1 To test a proportion of defective items one draws a sample Xn population with an unknown 0 € (0, 1). (₂) Give the likelihood function of the sample. (22) Find the m.l.e. and MLE of 0. Hint. (2) We can write the density f(x|0) of each r.v. X as 0² (10)¹-1(0.1) (0), ƒ (x|0) = {' {0² (1. 0, x = 0,1 otherwise. Thus, the likelihood function is fn (₂0) = 0° (1 - 0) 1 (0,1) (0) where = L' (0) = n(i=8); where is the sample mean of x₁,..., n. (X₁,..., Xn) from a Bernoulli 0 = = 2₁+ ... +än. (Explain why.) (22) The value of that maximizes the likelihood function is the same value that maximizes the log of fn (n) since the log function is strictly monotone. So, let L(0) : = Infn (xn|0) = oln0 + (n − o)ln(1 — 0). Then, show that
1 To test a proportion of defective items one draws a sample Xn population with an unknown 0 € (0, 1). (₂) Give the likelihood function of the sample. (22) Find the m.l.e. and MLE of 0. Hint. (2) We can write the density f(x|0) of each r.v. X as 0² (10)¹-1(0.1) (0), ƒ (x|0) = {' {0² (1. 0, x = 0,1 otherwise. Thus, the likelihood function is fn (₂0) = 0° (1 - 0) 1 (0,1) (0) where = L' (0) = n(i=8); where is the sample mean of x₁,..., n. (X₁,..., Xn) from a Bernoulli 0 = = 2₁+ ... +än. (Explain why.) (22) The value of that maximizes the likelihood function is the same value that maximizes the log of fn (n) since the log function is strictly monotone. So, let L(0) : = Infn (xn|0) = oln0 + (n − o)ln(1 — 0). Then, show that
MATLAB: An Introduction with Applications
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Author:Amos Gilat
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Chapter1: Starting With Matlab
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Transcribed Image Text:The text describes how the derivative of the likelihood function, denoted \(L'(\theta)\), changes sign from positive to negative as it passes through \(\bar{x}_n\). Here, \(\bar{x}_n\) is between 0 and 1, representing the empirical sample mean, which serves as the maximum likelihood estimator (MLE) of \(\theta\).
An additional note clarifies that a more detailed analysis, which distinguishes cases where \(\bar{x}_n = 0\) or \(\bar{x}_n = n\) from other values within (0, 1), yields the same result.
As a conclusion, the MLE of the sample is expressed as \(\hat{\theta} = \bar{X}_n\).
There are no graphs or diagrams to explain.
![**1**
To test a proportion of defective items, one draws a sample \( \mathbf{X_n} = (X_1, \ldots, X_n) \) from a Bernoulli population with an unknown \( \theta \in (0, 1) \).
(i)
- Give the likelihood function of the sample.
(ii)
- Find the m.l.e. and MLE of \( \theta \).
**Hint. (i)** We can write the density \( f(x|\theta) \) of each r.v. \( X_k \) as
\[
f(x|\theta) =
\begin{cases}
\theta^x (1-\theta)^{1-x} \mathbf{1}_{(0,1)}(\theta), & x = 0, 1 \\
0, & \text{otherwise}.
\end{cases}
\]
Thus, the likelihood function is
\[
f_n(\mathbf{x_n}|\theta) = \theta^\sigma (1-\theta)^{n-\sigma} \mathbf{1}_{(0,1)}(\theta)
\]
where
\[
\sigma := x_1 + \ldots + x_n. \quad (\text{Explain why.})
\]
(ii)
The value of \( \hat{\theta} \) that maximizes the likelihood function is the same value that maximizes the log of \( f_n(\mathbf{x_n}|\theta) \) since the log function is strictly monotone. So, let
\[
L(\theta) := \ln f_n(\mathbf{x_n}|\theta) = \sigma \ln \theta + (n - \sigma) \ln (1 - \theta).
\]
Then, show that
\[
L'(\theta) = n \frac{\bar{x_n} - \theta}{\theta (1 - \theta)};
\]
where \( \bar{x_n} \) is the sample mean of \( x_1, \ldots, x_n \).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fdb82eee3-c597-43b7-b98f-cffc18fbca72%2F4c55efb2-dc1e-4bf0-898b-11ebd4f95960%2Fo8687zr_processed.png&w=3840&q=75)
Transcribed Image Text:**1**
To test a proportion of defective items, one draws a sample \( \mathbf{X_n} = (X_1, \ldots, X_n) \) from a Bernoulli population with an unknown \( \theta \in (0, 1) \).
(i)
- Give the likelihood function of the sample.
(ii)
- Find the m.l.e. and MLE of \( \theta \).
**Hint. (i)** We can write the density \( f(x|\theta) \) of each r.v. \( X_k \) as
\[
f(x|\theta) =
\begin{cases}
\theta^x (1-\theta)^{1-x} \mathbf{1}_{(0,1)}(\theta), & x = 0, 1 \\
0, & \text{otherwise}.
\end{cases}
\]
Thus, the likelihood function is
\[
f_n(\mathbf{x_n}|\theta) = \theta^\sigma (1-\theta)^{n-\sigma} \mathbf{1}_{(0,1)}(\theta)
\]
where
\[
\sigma := x_1 + \ldots + x_n. \quad (\text{Explain why.})
\]
(ii)
The value of \( \hat{\theta} \) that maximizes the likelihood function is the same value that maximizes the log of \( f_n(\mathbf{x_n}|\theta) \) since the log function is strictly monotone. So, let
\[
L(\theta) := \ln f_n(\mathbf{x_n}|\theta) = \sigma \ln \theta + (n - \sigma) \ln (1 - \theta).
\]
Then, show that
\[
L'(\theta) = n \frac{\bar{x_n} - \theta}{\theta (1 - \theta)};
\]
where \( \bar{x_n} \) is the sample mean of \( x_1, \ldots, x_n \).
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