Let X₁, ..., Xn € [X], where X r.v. with pdf 0ï³−¹1(0,1)(x) w.r.t. the unknown parameter 0 > 0. Find the m.l.e. and MLE of 0. Hint. The likelihood function of the sample is 6* 3

MATLAB: An Introduction with Applications
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(We drop \( I_{(0,1)}-(x_1, \ldots, x_n) \) for convenience). Thus

\[
\frac{\partial}{\partial \theta} L(\theta) = \text{set it equal to } 0
\]

Finding \(\theta = -\frac{n}{\sum_{i=1}^{n} \ln x_i}\) as a critical point of \(L\). (Explain why \(\theta\) is positive.) Show that \( -\frac{n}{\sum_{i=1}^{n} \ln x_i}\) is a maximum point of \(L\) by checking the signs of \(L\) on the left (positive) and right (negative) of \(-\frac{n}{\sum_{i=1}^{n} \ln x_i}\). Therefore

\[
\hat{\theta} = -\frac{n}{\sum_{i=1}^{n} \ln x_i}
\]

and the MLE is

\[
\hat{\theta} = -\frac{n}{\sum_{i=1}^{n} \ln X_i}.
\]

[End of Text]
Transcribed Image Text:(We drop \( I_{(0,1)}-(x_1, \ldots, x_n) \) for convenience). Thus \[ \frac{\partial}{\partial \theta} L(\theta) = \text{set it equal to } 0 \] Finding \(\theta = -\frac{n}{\sum_{i=1}^{n} \ln x_i}\) as a critical point of \(L\). (Explain why \(\theta\) is positive.) Show that \( -\frac{n}{\sum_{i=1}^{n} \ln x_i}\) is a maximum point of \(L\) by checking the signs of \(L\) on the left (positive) and right (negative) of \(-\frac{n}{\sum_{i=1}^{n} \ln x_i}\). Therefore \[ \hat{\theta} = -\frac{n}{\sum_{i=1}^{n} \ln x_i} \] and the MLE is \[ \hat{\theta} = -\frac{n}{\sum_{i=1}^{n} \ln X_i}. \] [End of Text]
**Problem Statement:**

6*. Let \( X_1, \ldots, X_n \in [X] \), where \( X \) is a random variable with probability density function \( \theta x^{\theta - 1} 1_{(0,1)}(x) \) with respect to the unknown parameter \( \theta > 0 \). Find the maximum likelihood estimator (m.l.e.) and maximum likelihood estimate (MLE) of \( \theta \).

*Hint*: The likelihood function of the sample is

\[ f_n(x_1, \ldots, x_n | \theta) = \theta^n \left( \prod_{i=1}^{n} x_i \right)^{\theta - 1} 1_{(0,1)}(x_1, \ldots, x_n). \]

Denote \( L(\theta) = \ln f_n(x_1, \ldots, x_n | \theta) \). Then,

\[ L(\theta) = n \ln \theta + (\theta - 1) \sum_{i=1}^{n} \ln x_i. \]

(We drop \( 1_{(0,1)}(x_1, \ldots, x_n) \) for convenience). Thus, ...
Transcribed Image Text:**Problem Statement:** 6*. Let \( X_1, \ldots, X_n \in [X] \), where \( X \) is a random variable with probability density function \( \theta x^{\theta - 1} 1_{(0,1)}(x) \) with respect to the unknown parameter \( \theta > 0 \). Find the maximum likelihood estimator (m.l.e.) and maximum likelihood estimate (MLE) of \( \theta \). *Hint*: The likelihood function of the sample is \[ f_n(x_1, \ldots, x_n | \theta) = \theta^n \left( \prod_{i=1}^{n} x_i \right)^{\theta - 1} 1_{(0,1)}(x_1, \ldots, x_n). \] Denote \( L(\theta) = \ln f_n(x_1, \ldots, x_n | \theta) \). Then, \[ L(\theta) = n \ln \theta + (\theta - 1) \sum_{i=1}^{n} \ln x_i. \] (We drop \( 1_{(0,1)}(x_1, \ldots, x_n) \) for convenience). Thus, ...
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