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When we have a ratio of random variables, is their expectation/variance defined in the same way? That is, if we want to write out explicitly $E[\frac{X}{Y}]$ where X and Y are random variables, then

$$E[\frac{X}{Y}] = \int_\Omega \frac{X}{Y} p(x,y)dxdy$$

And similarly, is the variance of the ratio also defined the same way: $$V[\frac{X}{Y}] = \int_\Omega [\frac{X}{Y}-\mu_{\frac{x}{y}}]^2 p(x,y)dxdy$$

for some $\mu_{\frac{x}{y}}$?

This is a definitional question, I ask because I have not seen anyone write out the expectation of ratio of random variables explicitly.

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  • $\begingroup$ Seems like you might need to use an approximation? math.stackexchange.com/questions/40713/… and stat.cmu.edu/~hseltman/files/ratio.pdf $\endgroup$
    – information_interchange
    Commented Feb 13, 2020 at 2:48
  • $\begingroup$ 1. You should not have uppercase letters inside the integrals; in each case you should have $x/y$ where you have $X/Y$. 2. Since your posted questions ask about what something represents (and the title asks about notation) it's difficult to see how your comment is relevant to that issue. $\endgroup$
    – Glen_b
    Commented Feb 13, 2020 at 4:15
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    $\begingroup$ As for representing the expectation of the ratio, you have done so in your question (aside from the error I mentioned). However, evaluating it is a different question. Did you instead mean to ask about that? $\endgroup$
    – Glen_b
    Commented Feb 13, 2020 at 4:40
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    $\begingroup$ It's unclear what your expression means, because it mixes elements of the abstract theoretical integral (over $\Omega$) with the integral over the distribution. For a continuous bivariate distribution with density $p,$ for instance, the LOTUS tells you that $$E\left[\frac{X}{Y}\right]=\iint_{\mathbb{R}^2} \frac{x}{y}\,p(x,y)\,\mathrm{d}x\mathrm{d}y.$$ $\endgroup$
    – whuber ♦
    Commented Feb 14, 2020 at 21:06
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    $\begingroup$ Probability and statistics can be an extremely confusing subject. I have learned that when I'm unsure about the mathematical notation for something, there's probably a gap in my understanding so I go back to review it. That has been a helpful discipline. $\endgroup$
    – whuber ♦
    Commented Feb 14, 2020 at 21:43

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