Question:
$$\text {Let X be a Pareto random variable with } \lambda = 3. $$
$$ Find: E[X | X > 0] $$
The solution given skips some steps which makes it a bit unclear to me. It makes sense logically but I want to derive the solution using symbols.
I tried doing this:
$$ \sum_{x=0}^\infty x * \frac{Pr[X > 0, X=x]} {Pr [X>0]} $$
The solution is just this:
$$ E[X | X > 0] = \frac{E[X]}{Pr[X>0]} $$
which leads me to believe that my expression simplifies to this:
$$ \frac{1}{Pr[X>0]}\sum_{x=0}^\infty x * Pr[X=x] = \frac{E[X]}{Pr[X>0]} $$
What is the logic that makes $$ Pr[X>0, X = x] = Pr[X=x]? $$
I tried doing this also using the Bayes rule as well, like this:
$$ \sum_{x=0}^\infty x * \frac{Pr[X>0 | X = x] * Pr[X = x]}{Pr[X>0]} $$
which would imply that $$ Pr[X>0 | X = x] = 1 $$ to give us the same answer as above.
Can somebody explain this logic to me? For some reason this escapes me and losing sleep over this sucks. Thanks for any help.
$$\mathbb{E}[X]=\mathbb{E}[X|X>0]Pr(X>0)+\mathbb{E}[X|X=0]Pr(X=0)$$
Hence
$$\mathbb{E}[X]=\mathbb{E}[X|X>0]Pr(X>0)+0$$
That is $$\mathbb{E}[X]=\mathbb{E}[X|X>0]Pr(X>0)$$
Final conclusion:
$$\mathbb{E}[X|X>0]=\frac{\mathbb{E}[X]}{Pr(X>0)}$$
Remark:
$$Pr(X>0, X=x)=Pr(X=x)$$ is not true when $x=0$.