Question:
Let \( \{0.90, 0.50, 0.01, 0.95\} \) be a realization of a random sample of size 4 from the probability density function
\[
f(x) = \begin{cases}
\frac{\theta}{(1-\theta)} x^{(2\theta-1)/(1-\theta)}, & 0<x<1 \\
0, & \text{otherwise}
\end{cases}
\]
where \( 0.5 \leq \theta<1 \). Then the maximum likelihood estimate of \( \theta \) based on the observed sample equals
\[
\underline{\hspace{2cm}}
\]
(round off to 2 decimal places).
Let \( \{0.90, 0.50, 0.01, 0.95\} \) be a realization of a random sample of size 4 from the probability density function
\[ f(x) = \begin{cases} \frac{\theta}{(1-\theta)} x^{(2\theta-1)/(1-\theta)}, & 0<x<1 \\ 0, & \text{otherwise} \end{cases} \]
where \( 0.5 \leq \theta<1 \). Then the maximum likelihood estimate of \( \theta \) based on the observed sample equals\[ \underline{\hspace{2cm}} \]
(round off to 2 decimal places).Show Hint
To estimate parameters using the maximum likelihood method, construct the likelihood function and maximize it with respect to the parameter of interest.
Updated On: Dec 29, 2025
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Correct Answer: 0.5
Solution and Explanation
The likelihood function is constructed using the given probability density function. Taking the log-likelihood and maximizing it, we find the maximum likelihood estimate of \( \theta \):
\[
\hat{\theta} \approx 0.50.
\]
Thus, the value is \( 0.50 \).
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