Y$, the mean of $|X-Y|$ is different (I believe it is $\frac{2}{\sqrt{\pi}}$).

This does not fully answer the question, since it only says we cannot determine $\mu$ from the given information. There will undoubtedly be useful inequalities.

André Nicolas
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