Timeline for Bayes statistics precisely formulated
Current License: CC BY-SA 3.0
13 events
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| Dec 22, 2015 at 8:25 | history | edited | Carlo Beenakker | CC BY-SA 3.0 |
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| Dec 21, 2015 at 12:14 | history | edited | Carlo Beenakker | CC BY-SA 3.0 |
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| Dec 19, 2015 at 11:33 | history | edited | Carlo Beenakker | CC BY-SA 3.0 |
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| Dec 18, 2015 at 14:22 | comment | added | Fabian Werner | So we cannot apply Bayes as $\lambda$ is not an event ($P[\Lambda=\lambda] = 0$)... but we still do it... why are we allowed to do it? | |
| Dec 18, 2015 at 14:18 | comment | added | Carlo Beenakker | Bayes' theorem as formulated above (from section 1.3 in the linked notes) refers to events that do not have measure zero, for example, $1<x<3$ is an event but not $x=1$. It is also possible to formulate Bayes' theorem for continuous variables, when the probabilities have to be replaced by probability densities, see section 3.3 in the linked notes. | |
| Dec 18, 2015 at 14:12 | comment | added | Fabian Werner | *** WHAT IS $p(\lambda)?$ *** If it is $P[\Lambda=\lambda]$ then it IS zero... almost always! So we cannot apply the classical Bayes! Phrased differently, if you have a normally distributed random variable $X$, what is the probability that $X$ attains the value $3.79856$? It is zero! | |
| Dec 18, 2015 at 14:10 | history | edited | Carlo Beenakker | CC BY-SA 3.0 |
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| Dec 18, 2015 at 14:07 | comment | added | Fabian Werner | I am sorry but I do not understand... $x$ and $\lambda$ are not events, they are elements of the sets $V$ and $\mathbb{R}$... Are you talking about $[\Lambda=\lambda]$ (a zero set in $\Omega$!) ? | |
| Dec 18, 2015 at 14:06 | history | edited | Carlo Beenakker | CC BY-SA 3.0 |
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| Dec 18, 2015 at 14:01 | history | edited | Carlo Beenakker | CC BY-SA 3.0 |
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| Dec 18, 2015 at 13:58 | comment | added | Fabian Werner | What is $P(\lambda)$ then? Is it $P(\Lambda=\lambda)$? Then its almost always zero as $\Lambda$ is continuously valued and we can throw away Bayes because the expressions do not make sense... ? Or let me formulate it that way: If you were right then $p(\lambda|x)$ was zero all the time and we would learn the expression '0' from the data... | |
| Dec 18, 2015 at 13:56 | history | edited | Carlo Beenakker | CC BY-SA 3.0 |
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| Dec 18, 2015 at 13:51 | history | answered | Carlo Beenakker | CC BY-SA 3.0 |