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.

  • 2
    $\begingroup$ You can't write $\prod_{i=1}^n(1-Y_i)^{\theta-1}=(1-Y_i)^{n(\theta-1)}$, different $i$ indicates different values. $\endgroup$
    – Francis
    Nov 1, 2017 at 5:10

1 Answer 1


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


I will not go ahead from here.

  • $\begingroup$ 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? $\endgroup$
    – KirkD_CO
    Nov 1, 2017 at 5:11
  • $\begingroup$ "Together" should be "different" on that last comment. $\endgroup$
    – KirkD_CO
    Nov 1, 2017 at 5:14
  • $\begingroup$ Ah-ha! I also pulled the Gamma functions out without bringing along the exponent n. $\endgroup$
    – KirkD_CO
    Nov 1, 2017 at 5:21
  • $\begingroup$ 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. $\endgroup$
    – KirkD_CO
    Nov 2, 2017 at 3:59
  • $\begingroup$ Not need, since $ \frac{\Gamma(1+\theta)}{\Gamma(\theta)}=\frac{\theta !}{(\theta-1)!}=\theta$ $\endgroup$
    – Deep North
    Nov 2, 2017 at 4:12

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