# A question about the likelihood function

$X_1,\ldots,X_n$ $n$ sampled are random variable $f(x|\theta)=(-1)\theta \ln\theta 1_{x>\theta}$ where $1$ is the indicator function. What is the log likelihood function? It seems easy to multiply but i am not sure how to handle the term $(-1)^n$ and $1^n{x>\theta}$

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Some editing is needed to make it clear exactly what $f(x|\theta)$ is. – Dilip Sarwate Mar 16 '12 at 1:06

$1_{x\gt \theta}$ means there is only a non-zero density or likelihood when $X_i$ is greater than $\theta$. So for the whole sample you require all the $X_i$ to be greater than $\theta$ or, in other words, $\theta \lt \min_i \{X_i\}$.

The $(-1)$ could be there because $\ln \theta$ would be negative if $0 \lt \theta \lt 1$, but a probability density must be non-negative.

I suspect there is an error in what you have written for the conditional density or at least something is missing, as integrating it over $x$ does not obviously give $1$. Something like

$$f(x|\theta) = \frac{(-1)}{x \ln \theta} \; 1_{\theta \lt x \lt 1}$$

might be what you intended.

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But i think the existence of (-1) is enough to show that no log likelihood function exist right?? – Mathematics Mar 16 '12 at 3:14
No. If $0 \lt \theta \lt 1$ then $-\log(\theta)$ is positive and so might be a likelihood and so $\log( -\log(\theta))$ is real and so might be a log-likelihood. – Henry Mar 16 '12 at 7:13