The number of connections to a wrong phone number is often modeled by a Poisson distribution. To estimate the unknown parameter λ of that distribution (the mean number of wrong connections) one observes a sample ₁,..., n of wrong connections on n different days. Assum- ing that σ = x₁ + ... + în > 0, (i)give the likelihood function of the sample; (ii) find the m.l.e. and MLE of X. Hint. (i) If f(x|X) = e, then show that the likelihood function is 2 fn(Xn|X) = en\X° /[[2j!(o= 21+ .tin) j=1 22 (ii) L(X) = lnfn (Xñ |A) = get it. After finding L'(X), show that L'(X) = 0 iff λ = o/n = πn. Now, σ/n is a critical point of L(A). But we can rewrite L' as L'(X) = (n − X) showing that Thus, î= L' is positive for > < xn, L' is equal zero when λ = n, L'is negative when > > n. It proves that în is a local and hence the global maximum point of L and therefore of f(x₂|X). ... = n is the m.l.e. of X and  = Xn is the MLE of X.
The number of connections to a wrong phone number is often modeled by a Poisson distribution. To estimate the unknown parameter λ of that distribution (the mean number of wrong connections) one observes a sample ₁,..., n of wrong connections on n different days. Assum- ing that σ = x₁ + ... + în > 0, (i)give the likelihood function of the sample; (ii) find the m.l.e. and MLE of X. Hint. (i) If f(x|X) = e, then show that the likelihood function is 2 fn(Xn|X) = en\X° /[[2j!(o= 21+ .tin) j=1 22 (ii) L(X) = lnfn (Xñ |A) = get it. After finding L'(X), show that L'(X) = 0 iff λ = o/n = πn. Now, σ/n is a critical point of L(A). But we can rewrite L' as L'(X) = (n − X) showing that Thus, î= L' is positive for > < xn, L' is equal zero when λ = n, L'is negative when > > n. It proves that în is a local and hence the global maximum point of L and therefore of f(x₂|X). ... = n is the m.l.e. of X and  = Xn is the MLE of X.
MATLAB: An Introduction with Applications
6th Edition
ISBN:9781119256830
Author:Amos Gilat
Publisher:Amos Gilat
Chapter1: Starting With Matlab
Section: Chapter Questions
Problem 1P
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For the image above, a question and process for solving it are included. Please show work and explain how the answer is found for anything in the image that states "explain." The desired answer should be shown at the end of the work please.
![2. The number of connections to a wrong phone number is often modeled by a Poisson distribution. To estimate the unknown parameter \(\lambda\) of that distribution (the mean number of wrong connections) one observes a sample \(x_1, \ldots, x_n\) of wrong connections on \(n\) different days. Assuming that \(\sigma = x_1 + \ldots + x_n > 0\), (i) give the likelihood function of the sample; (ii) find the m.l.e. and MLE of \(\lambda\).
**Hint**. (i) If \(f(x|\lambda) = e^{-\lambda} \frac{\lambda^x}{x!}\), then show that the likelihood function is
\[
f_n(x_n|\lambda) = e^{-n\lambda} \frac{\lambda^\sigma}{\prod_{j=1}^n x_j!}(\sigma = x_1 + \ldots + x_n)
\]
(ii) \(L(\lambda) = \ln f_n(x_n|\lambda) = \text{get it.}\)
After finding \(L'(\lambda)\), show that \(L'(\lambda) = 0\) iff \(\lambda = \sigma/n = \bar{x}_n\).
Now, \(\sigma/n\) is a critical point of \(L(\lambda)\). But we can rewrite \(L'\) as \(L'(\lambda) = \frac{1}{\lambda_n}(\bar{x}_n - \lambda)\) showing that
\[
\begin{align*}
L' \text{ is positive for } \lambda < \bar{x}_n, \\
L' \text{ is equal to zero when } \lambda = \bar{x}_n, \\
L' \text{ is negative when } \lambda > \bar{x}_n.
\end{align*}
\]
It proves that \(\bar{x}_n\) is a local and hence the global maximum point of \(L\) and therefore of \(f_n(x_n|\lambda)\).
Thus, \(\hat{\lambda} = \bar{x}_n\) is the m.l.e. of \(\lambda\) and
\[
\hat{\Lambda} = X_n
\]
is the MLE of \(\lambda\).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fdb82eee3-c597-43b7-b98f-cffc18fbca72%2F64eae050-45c6-4437-89f4-80b8ed083e71%2Fxisyy8a_processed.png&w=3840&q=75)
Transcribed Image Text:2. The number of connections to a wrong phone number is often modeled by a Poisson distribution. To estimate the unknown parameter \(\lambda\) of that distribution (the mean number of wrong connections) one observes a sample \(x_1, \ldots, x_n\) of wrong connections on \(n\) different days. Assuming that \(\sigma = x_1 + \ldots + x_n > 0\), (i) give the likelihood function of the sample; (ii) find the m.l.e. and MLE of \(\lambda\).
**Hint**. (i) If \(f(x|\lambda) = e^{-\lambda} \frac{\lambda^x}{x!}\), then show that the likelihood function is
\[
f_n(x_n|\lambda) = e^{-n\lambda} \frac{\lambda^\sigma}{\prod_{j=1}^n x_j!}(\sigma = x_1 + \ldots + x_n)
\]
(ii) \(L(\lambda) = \ln f_n(x_n|\lambda) = \text{get it.}\)
After finding \(L'(\lambda)\), show that \(L'(\lambda) = 0\) iff \(\lambda = \sigma/n = \bar{x}_n\).
Now, \(\sigma/n\) is a critical point of \(L(\lambda)\). But we can rewrite \(L'\) as \(L'(\lambda) = \frac{1}{\lambda_n}(\bar{x}_n - \lambda)\) showing that
\[
\begin{align*}
L' \text{ is positive for } \lambda < \bar{x}_n, \\
L' \text{ is equal to zero when } \lambda = \bar{x}_n, \\
L' \text{ is negative when } \lambda > \bar{x}_n.
\end{align*}
\]
It proves that \(\bar{x}_n\) is a local and hence the global maximum point of \(L\) and therefore of \(f_n(x_n|\lambda)\).
Thus, \(\hat{\lambda} = \bar{x}_n\) is the m.l.e. of \(\lambda\) and
\[
\hat{\Lambda} = X_n
\]
is the MLE of \(\lambda\).
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