Question

Show that the average value of a random variable \(n\) whose probability function is the Poisson distribution (9.8) is the number \(\mu\) in \((9.8) .\) Also show that the standard deviation of the random variable is \(\sqrt{\mu} .\) Hint. Write the infinite series for \(e^{x}\), differentiate it and multiply by \(x\) to get \(x e^{x}=\sum\left(n x^{n} / n !\right) ;\) put \(x=\mu .\) To find \(\sigma^{2}\) differentiate the \(x e^{x}\) series again, etc.

Short answer

The average value is \( \mu \) and the standard deviation is \( \sqrt{\mu} \).

Step-by-step solution

Understand the Poisson Distribution

The Poisson distribution is defined by its probability mass function (pmf): \[ P(X = k) = \frac{\mu^k e^{-\mu}}{k!} \] where \( \mu \) is the mean (or average) of the distribution and \( k \) is a non-negative integer.

Write the Infinite Series for \( e^x \)

The exponential function \( e^x \) can be expressed as an infinite series: \[ e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!} \]

Differentiate the Series and Multiply by \( x \)

Differentiate both sides of the series with respect to \( x \) and then multiply by \( x \): \[ \frac{d}{dx}(e^x) = e^x \] Multiplying by \( x \): \[ x e^x = x \sum_{n=0}^{\infty} \frac{x^n}{n!} = \sum_{n=1}^{\infty} \frac{x^n}{(n-1)!} \] Notice how the summation index shifts after differentiation.

Rewrite the Series and Substitute \( x = \mu \)

Rewrite the series with the correct summation limit: \[ x e^x = \sum_{n=1}^{\infty} \frac{x^n}{(n-1)!} = \sum_{n=0}^{\infty} \frac{x^{n+1}}{n!} \] Then, substitute \( x = \mu \): \[ \mu e^{\mu} = \sum_{n=0}^{\infty} \frac{\mu^{n+1}}{n!} \]

Show the Mean \( \mu \)

From the weighted sum of the Poisson pmf: \[ \mu = E(X) = \sum_{n=0}^{\infty} n \frac{\mu^n e^{-\mu}}{n!} = \mu e^{-\mu} \sum_{n=1}^{\infty} \frac{\mu^{n-1}}{(n-1)!} = \mu e^{-\mu} \cdot e^{\mu} = \mu \]

Find the Standard Deviation

The variance of a Poisson-distributed random variable is the same as its mean: \[ \sigma^2 = Var(X) = E(X^2) - (E(X))^2 = \mu \] Therefore, the standard deviation \( \sigma \) is: \[ \sigma = \sqrt{\mu} \]

Key concepts

Average Value of Poisson Distribution

The Poisson distribution describes the probability of a given number of events occurring within a fixed interval of time or space. It is characterized by the parameter \( \mu \), which represents the average number of events. The average value of a random variable in a Poisson distribution is also called the mean. To find this average value, denote the random variable by \( X \) and use its probability mass function (pmf): \( P(X = k) = \frac{\mu^k e^{-\mu}}{k!} \).
The expected value, or mean, of the Poisson distribution can be found by: \( E(X) = \sum_{n=0}^{\infty} n \frac{\mu^n e^{-\mu}}{n!} \).
Through careful computation and recognizing how the sums interact with the exponential function, one finds that \(E(X) = \mu\).
This result makes intuitive sense because in a Poisson process with average rate \(\mu\), \(\mu\) is the expected number of occurrences.

Standard Deviation of Poisson Distribution

The standard deviation is a measure of the amount of variation or dispersion in a set of values. For the Poisson distribution, the variance and the mean are equal. This can be derived from properties of the Poisson distribution:
The variance of a Poisson-distributed random variable X is given by: \( \sigma^2 = Var(X) = E(X^2) - (E(X))^2 = \mu \).
Therefore, the standard deviation \( \sigma \) is the square root of the variance: \( \sigma = \sqrt{\mu} \).
This indicates that as \(\mu\) increases, both the mean and the variation around that mean also increase.

Probability Mass Function

The probability mass function (pmf) of a discrete random variable gives the probability that the variable is exactly equal to some value. For the Poisson distribution, the pmf is given by:
\( P(X = k) = \frac{\mu^k e^{-\mu}}{k!} \), where \( k \) is a non-negative integer.
This formula shows how the probability of observing exactly \( k \) events is determined by the mean number of events \( \mu \). The exponential function \( e^{-\mu} \) decays, ensuring that the probabilities sum to 1 over all possible values of \( k \). The factorial term \( k! \) adjusts for the number of ways \( k \) events can occur.

Exponential Function Series

The exponential function \( e^x \) can be expressed as an infinite series, a key concept in finding moments of a Poisson distribution. The series for \( e^x \) is:
\( e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!} \).
This representation is powerful because it allows us to manipulate and differentiate the function to obtain necessary identities. For example, differentiating \( e^x \) gives it back, as \( \frac{d}{dx}(e^x) = e^x \). Multiplying both sides by \( x \) lets us connect the series to \( \mu \) in the Poisson distribution.

Statistical Mean and Variance

In statistics, the mean is the average value of a random variable, while the variance measures the spread of its values. For the Poisson distribution, calculating the mean and variance involves using its pmf and properties of the exponential function series. The mean \( \mu \) is given by the expected value: \( E(X) = \sum_{n=0}^{\infty} n \frac{\mu^n e^{-\mu}}{n!} = \mu \).
The variance is: \( Var(X) = \sigma^2 = E(X^2) - (E(X))^2 = \mu \)
Consequently, these results confirm that both the mean and the variance for the Poisson distribution are equal to \( \mu \).