Is the expected value of this density function $Z$ finite?

Question

Is the expected value of this density function $Z$ finite?

$$f_Z(z) = \begin{cases} {\dfrac{\ln(z)}{z^2}}, & \text{for $z \ge 1$} \\[2ex] 0, & \text{for $z \lt 1$} \end{cases}$$

I know expected values can be infinite and when I graph this it looks like it has a horizontal asymptote at $y=0$, so it looks like the expected value can be infinite, but I am not sure? Can someone explain theoretically whether the expected value of this density is finite or infinite?

Answer 1

\begin{align} E(Z)&=\int_{\mathbb R} z f_Z(z) dz\\ &=\int_{-\infty}^\infty z \left[0 \times 1_{z < 1} + \frac{\ln z}{z^2} \times 1_{z \ge 1} dz\right]\\ &=\int_{-\infty}^\infty z [0 \times 1_{z < 1}] dz + \int_{-\infty}^\infty \left[\frac{\ln z}{z^2} \times 1_{z \ge 1} dz\right]\\ &=\int_{-\infty}^\infty \left[\frac{\ln z}{z^2} \times 1_{z \ge 1} dz\right]\\ &=\int_1^\infty z \frac{\ln z}{z^2} dz\\ &=\int_1^\infty \frac{\ln z}{z} dz\\ &= \frac{(\ln z)^2}{2}\biggr\rvert_1^\infty\\ &= \lim_{a \to \infty} \frac{(\ln z)^2}{2}\biggr\rvert_1^a\\ &= \lim_{a \to \infty} \frac{(\ln a)^2}{2}\\ &= \infty\\ \end{align} or $\displaystyle \lim_{a \to \infty} \frac{(\ln a)^2}{2}$ dne $\displaystyle \because \frac{(\ln a)^2}{2} \to \infty$ as $a \to \infty$

Answer 2

The expected value does not exist, or if you prefer is infinite. For the expression for $E(Z)$ is $$E(Z)=\int_1^\infty z\cdot \frac{\ln z}{z^2}\,dz.$$ For $z\ge e$, we have $\ln z\ge 1$. Thus by Comparison with $\int_e^\infty \frac{1}{z}\,dz$ our integral diverges.

Answer 3

Use integration by parts to see that the expected value is infinite. $\int_1^\infty \frac{ln(z)}{z}\,dz=\int_0^\infty u\,du$, where $u=ln(z)$.