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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!

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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.

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How do we differentiate gamma$\alpha$ to get the value of $\alpha$ ? – idpd15 Sep 3 '15 at 12:27

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