I have a question regarding conditional probability. What is the probability of the sum being smaller than Y, while one of the components of the sum is greater than Z (Z < Y, both constants). So:

$P(X_1+X_2<Y|X_1>Z)$ with $Z<Y$

Assume $X_1$ and $X_2$ are independently and identically distributed and their distributions are known. Also the joint distribution of $X_1+X_2$ is known.

Thanks in advance!

  • $\begingroup$ You can't calculate it without knowing the joint distribution of $X_1$ and $X_2$. (In fact what you really want here is the joint distribution of $X_1+X_2$ and $X_1$, but that can be derived from the joint distribution of $X_1$ and $X_2$.) $\endgroup$ – Ian Aug 5 '15 at 12:04
  • $\begingroup$ Suppose I know the joint distribution of $X_1+X_2$ and I know the distribution of $X_1$. Assume that both distributions are non-negative and continuous. Can it be calculated now? $\endgroup$ – Tschilpje Aug 5 '15 at 12:08
  • $\begingroup$ Are the X's independent, and are Y and Z constants? $\endgroup$ – David Quinn Aug 5 '15 at 14:09
  • $\begingroup$ $X_1$ and $X_2$ are independently and identically distributed. Both Y and Z are constants. $\endgroup$ – Tschilpje Aug 5 '15 at 14:23

Well, I can start this one off, at least, and get a form for you. By conditional probability:

$P(X_1 + X_1 < Y \mid X_1 > Z) = \displaystyle{\frac{P(X_1 + X_2 < Y, X_1 > Z)}{P(X_1 > Z)}}$

Then rearranging the numerator:

= $\displaystyle{\frac{P(X_1 < Y - X_2, X_1 > Z)}{P(X_1 > Z)}}$

= $\displaystyle{\frac{P(Z < X_1 < Y - X_2)}{P(X_1 > Z)}}$

which, by definition of cumulative distribution function (CDF), is:

= $\displaystyle{\frac{P(X_1 < Y - X_2) - P(X_1 < Z)}{P(X_1 > Z)}}$

and now rearranging the denominator using the law of total probability:

= $\displaystyle{\frac{P(X_1 < Y - X_2) - P(X_1 < Z)}{1 - P(X_1 \le Z)}}$

And since you know the distribution of $X_1$, you should be able to calculate this just using its CDF. Does anyone see an error here?

EDIT: the last two equations were edited to fix a typo in the numerator.

  • $\begingroup$ Thanks for your thoughts. One remark. Isn't it given the definition of the CDF: = $(P(X_1 < Y - X_2) - P(X_1 < Z) )/ (P(X_1 > Z)$ $\endgroup$ – Tschilpje Aug 6 '15 at 6:05
  • $\begingroup$ Then, is it possible to substitute $P(X_1<Y-X_2)$ by $P(X_1+X_2 < Y)$ in the final equation? $\endgroup$ – Tschilpje Aug 6 '15 at 11:01
  • $\begingroup$ Yes, you're absolutely right. Sorry for the typo! I fixed the last two equations. Yes, you can make that substitution. I put everything in terms of $X_1$ because I thought it would be easier. $\endgroup$ – StatsSorceress Aug 6 '15 at 12:38

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