Timeline for Proof that the probability of one RV being larger than $n-1$ others is $\frac{1}{n}$
Current License: CC BY-SA 3.0
14 events
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| Apr 13, 2017 at 12:44 | history | edited | CommunityBot |
replaced http://stats.stackexchange.com/ with https://stats.stackexchange.com/
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| Jan 16, 2015 at 10:47 | history | tweeted | twitter.com/#!/StackStats/status/556039986710335488 | ||
| Jan 15, 2015 at 22:48 | answer | added | Alecos Papadopoulos | timeline score: 8 | |
| Jan 15, 2015 at 22:38 | comment | added | REFlint | Thanks for picking that up Alecos, bad notation! I've fixed the question. | |
| Jan 15, 2015 at 22:38 | history | edited | REFlint | CC BY-SA 3.0 |
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| Jan 15, 2015 at 22:31 | answer | added | Danica | timeline score: 10 | |
| Jan 15, 2015 at 22:24 | comment | added | Glen_b | REFlint -- for continuous D, in which ways do you feel the symmetry argument outlined in my comment starting "Easy..." on your other question would fall short of a proof? Perhaps we would be able to provide you with some reassurance or details on what you feel it lacks. | |
| Jan 15, 2015 at 22:15 | answer | added | Aksakal | timeline score: 4 | |
| Jan 15, 2015 at 22:14 | comment | added | Alecos Papadopoulos | Looking at the OP's other question, it appears that he does not mean the product but just a collection, and whether $X_n$ is larger than the maximum of the $n-1$ collection. But it obviously created confusion here. | |
| Jan 15, 2015 at 22:06 | comment | added | whuber♦ | One might try to assume the $X_i$ are independent and $D$ is continuous, but even this will not suffice. Take, for instance, negative random variables and let $n$ be any odd number. Then the product $X_1\cdots X_{n-1}$ must be positive, whence $\Pr(X_n\gt X_1\cdots X_{n-1})=0$. Is it possible you meant something other than a product where you wrote "$X_1\ldots X_{n-1}$"? If this is intended as a shorthand for "$X_n\gt X_1$ and $X_n\gt X_2$ and ... and $X_n\gt X_{n-1}$," then ask @Glen_b to post his comment to your previous question as an answer, because it will serve just fine. | |
| Jan 15, 2015 at 21:53 | comment | added | kolonel | you need to put more assumptions on the distribution $D$. | |
| Jan 15, 2015 at 21:51 | comment | added | Yair Daon | This is not necessarily true. If $X_i = 1$ identically then what you're trying to prove is false. | |
| Jan 15, 2015 at 21:45 | history | edited | Alecos Papadopoulos | CC BY-SA 3.0 |
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| Jan 15, 2015 at 21:26 | history | asked | REFlint | CC BY-SA 3.0 |