I came across a probability distribution function in my work, it is however difficult to find in closed form, therefore I am looking to either upper bound or lower bound it.

Assuming $a,b,T$ are non-negative constants, and $X>0,Y>0$ are Random variables .

What is the direction of the inequality below?

$$\mathbb{P}\left( a \ X + b \ Y \geq T\right) \stackrel{\leq}{\geq} \mathbb{P}\left( X + Y \geq \frac{T}{a+b}\right) $$

i.e am I upper bounding or lower bounding ?

The reason I am doing that is because the distribution of $a X+bY $ is not easy to derive but it is with $X+Y$..


  • $\begingroup$ Seems like $\geq$ since everything is nonnegative. $\endgroup$ – Jihad Dec 18 '14 at 16:30
  • $\begingroup$ Thank Jihad, I understand your point however I still don't have a good reasoning.. @Jihad $\endgroup$ – Henry Dec 18 '14 at 16:37
  • $\begingroup$ @Jihad I have edited the question to make it more obvious, in this case I think you have to change the sign on inequality in your comment, sorry about that... $\endgroup$ – Henry Dec 18 '14 at 16:39
  • $\begingroup$ Ok, I think it can be in both directions. If $X$ and $Y$ are non-negative then it is obvious. The interesting case when one of them is positive and the second one is negative for instance. $\endgroup$ – Jihad Dec 18 '14 at 17:43
  • $\begingroup$ Thanks, Jihad I do have that everything is non-negative, so you are saying there is no definite direction? @Jihad $\endgroup$ – Henry Dec 18 '14 at 17:46

If we multiply $X + Y \ge \frac T{a+b}$ by $a+b$ we get $aX + bX + aY + bY \ge T$ This will be more likely to be true than $aX+bY \ge T$ because we are adding positive terms to the left. So in your question the sign should be $\lt$

  • $\begingroup$ Thanks, I don't understand what you mean by " this will be more likely to be true.." Do you mean I can't tell where it is an upperbound or lower bound? $\endgroup$ – Henry Dec 18 '14 at 21:56
  • $\begingroup$ You are comparing two probabilities. Every time the left side in your question is true, the right side is true, but there will presumably be times when the left side is false but the extra terms $bX+aY$ in the way I express it make he right side true. This means the right side is an upper bound for the left. $\endgroup$ – Ross Millikan Dec 18 '14 at 22:36
  • $\begingroup$ If you don't mind I would like to ask you a question regarding this, I totally understand your answer provided here, but what if $a=b=T=1$ then the inequality is not correct, how come it doesn't hold? $\endgroup$ – Henry Feb 14 '15 at 21:18
  • $\begingroup$ The inequality of the probabilities is still correct in that case. We then have $\Bbb P(x+y\ge T) \lt Bbb P(x+y \ge \frac T2)$ The right will be true all the time the left is, and as well when $\frac T2 \le x+y \lt T$ $\endgroup$ – Ross Millikan Feb 14 '15 at 22:31
  • $\begingroup$ if $X=Y=4/9$ and $a=T=b=1$ does the inequality hold? $\endgroup$ – Henry Feb 14 '15 at 22:33

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