Finding mean of Reciprocal of Uniform Distribution

Question

Given that Y is uniformly distributed on the interval (1,2), I am required to find $E(\frac{1}{Y})$
I am trying to solve this by finding the pdf for a random variable $Z = \frac{1}{Y}$ first, but I'm stuck. Is this the right way or are there any better way?

Also, I'm wondering if $E(\frac{1}{Y}) = \frac{1}{E(Y)}$?
Can anyone help me with this? Thank you!

Answer 1

There is no need to find the pdf of $Z$. $E\frac 1 Y=\int_1^{2} \frac 1 x dx=ln 2 -\ln 1 =\ln 2$.

It is not true that $E\frac 1 Y =\frac 1 {EY}$.

In fact if $Y$ is a positive random variable and $E\frac 1Y =\frac 1 {EY}$ then we can show that $Y$ must be a constant. The inequality $E\frac 1Y \geq \frac 1 {EY}$ is true for any positive random variable $Y$.

Answer 2

Recall that if $f$ is the pdf associated to $Y$ and $Z = g(Y)$, then $$\mathbb E[Z] = \int_{-\infty}^\infty g(y)f(y)dy.$$

In this case, $g(y) = 1/y$ and $f(y) = \boldsymbol{1}_{[1,2]}(y)$. Hence, you have $$\mathbb E[Z] = \int_1^2\frac{dy}{y} = \log 2.$$

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PS. Exchanging the order of $\mathbb E$ and $g$ is usually only valid if $g$ is a linear function. Moreover, if $g$ is convex, you have Jensen's inequality.