Timeline for Irwin-Hall Distribution relationship between two sets of events
Current License: CC BY-SA 4.0
6 events
| when toggle format | what | by | license | comment | |
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| Jul 18, 2019 at 19:12 | comment | added | Iosif Pinelis | @user1246462 : The equality $P(X+Z\le x,X+A\le x)=P(X+Z\le x,Y+A\le x|X=Y)$ will not hold in general even if $X$ and $Y$ are discrete. Indeed, $P(X+Z\le x,X+A\le x)=\sum_uf(u)F(u)^2$, where $f$ is the pmf of $X$ and $F$ is the cdf of $Z$, whereas $P(X+Z\le x,Y+A\le x|X=Y)=P(X+Z\le x,X+A\le x|X=Y)=\sum_uf(u)^2F(u)^2/\sum_uf(u)^2$ | |
| Jul 18, 2019 at 17:39 | comment | added | user1246462 | ah yes, I'd forgotten that the variables are continuous, not discrete, so the event $X = Y$ has zero-probability of happening. If all the variables are discrete, then we can show $P(X + Z \leq x, X + A \leq x) \geq P(X + Z \leq x, Y + A \leq x)$ by using the fact that $P(X + Z \leq x, X + A \leq x) = P(X + Z \leq x, Y + A \leq x | X = Y)$. | |
| Jul 18, 2019 at 17:08 | comment | added | Iosif Pinelis | @user1246462 : I am glad you liked this answer. Concerning the question in your comment: You should clarify what you mean by conditioning on the zero-probability event $\{X=Y\}$. Anyhow, in general, conditioning does not preserve the probability. | |
| Jul 18, 2019 at 14:26 | comment | added | user1246462 | Thanks for the response, it's very nice! I have one more question. Would it be correct to say $\Pr[(X \leq x) \cap (Z \leq x - X) \cap (A \leq x - X)] = \Pr[(X \leq x) \cap (Y \leq x) \cap (Z \leq x - X) \cap (A \leq x - Y) | X = Y]$ and the inequality follows directly from that? | |
| Jul 18, 2019 at 14:14 | vote | accept | user1246462 | ||
| Jul 18, 2019 at 0:07 | history | answered | Iosif Pinelis | CC BY-SA 4.0 |