expected value of non-negative random variables

Question

Let $P(Y \ge y) = {1 \over y \log(y)}$ and $y \ge e$, then $F(y) = 1- {1 \over y \log(y)}$ and $F' = {\log(y) + 1 \over (y \log (y))^2}$. Assuming that $Y$ is non-negative, you can get the expected value by : $$EY = \int_e^{\infty} P(Y \ge y) dy \quad \text{or} \quad EY = \int_e^\infty y dF = \int_e^\infty y F'(y) dy$$ The first integral will get me $\int_e^\infty {1 \over y \log(y)} dy$ but the second integral gives me $\int_e^\infty {1 \over y \log(y)}dy + \int_e^\infty {1 \over y(\log(y))^2}dy$.

What am I missing here? Why am I getting different results?

Answer 1

The random variable is a mixed rv, not absolutely continuous. Thus its density (mixed density, discrete in $y=e$ and continuous in $y>e$) is the following

$$f_Y(y) = \begin{cases} \frac{e-1}{e}, & \text{if $y=e$ } \\ \frac{logy+1}{y^2log^2y}, & \text{if $y>e$}\\ 0, & \text{elsewhere} \end{cases}$$

You can realize it observing that $F_Y(e)=\frac{e-1}{e}$ and $F_Y(e^-)=0$

The CDF has a jump in $y=e$ thus in this point the rv is discrete and the mass probability function must be calculated as follows

$$f_Y(e)=F_Y(e)-F_Y(e^-)$$

Answer 2

The random variable does not have density. $P(Y=e)=1-\frac 1 e$. So when you compute $EY$ you have to add $eP(Y=e)=e-1$ to $\int_{(e,\infty)} yF'(y)dy$.

Note that $\int_e^{\infty} \frac 1 {y \ln y} dy=\infty$. So $EY=\infty$.