Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I missed the day of class where we went over likelihood functions, and I had a quick question. If $X_1,...X_n$ are i.i.d. $ {\Gamma}(\alpha,\beta)$ r.v.s, I'm trying to find the likelihood function for $\alpha$ and $\beta$. From what I can tell, the likelihood function is defined as $f(x_1;\alpha,\beta)*...*f(x_1;\alpha,\beta)$. If i'm not mistaking, $f(x_1;\alpha,\beta)$ should be the same as the pdf for the gamma distribution (although not a pdf, on account of x being a fixed value here?), so would the likelihood function in this case be ${(\text{pdf-of-}\Gamma)}^n$? Further, I'm a bit confused on what the support for this function should be; in this case x is fixed, am I correct in supposing that the support is then $\alpha>0, \beta>0$? I hope this makes sense - I get the feeling I might be very confused about the whole thing. Thank you for any help!

share|improve this question

1 Answer 1

up vote 0 down vote accepted

If $X$ follows a gamma distribution with shape $\alpha$ and scale $\beta$, then its probability density is

$$p_{\alpha, \beta}(x) = \frac{ x^{\alpha-1} e^{-x/\beta}}{\Gamma(\alpha) \beta^\alpha }$$

Sometimes this is re-parameterized with $\beta^{\star} = 1/\beta$, in which case you will need to change this accordingly.

The likelihood function is just the density viewed as a function of the parameters. So, the log-likelihood function for an IID sample $X_1, ..., X_n$ from this distribution with realized values $x_1, ..., x_n$ is

$$ L(\alpha, \beta) = \sum_{i=1}^{n} \log \big( p_{\alpha, \beta}(x_i) \big) = (\alpha-1) \sum_{i=1}^n \log(x_i) - \frac{1}{\beta} \sum_{i=1}^{n}x_i - n\alpha \log(\beta) - n\log( \Gamma(\alpha) )$$

which can be maximized jointly as a function of $\alpha, \beta$ to get the MLE.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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