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If X is U[$0$,$\theta$], then the likelihood is given by $f(X,\theta) = \dfrac{1}{\theta}\mathbb{1}\{0\leq x \leq \theta\}$. The definition of Fisher information is $I(\theta) = \mathbb{E} \left[ \left(\dfrac{d \log(f(X,\theta))}{d\theta} \right)^2 \right]$. How can this be calculated when $\log f(X,\theta) $ is not defined for $\theta < X$? I understand that we also have $f(X,\theta) = 0$ for $\theta < X$ but can we ignore this when taking the expectation? If so, why?

glS
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bri
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    I'm not sure, but I think one chooses to define the log of the density only on the support of the density. – Shashi Nov 13 '17 at 22:41
  • I think that makes sense. I suppose we can see the random variable $X$ as a function from $X: \Omega \rightarrow [0,\theta]$, in which case $\log f(X,\theta)$ is well defined. Does that work? – bri Nov 15 '17 at 10:48
  • Yes that is one thing you can do. – Shashi Nov 15 '17 at 13:44
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    See this answer for why Fisher information is not defined here in the usual sense. – StubbornAtom May 15 '20 at 16:17
  • Thanks for the pointer. So as I've defined it, $I(\theta)$ does exist but it is not an 'interesting' quantity for the reasons outlined in that answer. – bri May 18 '20 at 08:31

1 Answers1

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it is $n^2/\theta$.

We get this from calculating the log-likelihood first which is $-n \log(\theta)$, then taking its derivative, we will get $\frac{-n}{\theta}$. squaring it and take its expectations we will have $\frac{n^2}{\theta^2}$

PrincessEev
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user656381
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    Fisher information does not exist for distributions with parameter-dependent supports. Using different formulae for the information function, you arrive at different answers. – StubbornAtom Mar 21 '19 at 08:30
  • @StubbornAtom Can you give an example of the argument you just gave? – Daniel Ordoñez Apr 16 '19 at 18:24
  • @DanielOrdoñez Fisher information is defined for distributions under some 'regularity conditions'. One of the conditions is that support of distribution should be independent of parameter. That is the main argument... – StubbornAtom Apr 16 '19 at 18:50
  • Yes, clearly it happens because you can't change the integral and the differential in this case. But isn't the Fisher Information defined as the first expression you gave? (the second one being a corollary when you can switch the differential and the integral) – Daniel Ordoñez Apr 16 '19 at 18:58
  • @DanielOrdoñez That is correct. The problem is that the information is defined under assumptions not all of which hold here. (use @ while replying so that we get pinged) – StubbornAtom Apr 16 '19 at 21:22