Let $X = (X_1,\dots, X_n)$ and $X_1,\dots, X_n$ be i.i.d Gamma($p,a,A$) random variables where $p$ and $a$ are known. Find the MLE of $\theta =A$.

We have \begin{align*} f_{\theta}(x) &=\prod_{i =1}^n \frac{1}{\Gamma(p)}x_i^{p-1}\frac{1}{a^{p-1}}\exp\Big\{-\frac{x_i}{a}\Big\}{1}_{[A,\infty]}(x_i)\\ &=\Bigg(\frac{1}{\Gamma(p)^na^{n(p-1)}}\exp\Big\{-\frac{1}{a}\sum_{i = 1}^nx_i\Big\} \prod_{i = 1}^nx_i^{p-1} \Bigg){1}_{[A,\infty]}(x_{[1]}) \end{align*} Thus \begin{align*} L(\theta;x) = \Bigg(\frac{1}{\Gamma(p)^na^{n(p-1)}}\exp\Big\{-\frac{1}{a}\sum_{i = 1}^nx_i\Big\} \prod_{i = 1}^nx_i^{p-1} \Bigg){1}_{[-\infty,x_{[n]}]}(A) \end{align*} Dont know where to go from here

  • $\begingroup$ You went astray in your first line. The likelihood you wrote is not that of a shifted gamma; you don't get it by simply replacing $\mathbf{1}_{0,\infty}$ by $\mathbf{1}_{A,\infty}$. Before you fix the shifting part, though, double check you know the density for an ordinary (unshifted) shape-scale gamma (e.g. check Wikipedia) $\endgroup$
    – Glen_b
    Commented Oct 14, 2019 at 3:20
  • $\begingroup$ that density was given as part of the question in the book $\endgroup$
    – yf297
    Commented Oct 14, 2019 at 3:25
  • $\begingroup$ 1. And you're 100% certain you copied it exactly? 2. If so, which book? (because if that's what it says, it's wrong) 3. Is this an exercise for some class? $\endgroup$
    – Glen_b
    Commented Oct 14, 2019 at 3:26
  • $\begingroup$ introduction to the theory of statistical inference by liero and zwanzig page 82. Im just self studying $\endgroup$
    – yf297
    Commented Oct 14, 2019 at 3:27
  • $\begingroup$ In any case isnt the likelihood function that i wrote down just a constant function of A? $\endgroup$
    – yf297
    Commented Oct 14, 2019 at 3:31

1 Answer 1


Let's first write down a density for a scale-shape parameterization for a Gamma and then shift it. Taking the density from Wikipedia (which has it correct), but making the variable $z$ rather than $x$:

$$f(z;\alpha ,\theta )=\frac{z^{\alpha -1}e^{-z/\theta }}{\theta ^{\alpha }\Gamma (\alpha )}\quad {\text{ for }}z>0,\quad \alpha ,\theta >0$$

Translating to the parameterization and notation your book is using (aside from the variable):

$$f(z;p ,a )=\frac{z^{p -1}e^{-z/a }}{a ^{p }\Gamma (p )}\, \mathbb{1}_{[0,\infty]},\quad p,a >0$$

You can immediately see that the book has the ordinary gamma wrong by simply putting $A=0$, and discovering they're missing a factor of $a$ on the denominator. This arises because of the Jacobian when the scale parameter is introduced; if you omit it the thing doesn't integrate to 1. You should check my assertion for yourself by integrating it (take it back to the form of an ordinary gamma integral, which you know the value of).

Now let's consider what happens when shift it up by $\delta$ (their $A$). Let $X = Z+\delta$, so $Z=X-\delta$ and $dx = dz$. The density therefore becomes:

$$f(x;p ,a ,\delta)=\frac{(x-\delta)^{p -1}e^{-(x-\delta)/a }}{a ^{p }\Gamma (p )}\, \mathbb{1}_{[\delta,\infty]},\quad p,a >0,\: \delta \in \mathbb{R}$$

Which now differs from the book in several respects. The book is wrong. (A little experience with location-scale families will tell you so at a glance.)

If you make sure to start with a correctly specified density (i.e. by starting with the phrase "shifted gamma" and working out what the density of that must be), you will get more out of the exercise. Don't rely on the book to be correct.


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