Suppose A Weather Reporting Station Has An Electronic Wind S
Suppose A Weather Reporting Station Has an Electronic Wind Speed Monit
Suppose a weather-reporting station has an electronic wind speed monitor whose time to failure is known to follow an exponential model with the probability density function (pdf) \(f_{X}(x; \theta) = \frac{1}{\theta} e^{-x/\theta}\), for \(x > 0\) and \( \theta > 0\). The station also has a backup monitor; the time until this second monitor fails is similarly modeled, and the total time until the monitoring system is out of commission is the sum of two independent exponential variables, \(Y_1\) and \(Y_2\). Consequently, the total time \(Y = Y_1 + Y_2\) follows a gamma distribution with the pdf \(f_{Y}(y; \theta) = \frac{1}{\theta^2} y e^{- y / \theta}\) for \(y \geq 0\), with \(0
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The primary goal of this analysis is to estimate the parameter \(\theta\) of the gamma distribution that models the total failure time \(Y\) of a weather monitoring station's system comprising two independent exponential monitors. By deriving the maximum likelihood estimate (MLE) of \(\theta\), we aim to quantify the typical failure duration and improve maintenance scheduling and reliability assessments for such systems.
Firstly, understanding the distribution governing the sum of two independent exponential variables is essential. The sum of two exponential variables with identical scale parameters \(\theta\) results in a gamma distribution with shape parameter 2 and scale parameter \(\theta\). Specifically, the pdf is given by:
\[f_Y(y; \theta) = \frac{y^{2 - 1} e^{- y / \theta}}{\theta^2 \Gamma(2)} = \frac{1}{\theta^2} y e^{- y / \theta}, \quad y \geq 0.\]
Given the observed data points \( y_1 = 9.2 \), \( y_2 = 5.6 \), \( y_3 = 18.4 \), \( y_4 = 12.1 \), and \( y_5 = 10.7 \), the likelihood function \(L(\theta)\) for these observations, assuming their independence, is the product of individual densities:
\[
L(\theta) = \prod_{i=1}^5 f_Y(y_i; \theta) = \prod_{i=1}^5 \frac{1}{\theta^2} y_i e^{- y_i / \theta}.
\]
Expressed more explicitly, the likelihood becomes:
\[
L(\theta) = \left(\frac{1}{\theta^2}\right)^5 \prod_{i=1}^5 y_i \times e^{-\sum_{i=1}^5 y_i / \theta} = \frac{\prod_{i=1}^5 y_i}{\theta^{10}} e^{-\left(\sum_{i=1}^5 y_i\right) / \theta}.
\]
To simplify the maximization process, it is standard to work with the log-likelihood function \( \ell(\theta) \):
\[
\ell(\theta) = \ln L(\theta) = \sum_{i=1}^5 \left[\ln y_i - 2 \ln \theta \right] - \frac{\sum y_i}{\theta} = \left(\sum_{i=1}^5 \ln y_i \right) - 10 \ln \theta - \frac{\sum y_i}{\theta}.
\]
Calculating the sum of the observed data points and the sum of their logarithms is necessary:
\[
\sum y_i = 9.2 + 5.6 + 18.4 + 12.1 + 10.7 = 56.0,
\]
\[
\sum \ln y_i = \ln 9.2 + \ln 5.6 + \ln 18.4 + \ln 12.1 + \ln 10.7 \approx 2.219 + 1.722 + 2.913 + 2.491 + 2.369 = 11.714.
\]
Thus, the log-likelihood function simplifies to:
\[
\ell(\theta) = 11.714 - 10 \ln \theta - \frac{56.0}{\theta}.
\]
To find the maximum likelihood estimate, differentiate with respect to \(\theta\) and set the derivative to zero:
\[
\frac{d\ell}{d\theta} = -\frac{10}{\theta} + \frac{56.0}{\theta^2} = 0.
\]
Multiplying through by \(\theta^2\) to clear denominators:
\[
-10 \theta + 56.0 = 0,
\]
\[
10 \theta = 56.0,
\]
\[
\theta = \frac{56.0}{10} = 5.6.
\]
This critical point corresponds to the maximum of the likelihood (since the likelihood function is concave in \(\theta\)), and therefore, the maximum likelihood estimate (MLE) of \(\theta\) is approximately 5.6.
In conclusion, based on the available data and the gamma distribution model for the total failure time, the estimated \(\theta\) parameter—which signifies the scale parameter of the underlying exponential distributions—is 5.6. This estimate can be employed to assess the reliability and expected failure times of similar systems, guiding maintenance planning and system improvements.
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