Maximum Likelihood Estimator - Beta Distribution

I'm taking a Mathematical Statistics course and trying to work through a homework problem that reads:

Let Y1, ..., Yn be a random sample from a Beta(1,$\theta$) population. Derive the MLE for $\theta$.

Getting started, I set up the likelihood function as:

$$L(\theta) = \frac{\Gamma(1+\theta)}{\Gamma(1)\Gamma(\theta)}\prod_{i=1}^n(1-Y_i)^{\theta-1}$$

This simplifies to:

$$L(\theta) = \theta(1-Y_i)^{n(\theta-1)}$$

Taking the log:

$$l(L) = log(\theta) + n(\theta-1)log(1-Y_i)$$

Take the derivative:

$$\frac{dl}{d\theta} = \frac{1}{\theta}+nlog(1-Y_i)$$

Solve for $\theta$:

$$\theta = \frac{-1}{nlog(1-Y_i)}$$

But this can't be right as I've not dealt with the summation of $Y_i$. In most examples I've seen, this goes away as a result of the summation being divided by n,b ut in this case I can't find where I've gone wrong.

I appreciate any tips.

• You can't write $\prod_{i=1}^n(1-Y_i)^{\theta-1}=(1-Y_i)^{n(\theta-1)}$, different $i$ indicates different values. – Francis Nov 1 '17 at 5:10

I think your likelihood fucntion is wrong, for your Beta distrbution, the $pdf$ is

$$f(y)=\frac{\Gamma(1+\theta)}{\Gamma(1)\Gamma(\theta)}(1-y)^{\theta-1}$$ The likelihood function will be

$L(\theta)=\frac{\Gamma(1+\theta)}{\Gamma(1)\Gamma(\theta)}(1-y_1)^{\theta-1}\frac{\Gamma(1+\theta)}{\Gamma(1)\Gamma(\theta)}(1-y_2)^{\theta-1}...\frac{\Gamma(1+\theta)}{\Gamma(1)\Gamma(\theta)}(1-y_n)^{\theta-1}\\= (\frac{\Gamma(1+\theta)}{\Gamma(1)\Gamma(\theta)})^n\left [ \prod_{i=1}^n(1-y_i)\right]^{\theta-1}$

Now take the log

$l(\theta)=nlog(\frac{\Gamma(1+\theta)}{\Gamma(1)\Gamma(\theta)})+(\theta-1)\sum_{i=1}^nlog(1-y_i)$

I will not go ahead from here.

• Thanks for the response. I corrected an error in my original post, but I don't think it affects your response. I see the position of the exponent is together, and you have no theta to start - don't the Gamma functions simplify to theta? – KirkD_CO Nov 1 '17 at 5:11
• "Together" should be "different" on that last comment. – KirkD_CO Nov 1 '17 at 5:14
• Ah-ha! I also pulled the Gamma functions out without bringing along the exponent n. – KirkD_CO Nov 1 '17 at 5:21
• Is there any needed to keep the Gamma functions in this form? This reduces to theta, no? And then theta^n once pulled outside the product. – KirkD_CO Nov 2 '17 at 3:59
• Not need, since $\frac{\Gamma(1+\theta)}{\Gamma(\theta)}=\frac{\theta !}{(\theta-1)!}=\theta$ – Deep North Nov 2 '17 at 4:12