# CDF of sum of independent discrete random variables

I know some probability theory but I am still not very familiar with more advanced topics in this area. I was wondering whether anyone can help me with the following question.

Is there any way possible to show/prove that

$$\textrm{Pr}(X+Y\leq L) < \textrm{Pr}(X+Y+Z\leq L)\quad?$$

where $X$, $Y$, $Z$ are discrete independent random variables and the probabilities express essentially the CDF of their sums. $L$ is a constant known value.

If anyone has any thoughts/feedback, they would be of tremendous help!

• Do you know anything about X,Y,X besides the fact that they are discrete? – Tim Aug 25 '16 at 16:57
• I know that X, Y and Z attain 2 or 3 states/values, each of the states has an associated probability. So I know those probabilities as well. I know for example the Pr(X = x1), Pr(X = x2), and similarly for Y and Z. – user254769 Aug 25 '16 at 17:02
• I think as stated, this can't be true. What if Z is a random variable with a constant value ? Am I misunderstanding what you are saying ? – meh Aug 25 '16 at 17:03
• Clearly this inequality is not always true. My question is: under what conditions could it be true? – user254769 Aug 25 '16 at 17:09

In particular, if $Z< 0$, then
$$\{\omega\in \Omega: X(\omega) +Y(\omega) \leq L\}\subset \{\omega\in \Omega: X(\omega) +Y(\omega) + Z(\omega) \leq L\}.$$
Consequently $P(X+Y\leq L) < P(X+Y+Z\leq L)$. There is a more complex condition under which the inequallity is still true, but you need to analyze conditions on $Z$ such that $P(X+Y+Z\leq L) = \sum _z P(X+Y\leq L -z) P(Z=z) > P(X+Y\leq L)$ (law of total probability).