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Two different computer systems are monitored for a total of \(n\) weeks. Let \(X_{i}\) denote the number of breakdowns of the first system during the \(i\) th week, and suppose the \(X_{i}\) 's are independent and drawn from a Poisson distribution with parameter \(\lambda_{1}\). Similarly, let \(Y_{i}\) denote the number of breakdowns of the second system during the \(i\) th week, and assume independence with each \(Y_{i}\) Poisson with parameter \(\lambda_{2}\). Derive the mle's of \(\lambda_{1}, \lambda_{2}\), and \(\lambda_{1}-\lambda_{2}\). [Hint: Using independence, write the joint \(\mathrm{pmf}\) (likelihood) of the \(X_{i}\) 's and \(Y_{i}\) 's together.]

Short Answer

Expert verified
MLEs are \(\hat{\lambda}_1 = \frac{\sum x_i}{n}\), \(\hat{\lambda}_2 = \frac{\sum y_i}{n}\), and \(\widehat{\lambda_1 - \lambda_2} = \frac{\sum x_i}{n} - \frac{\sum y_i}{n}\).

Step by step solution

01

Understand the Problem

We need to find the maximum likelihood estimates (mles) for parameters \(\lambda_1\), \(\lambda_2\), and \(\lambda_1 - \lambda_2\) using the provided Poisson distribution model for breakdown counts of two systems over \(n\) weeks.
02

Define the Poisson Distribution

A random variable \(X_i\) that follows a Poisson distribution with parameter \(\lambda_1\) has the probability mass function (pmf): \[p(X_i = x_i) = \frac{e^{-\lambda_1} \lambda_1^{x_i}}{x_i!}\ \]Similarly for \(Y_i\), \[p(Y_i = y_i) = \frac{e^{-\lambda_2} \lambda_2^{y_i}}{y_i!}\ \] where \(x_i\) and \(y_i\) are the observed number of breakdowns in the \(i\)th week for the first and second systems, respectively.
03

Write the Joint pmf (Likelihood Function)

For \(n\) independent weeks, the likelihood function \(L(\lambda_1, \lambda_2)\) for all observed breakdowns \(X_i\)'s and \(Y_i\)'s is given by the product of individual probabilities:\[L(\lambda_1, \lambda_2) = \prod_{i=1}^n \frac{e^{-\lambda_1} \lambda_1^{x_i}}{x_i!} \times \prod_{i=1}^n \frac{e^{-\lambda_2} \lambda_2^{y_i}}{y_i!}\]which simplifies to:\[L(\lambda_1, \lambda_2) = e^{-n\lambda_1} \lambda_1^{\sum x_i} \times e^{-n\lambda_2} \lambda_2^{\sum y_i} \]ignoring the constant \(\prod_{i=1}^n \frac{1}{x_i!} \times \frac{1}{y_i!}\).
04

Find the Log-Likelihood Function

The log-likelihood function \(\ell(\lambda_1, \lambda_2)\) is obtained by taking the natural logarithm of the likelihood function:\[\ell(\lambda_1, \lambda_2) = -n\lambda_1 + \left(\sum x_i \right)\log\lambda_1 - n\lambda_2 + \left(\sum y_i \right)\log\lambda_2 \]
05

Differentiate and Solve for MLEs

To find the mles, take partial derivatives of \(\ell(\lambda_1, \lambda_2)\) with respect to \(\lambda_1\) and \(\lambda_2\), then set them to zero:- With respect to \(\lambda_1\): \[\frac{\partial \ell}{\partial \lambda_1} = -n + \frac{\sum x_i}{\lambda_1} = 0 \] Solving, we get \(\hat{\lambda}_1 = \frac{\sum x_i}{n}\).- With respect to \(\lambda_2\): \[\frac{\partial \ell}{\partial \lambda_2} = -n + \frac{\sum y_i}{\lambda_2} = 0 \] Solving, we get \(\hat{\lambda}_2 = \frac{\sum y_i}{n}\).
06

