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.

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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 \(\lambda\).
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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