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This question already has an answer here:

I am looking at the answer on this thread: Why likelihood is not always a density function?

Here as I understand that the likelihood function is given by:

$$ L(\theta) = \frac{1}{\theta} \quad \mathrm{in\ range}\ [0, \theta] $$

First, I am confused by the notation in the reply in the thread I posted. So the likelihood is defined as $U(0, \theta)$. I am confused as to why the parameter $\theta$ appears in the support of the uniform distribution. Should the support not be just constants like $U(0, +\infty)$?

[EDIT]: As pointed out below $U(0, +\infty)$ is not possible but I was still expecting something like $U(0, 1)$ for standard uniform distribution or something of the form $U(a, b)$. In any setting, $\theta$ is unknown so how can it be the parameter of the uniform distribution?

Second, when we observe $Y=1$, why does the likelihood become

$$ L(\theta) = \frac{1}{\theta} \quad \mathrm{for}\ \theta > 1 $$

Why now $\theta > 1$ instead of $\theta > 0$

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marked as duplicate by kjetil b halvorsen, Michael Chernick, John, mdewey, Sean Easter Aug 7 '17 at 21:06

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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    $\begingroup$ Parameters are constants! Note that you cannot have $U(0,\infty)$ $\endgroup$ – Glen_b Jun 24 '15 at 5:37
  • $\begingroup$ In the bayesian paradigm, are the parameters not RVs? Is that not why we have a distribution over them (the prior?) $\endgroup$ – user42140 Jun 24 '15 at 7:51
  • $\begingroup$ 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. $\endgroup$ – user42140 Jun 24 '15 at 8:08
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    $\begingroup$ 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. $\endgroup$ – whuber Jun 24 '15 at 13:36
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    $\begingroup$ 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? $\endgroup$ – user42140 Jun 24 '15 at 20:55

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