Timeline for Can a random variable be a deterministic function of other random variables yet be independent of them?
Current License: CC BY-SA 3.0
14 events
| when toggle format | what | by | license | comment | |
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| Mar 10, 2017 at 16:38 | history | edited | Pere | CC BY-SA 3.0 |
grammar
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| Sep 19, 2016 at 8:33 | history | edited | Pere | CC BY-SA 3.0 |
jointly independent
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| S Sep 19, 2016 at 8:28 | history | suggested | Therkel | CC BY-SA 3.0 |
Formatted the math with MathJax
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| Sep 19, 2016 at 7:12 | comment | added | Therkel | @Pere I took the liberty of adding MathJax to your (great) answer to improve readability. I hope that is okay - otherwise reject the edit! | |
| Sep 19, 2016 at 7:11 | review | Suggested edits | |||
| S Sep 19, 2016 at 8:28 | |||||
| Sep 18, 2016 at 19:44 | comment | added | Pere | Yes, I edited it again to apply "trivial" to the resulting random variable and not to the function. I'm not sure if there is a rigorous definition of trivial random variable, but I think it's clear enough - in fact, I'm not sure if there is a definition of random variable that includes something that always gives the same result (a deterministic random variable). However, since it points to your answer I think readers will easily understand what is the exception about. | |
| Sep 18, 2016 at 19:40 | history | edited | Pere | CC BY-SA 3.0 |
what's trivial is the random variable, not the function
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| Sep 18, 2016 at 19:39 | vote | accept | Christopher | ||
| Sep 18, 2016 at 19:39 | comment | added | Mark L. Stone | Well, the correctness of your edited statement depends on your definition of non-trivial function. Consider any function $f(x)$ which is one to one. Apply it to a random variable $Y$ which is a constant with probability one. Then $f(Y)$ is independent of $Y$. | |
| Sep 18, 2016 at 19:36 | history | edited | Pere | CC BY-SA 3.0 |
including the case of a trivial function
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| Sep 18, 2016 at 19:34 | comment | added | Christopher | @Patty , the counter example i provided was an XOR . Seems like this is correct . Yes , Pere . +1 (Pairwise independence does not imply joint independence ) . | |
| Sep 18, 2016 at 19:32 | history | edited | Pere | CC BY-SA 3.0 |
expanding proof
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| Sep 18, 2016 at 19:32 | comment | added | Mark L. Stone | Your statement "An univariate real random variable that is a deterministic function of another random variable is not independent of it. " is incorrect, as shown in my answer. | |
| Sep 18, 2016 at 19:23 | history | answered | Pere | CC BY-SA 3.0 |