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Suppose the random variables $X_1, X_2, . . .X_n$ are independent each with the distribution

$$f(x;\theta) = \frac{\theta \ x^{\theta-1}}{3^\theta} \ \ \text{ for} \ \ 0 \leq x \leq 3.$$

Find the Maximum Likelihood estimate for $\theta$.

Jean Marie
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user3026388
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  • The likelihood function is $L(\theta)=f(x_1,\theta)f(x_2,\theta)...f(x_n,\theta)=\theta^n(x_1x_2...x_n)^{\theta-1}/3^{n\theta}$.

    As it is often the case it is easier to maximize $$\ln\left(L(\theta)\right)=n\ln(\theta)+(\theta-1)\ln(x_1x_2...x_n)-n\theta\ln(3)$$

    Take derivatives and equate to zero

    $$0=\frac{n}{\theta}+\ln(x_1x_2...x_n)-n\ln(3)$$

    We get that $$\theta=\frac{n}{n\ln(3)-\ln(x_1x_2...x_n)}$$

     – OR. Jun 22 '17 at 13:01
  • You have been asking 5 questions in a 7 hours time slot... Don't you think it's time to slow down ? We do not help you on the long term by serving you solutions on a tray. In this example, it is very strange that you haven't reached the step $\ln\left(L(\theta)\right)=n\ln(\theta)+(\theta-1)\ln(x_1x_2...x_n)-n\theta\ln(3)$ given by Mlazhinka etc... – Jean Marie Jun 22 '17 at 13:10

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