Let the density function be given by

$$ f(x;a,b) = \frac{a + 2 b g(x) + (1-a-b) g(x)^2}{(1-x)(2 b g(x) + (1-a-b) g(x)^2)}$$

where $a$ and $b$ are parameters of interest and $g(x)$ is a known function.

I was told that using this density function in maximum likelihood, the parameters $a$ and $b$ are identified.

The concept of identification is clear to me, but what are the rigorous mathematical considerations to conclude identifiability here?


2 Answers 2


In Chapter of 14 [p. 2] of Greene's book it's stated that for a likelihood function:

The parameter vector $\theta$ is identified (estimable) if for any other parameter vector, $\theta^* \neq \theta$, for some data $y$, $L(\theta^∗|y)\neq L(\theta |y)$.

Based on that, it's not hard to check that for different values of $a$ and $b$ you'll have different values for the likelihood of the given density.

EDIT: When I say different values of $f$ for different values of $a$ and $b$ I should actually say the likelihood function must be a one-to-one correspondence (or bijective, if you prefer).

  • $\begingroup$ Starting with $f(x; a, b) = f(x; a', b')$, I obtain $a'(-2b + g(x) (b-1)) + a(2b' + g(x)(1-b))$, and then by comparing coefficients this implies that $a = a'$ and $b = b'$. Is that what you had in mind? $\endgroup$
    – bonifaz
    May 25, 2017 at 18:48
  • $\begingroup$ @bonifaz Exactly. $\endgroup$ May 25, 2017 at 19:04

I think a reasonable approach to a rigorous proof would be to assume the contrapositive -- that is, assume there exists some $a'$ and some $b'$ such that $$ f(x;a',b') = f(x;a,b)$$ and demonstrate based on this assumption that $a=a'$ and $b=b'$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.