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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!

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    $\begingroup$ Do you know anything about X,Y,X besides the fact that they are discrete? $\endgroup$ – Tim Aug 25 '16 at 16:57
  • $\begingroup$ 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. $\endgroup$ – user254769 Aug 25 '16 at 17:02
  • $\begingroup$ 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 ? $\endgroup$ – meh Aug 25 '16 at 17:03
  • $\begingroup$ Clearly this inequality is not always true. My question is: under what conditions could it be true? $\endgroup$ – user254769 Aug 25 '16 at 17:09
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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).

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  • $\begingroup$ RIPO thank you for your reply! But i'm a little confused: Z<0 ?? Z has to be negative for the inequality to hold? Could you elaborate more on this please. $\endgroup$ – user254769 Aug 25 '16 at 17:32

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