Is the summation below true?

[latex] \sum_{a}a * Pr[R=a] \leq \sum_{a\leq b}b * Pr[R=a]. [/latex]

 

Where R is a random variable and Pr[R=a] means that the probability that the random variable is equal to some number 'a'. You can ignore that part and replace Pr[R=a] with x:

[latex] \sum_{a}a * x \leq \sum_{a\leq b}b * x. [/latex]

 

The first summation provides all 'a' values, so the summation is over larger amount of terms. The right hand side sums only those a<=b, so the right hand sums over less amount of terms. But at the same time the right hand is multiplied by b>=a. So it is not clear which one is bigger. I am not extremely familiar with Riemann sum, so hopefully someone here knows what I am talking about. My first intuition is that the above is only true if the upper limit is finite, or that the above is true regardless since you can always have infinite plus 1, i.e. you can always have a number greater than the upper limit for [latex] \sum_{a}a * x [/latex] even if the upper limit for [latex] \sum_{a}a * x [/latex] is infinite. This is the gist of my question.

 

I am trying to prove Markov's Theorem for when Pr[R<=x].

Edited April 16, 201411 yr by DylsexicChciken

Without more precise descriptions of a and b, the inequality can't be decided.