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Expectation of ratio of 2 independent random variables ?

  1. Jan 27, 2014 #1

    i was wondering if the following is valid:

    E[x/y] = E[x] / E[y], given that {x,y} are non-negative and independent random variables and E[.] stands for the expectation operator.

  2. jcsd
  3. Jan 27, 2014 #2


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    No, this is not true. It is true that E[x/y] = E[x]*E[1/y] but it is not true that E[1/y] = 1/E[y]. For example if y is a uniform random variable taking values between 0 and 1,
    [tex] E[1/y] = \int_{0}^{1} \frac{1}{y} dy = \infty. [/tex]
    Even if you restrict yourself away from zero to avoid stupid division problems, if y is a uniform random variable between 1 and 2,
    [tex] E[1/y] = \int_{1}^{2} \frac{1}{y} dy = \ln(2) \neq 1.5[/tex]
  4. Jan 27, 2014 #3
    More can be said. Given [itex]y[/itex] is [itex]>0[/itex]-valued with finite expectation, you will always have [itex]\mathbb E\left[\dfrac1y\right] > \dfrac1{\mathbb E[y]}[/itex], except in the extreme case that [itex]y[/itex] exhibits no randomness. This follows from Jensen's inequality, since the function [itex]t\mapsto \dfrac1t[/itex] is strictly convex on [itex](0,\infty)[/itex].
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