I'm trying to find the maximum likelihood of this function. I have samples in this question as follows: (0.77, 0.82, 0.94, 0.92, 0.98)

$$f_Y(y;\theta)=\theta y^{\theta-1} ;, \quad 0 \le y \le 1 ;, 0 \lt \theta$$

$$L(\theta) = \prod\limits_{i=1}^{n} \theta y_i^{\theta-1} \\ = \theta^n\prod\limits_{i=1}^{n}y_i^{\theta-1}$$

From here i'm stuck. I'm not sure if I should take the log now or there is one more move before I take the log. The answer is 8.00. if I put the values it would probably make it easier for me, but it appears to me that I do not know a technique or identity from here to move on. Any help?

Continuing on:

$$\text{Let }\; T = \frac{\partial}{\partial \theta}\bigg(n\ln \theta + (\theta -1)\sum_{i=1}^n \ln y_i\bigg) = \frac{n}{\theta} + \sum_\limits{i=1}^{n}ln(y_i) \\ \text{Let } \; T = 0 = \frac{n}{\theta} -0.261 - 0.198 - 0.083 - 0.061 - 0.02 \\ \text{implies } \hat{\theta} = \frac{5}{0.623} = 8$$

  • 1
    $\begingroup$ The MLE of the $\text{Beta}(\theta,1)$ distribution was derived here. $\endgroup$ – StubbornAtom Oct 4 '18 at 20:38
  • $\begingroup$ Thankyou @StubbornAtom that was a helpful article on R and bootstrap too. $\endgroup$ – Bucephalus Oct 4 '18 at 21:36

Continuing from where you left it, it simply is :

$$ L(\theta) = \theta^n\prod\limits_{i=1}^{n}y_i^{\theta-1} \Rightarrow \ln (L(\theta)) = \ln\bigg(\theta^n\prod\limits_{i=1}^{n}y_i^{\theta-1} \bigg) \Leftrightarrow \ln(L(\theta)) = n\ln\theta + \ln\bigg(\prod_{i=1}^ny_i^{\theta-1}\bigg) $$


$$\ln(L(\theta)) = n \ln \theta + \sum_{i=1}^n \ln(y_i^{\theta-1}) \Leftrightarrow \ln(L(\theta)) = n\ln \theta + (\theta -1)\sum_{i=1}^n \ln y_i$$

To find the MLE for the given function, differentiate by $\theta$ and solve with respect to $\theta$ so you get the estimator $\hat{\theta}$:

$$\frac{\partial \ln(L(\theta))}{\partial \theta} = \frac{\partial}{\partial \theta}\bigg(n\ln \theta + (\theta -1)\sum_{i=1}^n \ln y_i\bigg) \Leftrightarrow \dots$$

Finally, calculate the MLE values for the given points you have but remember, the likelihood is yielded by using the $\ln$ function, which means that your final values should be exponanted to $e$.

  • 1
    $\begingroup$ $ \ln\bigg(\prod_{i=1}^ny_i^{\theta-1}\bigg) = \sum_{i=1}^n \ln(y_i^{\theta-1})$, good one. Thankyou @Rebellos $\endgroup$ – Bucephalus Oct 4 '18 at 11:02
  • 1
    $\begingroup$ I got that to the answer in the book, but what you're saying is, although that might satisfy the book in interrogating knowledge of the maximum likelihood process, the figure is not really useful in that form until it is brought back into the same "units". $\endgroup$ – Bucephalus Oct 4 '18 at 11:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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