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Consider a ratio distribution Z=X/Y. If 0 is in the domain of Y it seems intuitive that the mean of Z is undefined because Z itself will be undefined at many points.

However there are many results where the denominator distribution contains 0 and the resulting ratio has finite mean. I can't figure out why these ratio distributions would have finite means. Any help would be much appreciated.

GaussGaussGauss
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1 Answers1

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The denominator distribution may contains 0, but the appearance of 0 is rare enough that it may not pose problem.

  • for continuous random variables : the probability of drawing exactly 0 is 0. So the probability of having a diverging ratio is zero. Having finite means is not that surprising.

Note that having finite means usually have to do with bahevior around diverging point and at infinity, as it is an integration problem :

$$ \mu_Z = E[Z] = \int_{-\infty}^\infty z f_Z(z)dz$$

So the question is more about the behavior of $f_Z$ around 0, and at +/-$\infty$. You may encounter both case where $zf_Z(z)$ is integrable around 0 or not.

  • for discrete random variables : the result may empirically hold because the value 0 wasn't drawn. It may not hold for bigger (or unluckier) draws.
lcrmorin
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