Question:

Give step-by-step solution with explanation and final answer: Problem: 4: Exponential distribution: MLE, Fisher info, CI Solution: This problem involves parameter estimation for an exponential distribution using Type II censored data. A total of $n=12$ articles are tested, and the experiment stops after $r=9$ failures have been observed. The failure times are $t_1, \dots, t_9$. The remaining $n-r=3$ articles have lifetimes longer than the last observed failure time, $t_9 = 109$. The probability density function (PDF) of the exponential distribution with mean $\mu$ is given by $f(t; \mu) = \frac{1}{\mu} e^{-t/\mu}$ for $t \ge 0$. The corresponding survival function is $S(t; \mu) = P(T > t) = e^{-t/\mu}$. ### (a) Maximum Likelihood Estimate of Mean $\mu$ 1. **Construct the Likelihood Function** For Type II censored data, the likelihood function $L(\mu)$ is the product of the PDFs for the $r$ failed items and the survival probabilities for the $n-r$ items that did not fail. $$ L(\mu) = \left[ \prod_{i=1}^{r} f(t_i; \mu) \right] \left[ S(t_r; \mu) \right]^{n-r} $$ Substituting the expressions for the exponential distribution with $r=9$ and $n=12$: $\begin{aligned} L(\mu) &= \left[ \prod_{i=1}^{9} \frac{1}{\mu} e^{-t_i/\mu} \right] \left[ e^{-t_9/\mu} \right]^{12-9} \\ &= \left( \frac{1}{\mu^9} e^{-\frac{1}{\mu}\sum_{i=1}^{9} t_i} \right) \left( e^{-3t_9/\mu} \right) \\ &= \frac{1}{\mu^9} \exp\left( -\frac{1}{\mu} \left( \sum_{i=1}^{9} t_i + 3t_9 \right) \right) \end{aligned}$ 2. **Determine the Log-Likelihood Function** To simplify differentiation, we work with the natural logarithm of the likelihood function, $l(\mu) = \ln L(\mu)$. Let $T = \sum_{i=1}^{9} t_i + 3t_9$ be the total time on test. $$ l(\mu) = \ln\left(\frac{1}{\mu^9} e^{-T/\mu}\right) = -9 \ln(\mu) - \frac{T}{\mu} $$ 3. **Find the Maximum Likelihood Estimate (MLE)** Differentiate $l(\mu)$ with respect to $\mu$ and set the result to zero to find the MLE, $\hat{\mu}$. $$ \frac{dl}{d\mu} = -\frac{9}{\mu} + \frac{T}{\mu^2} $$ Setting the derivative to zero: $\begin{aligned} -\frac{9}{\hat{\mu}} + \frac{T}{\hat{\mu}^2} &= 0 \\ \frac{T}{\hat{\mu}^2} &= \frac{9}{\hat{\mu}} \\ \hat{\mu} &= \frac{T}{9} \end{aligned}$ 4. **Calculate the Numerical Value** The observed failure times are $8, 14, 23, 32, 46, 57, 69, 88, 109$. $\begin{aligned} \sum_{i=1}^{9} t_i &= 8 + 14 + 23 + 32 + 46 + 57 + 69 + 88 + 109 = 446 \\ T &= 446 + 3 \times 109 = 446 + 327 = 773 \\ \hat{\mu} &= \frac{773}{9} \approx 85.89 \end{aligned}$ $\boxed{\hat{\mu} = \frac{773}{9} \approx 85.89}$ ### (b) Fisher Information for $\hat{\mu}$ 1. **Derive the Fisher Information** The Fisher information, $I(\mu)$, is the negative expectation of the second derivative of the log-likelihood. First, find the second derivative: $$ \frac{d^2l}{d\mu^2} = \frac{d}{d\mu}\left(-\frac{9}{\mu} + \frac{T}{\mu^2}\right) = \frac{9}{\mu^2} - \frac{2T}{\mu^3} $$ For Type II censored data from an exponential distribution, the expected total time on test is $E[T] = r\mu = 9\mu$. $\begin{aligned} I(\mu) &= -E\left[\frac{d^2l}{d\mu^2}\right] = -E\left[\frac{9}{\mu^2} - \frac{2T}{\mu^3}\right] \\ &= \frac{2E[T]}{\mu^3} - \frac{9}{\mu^2} = \frac{2(9\mu)}{\mu^3} - \frac{9}{\mu^2} \\ &= \frac{18}{\mu^2} - \frac{9}{\mu^2} = \frac{9}{\mu^2} \end{aligned}$ 2. **Calculate the Observed Fisher Information** The question asks for the Fisher information for $\hat{\mu}$. This is typically interpreted as the observed Fisher information, $I(\hat{\mu})$, which is obtained by evaluating $I(\mu)$ at $\mu = \hat{\mu}$. $$ I(\hat{\mu}) = \frac{9}{\hat{\mu}^2} = \frac{9}{(773/9)^2} = \frac{9 \times 81}{773^2} = \frac{729}{597529} \approx 0.00122 $$ $\boxed{I(\hat{\mu}) = \frac{729}{597529} \approx 0.00122}$ ### (c) 90% Confidence Interval for $\mu$ 1. **Define the Pivot Quantity** We use the quantity $Z = (\hat{\mu} - \mu) / \text{se}(\hat{\mu})$, which is approximately standard normal for large $r$. The standard error of the estimate, $\text{se}(\hat{\mu})$, is the square root of the estimated variance of $\hat{\mu}$. $$ \text{Var}(\hat{\mu}) \approx \frac{1}{I(\mu)} = \frac{\mu^2}{9} $$ We estimate the variance by substituting $\hat{\mu}$ for $\mu$: $$ \widehat{\text{Var}}(\hat{\mu}) = \frac{\hat{\mu}^2}{9} $$ The standard error is: $$ \text{se}(\hat{\mu}) = \sqrt{\frac{\hat{\mu}^2}{9}} = \frac{\hat{\mu}}{3} $$ 2. **Construct the Confidence Interval** For a 90% confidence interval, the confidence level is $1-\alpha = 0.90$, so $\alpha = 0.10$. The critical value from the standard normal distribution is $z_{\alpha/2} = z_{0.05} \approx 1.645$. The confidence interval is given by: $$ \hat{\mu} \pm z_{\alpha/2} \cdot \text{se}(\hat{\mu}) $$ 3. **Calculate the Numerical Interval** First, calculate the standard error: $$ \text{se}(\hat{\mu}) = \frac{773/9}{3} = \frac{773}{27} \approx 28.630 $$ Next, calculate the margin of error: $$ E = z_{0.05} \cdot \text{se}(\hat{\mu}) \approx 1.645 \times \frac{773}{27} \approx 47.096 $$ The 90% confidence interval is: $\begin{aligned} \text{Lower bound} &= \hat{\mu} - E \approx 85.889 - 47.096 = 38.793 \\ \text{Upper bound} &= \hat{\mu} + E \approx 85.889 + 47.096 = 132.985 \end{aligned}$ $\boxed{(38.79, 132.98)}$ Act like chegg expert tutor and rewrite the same solution as it is