Problem 263 - An Introduction to Mathematical Statistics and Its Applications

Question

The random variable \(X\) has a Poisson distribution \(p_{X}(k)=e^{-\lambda} \lambda^{k} / k !, k=0,1,2, \ldots\). Find the momentgenerating function for a Poisson random variable. Recall that $$ e^{r}=\sum_{k=0}^{\infty} \frac{r^{k}}{k !} $$

Step-by-Step Solution

Answer

The moment-generating function for a Poisson random variable is \(M_{X}(t) = e^{\lambda (t - 1)}\).

Step 1: Define Moment Generating Function

Start by defining the moment-generating function (MGF) of a random variable. The MGF is given by \(M_{X}(t) = E(e^{tX})\), which signifies taking the expectation of \(e^{tX}\). In this case for a discrete random variable like a Poisson, this equates to the sum over all possible values k of \(e^{tk} * p_X(k)\).

Step 2: Substituting into the MGF

Substitute the Poisson PMF into the definition of the MGF: \(M_{X}(t) = \sum_{k=0}^{\infty} e^{tk} * e^{-\lambda} * \lambda^{k} / k!\). The goal is to simplify this expression and eventually get it in the form of \(e^{r}\). To do this, rewrite the expression as \(M_{X}(t) = \sum_{k=0}^{\infty} e^{(t\lambda-\lambda)} * \lambda^{k} / k!\). much importance should be paid on making collective terms under exponent of e, and also rearranging elements to match the given taylor series.

Step 3: Comparing to Taylor Series

This expression can now be recognized as the Taylor Series for \(e^{r}\), except \(r = (t\lambda-\lambda)\). So \(\sum_{k=0}^{\infty} r^{k} / k!\) equals \(e^{r}\), hence \(M_{X}(t) = e^{(t\lambda-\lambda)}\).

Step 4: Simplify using Laws of Exponents

Finally, use the law of exponents on the expression \(M_{X}(t) = e^{(t\lambda-\lambda)}\) to see that it equals \(e^{t\lambda} * e^{-\lambda} = e^{\lambda (t - 1)}\).

Key Concepts

Moment-Generating Function

The concept of a moment-generating function (MGF) is fundamental in probability theory. It is a valuable tool for characterizing the distribution of a random variable. For a random variable \(X\), the MGF, denoted as \(M_X(t)\), is defined as the expected value of the exponential function of \(tX\):

• \( M_X(t) = E(e^{tX}) \)

This expression means that we take the sum of the product of \(e^{tX}\) and the probability distribution of \(X\). When dealing with discrete variables, like those with a Poisson distribution, this involves summing over all possible values of the random variable.
The beauty of the MGF is that it uniquely determines the probability distribution if the MGF exists in an open neighborhood of \(t = 0\). More practically, MGFs are used to find the expectations and variances of distributions by differentiating the function.
For the Poisson distribution, where the probability mass function (PMF) is \(p_{X}(k) = e^{-\lambda} \frac{\lambda^k}{k!}, k = 0, 1, 2, \ldots\), plugging this into the MGF definition involves substituting \(p_{X}(k)\) into the sum in the expected value expression.

Expectation in Probability

Expectation is a cornerstone concept in probability, reflecting the average or "expected" value a random variable can take. Also known as the expected value or mean, for a discrete random variable \(X\), it is calculated as:

• \( E(X) = \sum_{k} k \cdot p_{X}(k) \)

This formula sums the products of each outcome \(k\) of the random variable and its associated probability \(p_{X}(k)\). In essence, expectation assesses the center of distribution or the long-run average of outcomes.
For a Poisson random variable with parameter \(\lambda\), the expectation is neatly tied to this parameter, given by \(E(X) = \lambda\). This indicates that over many trials, the average number of successes in a fixed interval is \(\lambda\).
In terms of MGFs, the first derivative of the moment-generating function \(M_X(t)\) with respect to \(t\), evaluated at \(t = 0\), gives the expectation of \(X\). This derivative essentially measures how much the MGF changes near the origin, which is related to the average value of the random variable.

Probability Mass Function

The probability mass function (PMF) is a fundamental function that gives the probability that a discrete random variable is exactly equal to some value. For a Poisson distributed random variable \(X\), the PMF is defined as follows:

• \( p_{X}(k) = \frac{e^{-\lambda} \lambda^k}{k!} \)

for \(k = 0, 1, 2, \ldots\). This function is derived from the Poisson process, which is commonly used to model the number of events occurring within a fixed period of time or space.
The PMF has several important properties:

• It sums to 1 over all possible values, satisfying the total probability law.
• It is non-negative, as probabilities cannot be negative.
• For Poisson, the shape of the PMF is determined by \(\lambda\), which acts as both the mean and the variance of the distribution.

Understanding the PMF is crucial when studying random variables since it forms the basis from which expectations, MGFs, and other probabilities are calculated. By mastering the PMF, one gains insight into the behavior and characteristics of the associated random variable.