# Calculate expected value i.i.d. from distribution function

Let random variables $X_n, n\in \mathbb N$ i.i.d. with distribution function $F_{X_i}(x)=(1-x^{-\lambda})1_{x > 1} \, , \lambda>0$ I want to calculate the expected value and variance. I am not sure about my attempt. I want to determine the derivative to get the density and then simply use formula to calculate the expected value. The derivative(density) is simply :$$\lambda x^{-\lambda -1}$$This does not seem correct to me, but how else could one calculate the expected value? Any help is much appreciated!

I don't really see anything wrong with your approach. We get: $$\int_1^{\infty}\lambda x^{1-\lambda}dx=\underset{d\to\infty}\lim \frac {\lambda d^{2-\lambda}}{2-\lambda}-\frac{\lambda}{2-\lambda},$$ which is \begin{align} \infty&,~~~\text{for }\lambda\in(0,2)\\ \text{undefined}&,~~~\text{for }\lambda=2\\ -\frac{\lambda}{2-\lambda}&,~~~\text{for }\lambda\in(2,\infty)\\ \end{align} In particular, the expected value only exists for $\lambda>2$. You can calculate the variance similarly.
• $\int_1^{\infty}\lambda x^{-\lambda}dx=\underset{d\to\infty}\lim \frac {\lambda d^{1-\lambda}}{1-\lambda}-\frac{\lambda}{1-\lambda}$ so it is undefined for $\lambda =1$ and $\frac{\lambda}{1-\lambda}, \lambda \in (1,\infty)$ – user409387 Jun 5 '17 at 12:11