I am working on a problem and am a bit stuck.
Problem: For $X$ with $p(x) = x^{-2}$, $x\ge1$, find $E(X)$ and $P(X\le2)$
What I have done so far:
$$E(X) = \int_{1}^{\infty}xp(x)d(x) = \int_{1}^{\infty}x^{-1}d(x) = \infty.$$
But I am a little bit confused on how to find the probability.
Ok for the evaluation of $E(X)$, except for that $2$. By the way, it is $+\infty$. For the probability:
$$P(X \leq 2) = \int_1^2 x^{-2} dx = \left.-x^{-1}\right|_{x=1}^{x=2} = -\frac{1}{2} + \frac{1}{1} = \frac{1}{2}.$$