Suppose $X$ is a discrete random variable such that $P(X \geq a) = 1$. Then, $E[X] \geq a$.
I feel that this proof is fairly simple, but I am having a bit of trouble with it. I think that what I want to get to is showing $(a)P(X \geq a) = (a)(1) = a$, but I am not sure how to get from $E[X] = \sum x_ip_i$ to $E[X] \geq (a)P(X \geq a)$.
Since we don't know the range of values of $X$, I am finding it hard to connect these two ideas. I was thinking $\sum x_ip_i \geq (a)P(X = a)$, but I am first of all not sure that this is true (what if every value other than $a$ is negative?) and I am not sure how it relates to $(A)P(X=a)$.
It may be that I am missing something simple, but I am not sure how to connect these ideas, or if this is the correct way to do this proof.
Let $(x_i)_{i\in \mathbb N}$ be the support of $X$, ie $\forall i, P(X=x_i)>0$ and $\sum_i P(X=x_i)=1$.
Note that $\forall i, x_i\geq a$. Indeed, if $x_j<a$, then $(X=x_j)\subset (X<a)$ and $P(X=x_j)=0$.
Then $E(X) = \sum_i P(X=x_i) x_i \geq \sum_i P(X=x_i)a = a $
If you know measure theory, $X-a\geq 0$ $P$-a.s, hence $\int (X(w)-a) dP(w)\geq 0$, i.e $E(X)\geq a$.