MHB Maximizing Likelihood Estimator of β

[jmorgan]

Assuming α is known, find the maximum likelihood estimator of β

f(x;α,β) = , 1 ,,,,,,, .(x^α.e^-x/β)
,,,,,, ,,,,,,α!β^α+1

I know that firstly you must take the likelihood of L(β). But unsure if I have done it correctly. I came out with the answer below, please can someone tell me where/if I have gone wrong.

L(β)= (α!β^α+1)^-n.Σxi^α.e^Σxi/βn

 

[HallsofIvy]

I don't understand your question. The "maximum Likelihood" estimator for a parameter is the value of the parameter that makes a given outcome most likely. But you have not given an "outcome" here.

 

[Siron]

I think that you're going in the right direction. However, your calculation is not entirely correct. Suppose that we have given observations $x_1,\ldots,x_n$ from the given distribution. The likelihood is then given by
$$\mathcal{L}(x_1,\ldots,x_n,\alpha,\beta) = \prod_{i=1}^{n} \frac{1}{\alpha ! \beta^{\alpha+1}} x_i^{\alpha}e^{-x_i/\beta}.$$
We wish to find the value of $\beta$ that maximizes the likelihood. Since it is quite common to work with the logarithm, let us first take the log of both sides:
$$\log \mathcal{L}(x_1,\ldots,x_n,\alpha,\beta) = -n \log(\alpha) - n (\alpha+1) \log(\beta)+ \alpha \sum_{i=1}^{n} \log(x_i) - \frac{\sum_{i=1}^{n} x_i}{\beta}.$$
Taking the derivative w.r.t $\beta$, we obtain
$$\frac{\partial \log \mathcal{L}(x_1,\ldots,x_n,\alpha,\beta)}{d\beta} = -n(\alpha+1)\frac{1}{\beta} - \frac{1}{\beta^2} \sum_{i=1}^{n} x_i.$$
To proceed, set the RHS equal to $0$ and solve for $\beta$. This is the required MLE.