MLE of \(\lambda_1 - \lambda_2\)

The mle of \(\lambda_1 - \lambda_2\) is simply the difference between the mles of \(\lambda_1\) and \(\lambda_2\):\[\widehat{\lambda_1 - \lambda_2} = \hat{\lambda_1} - \hat{\lambda_2} = \frac{\sum x_i}{n} - \frac{\sum y_i}{n}\]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Poisson distribution
In statistical terms, a Poisson distribution is a probability distribution that describes the number of events that will occur within a fixed period or space, assuming that these events happen at a constant mean rate and independently of the time since the last event. This is especially useful in modeling situations where you are counting the number of occurrences, such as breakdowns of computer systems, as given in the original problem.
The probability mass function (pmf) for a Poisson-distributed random variable is given by the formula:
  • For a random variable \(X\) with parameter \(\lambda\) (average rate of occurrence):
  • \[ p(X = k) = \frac{e^{-\lambda} \lambda^k}{k!} \]
Here, \(k\) represents the number of events (breakdowns), \(\lambda\) is the average rate, and \(e\) is the base of the natural logarithm.
In the exercise, both computer systems follow a Poisson distribution with parameters \(\lambda_1\) and \(\lambda_2\) for the two systems, respectively. Each represents the average number of breakdowns expected in one week.
independence of random variables
Independence among random variables is a fundamental concept in probability theory. It implies that the occurrence of one event or outcome does not affect the occurrence of another. In mathematical terms, two random variables \(X\) and \(Y\) are independent if:
  • The joint probability distribution can be expressed as the product of their individual probability distributions:
  • \[P(X = x, Y = y) = P(X = x) \cdot P(Y = y)\]
In the provided exercise, the breakdowns \(X_i\) of the first system and \(Y_i\) of the second system are independent. This means that knowing the number of breakdowns in the first system offers no information about breakdowns in the second system, and vice versa. The independence assumption allows us to calculate the joint probability, or likelihood, as a product of the individual probabilities, which simplifies the computation of parameter estimates.
log-likelihood function
To compute maximum likelihood estimates (MLEs), we utilize the log-likelihood function. This transformation is beneficial because taking the natural logarithm of the likelihood function transforms products into sums, simplifying differentiation and calculation.
In the exercise context, the log-likelihood function for the parameters \(\lambda_1\) and \(\lambda_2\) is derived from the joint likelihood function of the Poisson random variables. It is represented by:
  • \(\ell(\lambda_1, \lambda_2) = -n\lambda_1 + (\sum x_i) \log\lambda_1 - n\lambda_2 + (\sum y_i) \log\lambda_2\)
Where \(n\) is the number of weeks being observed, and \(\sum x_i\) and \(\sum y_i\) are the total breakdowns of the two systems across all \(n\) weeks. Taking derivatives of this function helps find the parameter values that maximize the likelihood, hence obtaining the MLEs.
parameter estimation
Parameter estimation involves finding numerical values for the parameters of a probability distribution that makes the observed data most likely. With maximum likelihood estimation (MLE), it involves maximizing the likelihood or log-likelihood function with respect to the parameters.In our exercise, the parameters in question are \(\lambda_1\) and \(\lambda_2\), which represent the average breakdown rate for each system. The MLE for each parameter is calculated by taking the derivative of the log-likelihood function with respect to the parameter, setting it to zero, and solving for the parameter. For this problem, we obtain:
  • \(\hat{\lambda}_1 = \frac{\sum x_i}{n}\)
  • \(\hat{\lambda}_2 = \frac{\sum y_i}{n}\)
These represent the estimated average number of breakdowns per week for each system. These estimates are intuitive since they are simply the average number of observed breakdowns over \(n\) weeks. Furthermore, the estimator for the difference \(\lambda_1 - \lambda_2\) is derived by subtracting the two estimates, giving insight into which system is more reliable.

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Most popular questions from this chapter

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