Expectation of ratio of 2 independent random variables ?

[nikozm]

Hi,

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.

Thanks

 

[Office_Shredder]

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,
$$E[1/y] = \int_{0}^{1} \frac{1}{y} dy = \infty.$$
Even if you restrict yourself away from zero to avoid stupid division problems, if y is a uniform random variable between 1 and 2,
$$E[1/y] = \int_{1}^{2} \frac{1}{y} dy = \ln(2) \neq 1.5$$

 

[economicsnerd]

More can be said. Given $y$ is $>0$-valued with finite expectation, you will always have $\mathbb E\left[\dfrac1y\right] > \dfrac1{\mathbb E[y]}$, except in the extreme case that $y$ exhibits no randomness. This follows from Jensen's inequality, since the function $t\mapsto \dfrac1t$ is strictly convex on $(0,\infty)$.