Timeline for Conditional Expectation via Integral over Quantile Function
Current License: CC BY-SA 3.0
12 events
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| May 11, 2017 at 9:17 | answer | added | Stéphane Laurent | timeline score: 1 | |
| May 11, 2017 at 7:40 | history | edited | amoeba |
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| Apr 13, 2017 at 12:44 | history | edited | CommunityBot |
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| Mar 30, 2016 at 3:11 | history | tweeted | twitter.com/StackStats/status/715013573987209216 | ||
| Jun 20, 2014 at 19:19 | comment | added | Joz | See edit in the original question. It would be nice if someone could confirm the correctness of the solution. Thanks a lot for your help so far!! | |
| Jun 20, 2014 at 18:56 | history | edited | Joz | CC BY-SA 3.0 |
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| Jun 20, 2014 at 18:40 | history | edited | Joz | CC BY-SA 3.0 |
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| Jun 20, 2014 at 17:52 | history | edited | Joz | CC BY-SA 3.0 |
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| Jun 20, 2014 at 15:27 | history | edited | Joz | CC BY-SA 3.0 |
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| Jun 19, 2014 at 19:25 | comment | added | Avraham | Two notes. Firstly, $P(X=x)$ for any random variable $X$ which has a continuous distribution is 0. Secondly, I think you need the denominator in your first step. Specifically $ E\left[X\vert X<q_\theta\right] = \frac{\int_{-\infty}^\infty xf(x\vert x < q_\theta)dx}{\int_{-\infty}^\infty f(x\vert x < q_\theta)dx}$ The expectation is only over the viable area. | |
| Jun 19, 2014 at 19:02 | comment | added | whuber♦ | I believe that if you consult your favorite definition of conditional distribution, this question will resolve itself easily. In particular I am thinking of the relationship $\Pr(A|B)\Pr(B)=\Pr(A)$ where $A$ is the event $X\le x$ and $B$ is the event $X\le q_\theta$. | |
| Jun 19, 2014 at 18:32 | history | asked | Joz | CC BY-SA 3.0 |