Timeline for Question about the calculation of likelihood function [duplicate]
Current License: CC BY-SA 3.0
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| Aug 7, 2017 at 21:06 | history | closed |
kjetil b halvorsen♦ Michael R. Chernick John mdewey Sean Easter |
Duplicate of What is the difference between "likelihood" and "probability"? | |
| Aug 6, 2017 at 20:24 | review | Close votes | |||
| Aug 7, 2017 at 21:06 | |||||
| Apr 13, 2017 at 12:44 | history | edited | CommunityBot |
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| Jun 25, 2015 at 7:08 | history | edited | user42140 | CC BY-SA 3.0 |
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| Jun 24, 2015 at 20:57 | comment | added | user42140 | and coming back to my original question: In the thread that I referenced, I still cannot figure out why the likelihood is defined for $\theta > 1$ rather than the original domain of $[0, $\theta]$? I would comment on that thread but I do not have enough reputation points. | |
| Jun 24, 2015 at 20:55 | comment | added | user42140 | Ok, after reading that thread one thing that seems to be the case is that the model parameters are constant but unknown and the prior distribution does not model the variability of the parameter but rather the fact that we are uncertain about the parameter value. So, even though the parameter can take many values with non-zero probability, it is still a constant in the sense that there is a true parameter value which we can never know with full certainty. Is that the correct way to look at it? | |
| Jun 24, 2015 at 13:36 | comment | added | whuber♦ | Because this post concerns the meaning of likelihood and notation for describing it, may I suggest reading the thread on these subjects at stats.stackexchange.com/questions/2641? That may resolve many of the underlying issues reflected here. One issue it might not resolve concerns the last one, which comes down to the proper description of density functions. The density of $U(0,\theta)$ is not $1/\theta$: it is zero outside the interval $[0,\theta]$. It is essential to specify that when working with the density. | |
| Jun 24, 2015 at 8:41 | history | edited | user42140 | CC BY-SA 3.0 |
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| Jun 24, 2015 at 8:08 | comment | added | user42140 | Just to clarify, I thought in the frequentist way of thinking $\theta$ is a constant (although an unknown one) and in the Bayesian way $\theta$ is a latent RV. | |
| Jun 24, 2015 at 7:51 | comment | added | user42140 | In the bayesian paradigm, are the parameters not RVs? Is that not why we have a distribution over them (the prior?) | |
| Jun 24, 2015 at 5:37 | comment | added | Glen_b | Parameters are constants! Note that you cannot have $U(0,\infty)$ | |
| Jun 24, 2015 at 4:26 | history | asked | user42140 | CC BY-SA 3.0 |