Question

If \(X_{1}, X_{2}\) and \(X_{3}\) are three independent random variables having the Poisson distribution with the parameter \(\lambda\), show that $$ \hat{\lambda_{1}}=\frac{X_{1}+2 X_{2}+3 X_{3}}{6} $$ is an unbiased estimator of \(\lambda\). Also, compare the efficiency of \(\hat{\lambda_{1}}\) with that of the alternate estimator. $$ \hat{\lambda_{2}}=\frac{X_{1}+X_{2}+X_{3}}{3} $$.

Short answer

The estimator \(\hat{\lambda_{1}}\) is an unbiased estimator of \(\lambda\) as its expected value equals \(\lambda\). Comparing both estimators, it is found that \(\hat{\lambda_{1}}\) is less efficient than \(\hat{\lambda_{2}}\).

Step-by-step solution

Calculate the Expected value of \(\hat{\lambda_{1}}\)

The expected value E() of a function is equal to the sum of all possible values the random variable can take, multiplied by their respective probabilities. Considering that \(\hat{\lambda_{1}}=\frac{X_{1}+2 X_{2}+3 X_{3}}{6}\), it is needed to calculate the expected value of \(\hat{\lambda_{1}}\). We know that \(E(cX) = cE(X)\) and for Poisson distribution, \(E(X) = \lambda\), therefore \(E(\hat{\lambda_{1}}) = E(\frac{X_{1}+2 X_{2}+3 X_{3}}{6}) = \frac{1}{6}[E(X_{1}) + 2E(X_{2}) + 3E(X_{3})] = \frac{1}{6}[3\lambda] = \lambda\).

Calculate the Variance of \(\hat{\lambda_{1}}\)

Variance Var() of a random variable is the expected value of the square of the deviation of this variable from the expected value, more directly Var() = E\([(X - E(X))^2]\). Being so and knowing that the variance for a Poisson distribution Var(X) = \(\lambda\), we have: Var(\(\hat{\lambda_{1}}\)) = Var(\(\frac{X_{1}+2 X_{2}+3 X_{3}}{6}\)) which simplifies to \(\frac{1}{36}(Var(X_{1}) + 4Var(X_{2}) + 9Var(X_{3})) = \frac{1}{12}\lambda\).

Calculate the Expected value and Variance of \(\hat{\lambda_{2}}\)

Similarly, we calculate E(\(\hat{\lambda_{2}}\)) and Var(\(\hat{\lambda_{2}}\)). We know that \(\hat{\lambda_{2}}=\frac{X_{1}+X_{2}+X_{3}}{3}\), and using the same principles, we find E(\(\hat{\lambda_{2}}\)) = \(\lambda\) and Var(\(\hat{\lambda_{2}}\)) = \(\frac{1}{9}\lambda\).

Comparison of the Efficiency

Efficiency is a way of comparing estimators. It is calculated by taking the ratio of their variances. Therefore, if the efficiency is less than one, the numerator estimator is more efficient than the denominator estimator, and vice versa. The efficiency \(\frac{Var(\hat{\lambda_{1}})}{Var(\hat{\lambda_{2}})} = \frac{\frac{1}{12}\lambda}{\frac{1}{9}\lambda} = \frac{3}{4}\), which is less than one.

Key concepts

Poisson Distribution

The Poisson distribution is a discrete probability distribution that expresses the probability of a given number of events occurring in a fixed interval of time or space if these events occur with a known constant mean rate and independently of the time since the last event. This distribution is named after French mathematician Siméon Denis Poisson.

The Poisson distribution is widely used in fields such as physics, ecology, finance, and even traffic engineering. The key parameter of this distribution is \( \lambda \) which represents the average number of events in the interval. It's crucial to recognize that for a Poisson distribution, the mean and variance are both equal to \( \lambda \). This unique property is what enables us to handle problems involving Poisson processes.

Expected Value

In probability theory, the expected value (or mathematical expectation) of a random variable is a fundamental concept that provides a measure of the center of the distribution of the variable. Essentially, it is the average value you would expect to see over a large number of trials or occurrences of the stochastic process.

For a Poisson-distributed random variable \( X \) with a rate parameter \( \lambda \), its expected value is \( E(X) = \lambda \). In words, if we were to average out the counts of events over many intervals, we’d expect the count to be \( \lambda \) on average. When computing the expected value of a combination or function of independent Poisson random variables, we can use the linearity of expectation to simplify calculations, as shown in the exercise.

Variance of Estimator

The concept of variance in statistics quantifies how much a set of numbers is spread out. It’s one of the key measures of dispersion. When we're using an estimator - which is a function of random variables that estimates some parameter - the variance of the estimator gives us an idea of its precision; low variance corresponds to high precision, and high variance indicates less precision.

In the context of the Poisson distribution, the variance of an estimator, such as \( \hat{\lambda_{1}} \) or \( \hat{\lambda_{2}} \) in the exercise, tells us how much the estimator will typically deviate from the parameter \( \lambda \) it is trying to estimate. Calculating the variance of an estimator involves finding the expected value of the squared deviation of the estimator from its own expected value, and it provides insight into the reliability of the estimator.

Statistical Efficiency

Statistical efficiency is a measure of the performance of an estimator in terms of its variance. It is used to compare different estimators for the same parameter. An estimator that has smaller variance is said to be more efficient, as it is expected to be closer to the parameter it is estimating in repeated samples.

Efficiency is evaluated as a relative concept, usually by comparing the variance of two estimators. The estimator with the lower variance is deemed more efficient. In some cases, as with the exercise where the estimators \( \hat{\lambda_{1}} \) and \( \hat{\lambda_{2}} \) are compared, their efficiency can be quantitatively assessed by taking the ratio of their variances. Through such comparisons, statisticians can choose the estimator that offers the best trade-off between various aspects, including bias (or the lack thereof) and variance.