# Question about the calculation of likelihood function [duplicate]

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 $\theta > 0$

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

• Parameters are constants! Note that you cannot have $U(0,\infty)$ – Glen_b Jun 24 '15 at 5:37
• 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. – user42140 Jun 24 '15 at 8:08
• 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. – whuber Jun 24 '15 at 13:36