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:
$$\text{Let }\; T = \frac{\partial}{\partial \theta}\bigg(n\ln \theta + (\theta -1)\sum_{i=1}^n \ln y_i\bigg) = \frac{n}{\theta} + \sum_\limits{i=1}^{n}ln(y_i) \\ \text{Let } \; T = 0 = \frac{n}{\theta} -0.261 - 0.198 - 0.083 - 0.061 - 0.02 \\ \text{implies } \hat{\theta} = \frac{5}{0.623} = 8$$
Continuing from where you left it, it simply is :
$$ L(\theta) = \theta^n\prod\limits_{i=1}^{n}y_i^{\theta-1} \Rightarrow \ln (L(\theta)) = \ln\bigg(\theta^n\prod\limits_{i=1}^{n}y_i^{\theta-1} \bigg) \Leftrightarrow \ln(L(\theta)) = n\ln\theta + \ln\bigg(\prod_{i=1}^ny_i^{\theta-1}\bigg) $$
$$\Leftrightarrow$$
$$\ln(L(\theta)) = n \ln \theta + \sum_{i=1}^n \ln(y_i^{\theta-1}) \Leftrightarrow \ln(L(\theta)) = n\ln \theta + (\theta -1)\sum_{i=1}^n \ln y_i$$
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 $e$.