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Magazine Chapter 5: Problem 92
A random variable \(X\) has the Poisson distribution \(f(x)=\frac{e^{-\lambda} \lambda^{x}}{x !}, \quad x=0,1, \ldots\) (a) Show that the moment-generating function is \(M_{X}(t)=e^{\lambda\left(e^{t}-1\right)}\) (b) Use \(M_{X}(t)\) to find the mean and variance of the Poisson random variable.
Short Answer
Expert verified
The MGF of Poisson is \(e^{\lambda(e^t-1)}\); mean and variance are both \(\lambda\).
Step by step solution
01
Recall the moment-generating function of Poisson Distribution
The moment-generating function (MGF) of a random variable \(X\) is defined as \(M_X(t) = E[e^{tX}]\). For a Poisson distribution with parameter \(\lambda\), this function helps us find the moments such as mean and variance.
02
Derive the MGF of Poisson Distribution
For a Poisson random variable, we have:\[ M_X(t) = E[e^{tX}] = \sum_{x=0}^{\infty} e^{tx} \frac{e^{-\lambda} \lambda^x}{x!} \]Simplifying, we get:\[ M_X(t) = e^{-\lambda} \sum_{x=0}^{\infty} \frac{(\lambda e^t)^x}{x!} \]Recognize the series as the expansion of \(e^{\lambda e^t}\) to conclude that:\[ M_X(t) = e^{\lambda (e^t - 1)} \]
03
Differentiate to find the mean
To find the mean, compute the first derivative of the MGF and evaluate it at \(t = 0\):\[ \frac{d}{dt}M_X(t) = \lambda e^t e^{\lambda (e^t - 1)} \]Substitute \(t = 0\) into the derivative:\[ \left. \frac{d}{dt}M_X(t) \right|_{t=0} = \lambda e^{0} \times e^{0} = \lambda \]
04
Differentiate to find the variance
The second moment is found by the second derivative of the MGF at \(t = 0\), which involves applying the product rule:\[ \frac{d^2}{dt^2}M_X(t) = \lambda e^t (\lambda e^t + 1) e^{\lambda (e^t - 1)} \]Evaluate at \(t=0\):\[ \left. \frac{d^2}{dt^2}M_X(t) \right|_{t=0} = \lambda (\lambda + 1) \]The variance is the second moment minus the square of the mean \((\lambda^2)\):\[ \text{Variance} = (\lambda + \lambda^2) - \lambda^2 = \lambda \]
05
Conclusion
The mean of the Poisson distribution is \(\lambda\) and the variance is also \(\lambda\). The MGF method confirms these properties by taking the derivatives of the moment-generating function derived earlier.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Moment-Generating Function
The moment-generating function (MGF) is a powerful tool in probability theory. It simplifies the process of finding moments of a random variable, like the mean and variance. For the Poisson distribution, the MGF is defined as \( M_X(t) = E[e^{tX}] \). This means you are essentially finding the expected value of \( e^{tX} \) for a Poisson random variable \( X \) with a given parameter \( \lambda \).
One key benefit of the MGF is how it transforms complex expressions into more manageable forms. For Poisson distributions, calculating this function involves recognizing a series expansion for the exponential \( e^{\lambda e^t} \). This result helps to derive other important statistical properties easily.
One key benefit of the MGF is how it transforms complex expressions into more manageable forms. For Poisson distributions, calculating this function involves recognizing a series expansion for the exponential \( e^{\lambda e^t} \). This result helps to derive other important statistical properties easily.
Mean of Poisson Distribution
The mean, or expected value, of a Poisson distribution is denoted as \( \lambda \). This value indicates the average number of events that occur in a fixed interval of time or space given the rate \( \lambda \).
To derive the mean using the moment-generating function, take the first derivative of \( M_X(t) \) with respect to \( t \) and evaluate it at \( t = 0 \). In the Poisson distribution's case, this results in \( \lambda \), confirming that the mean directly equates to the parameter \( \lambda \).
This property underscores how the parameter \( \lambda \) not only represents the average rate but is central to the distribution's behavior.
To derive the mean using the moment-generating function, take the first derivative of \( M_X(t) \) with respect to \( t \) and evaluate it at \( t = 0 \). In the Poisson distribution's case, this results in \( \lambda \), confirming that the mean directly equates to the parameter \( \lambda \).
This property underscores how the parameter \( \lambda \) not only represents the average rate but is central to the distribution's behavior.
Variance of Poisson Distribution
For the Poisson distribution, the variance is also \( \lambda \). This implies that the distribution's variability, like its mean, is directly tied to the \( \lambda \) parameter.
To find this using the MGF, compute the second derivative of \( M_X(t) \) and evaluate at \( t = 0 \). The procedure involves using the product rule and simplifies to \( \lambda (\lambda + 1) \). From here, subtract the square of the mean \( \lambda^2 \) to get \( \lambda \) once again.
This result highlights that the mean and variance of the Poisson distribution are equal, a unique and defining characteristic.
To find this using the MGF, compute the second derivative of \( M_X(t) \) and evaluate at \( t = 0 \). The procedure involves using the product rule and simplifies to \( \lambda (\lambda + 1) \). From here, subtract the square of the mean \( \lambda^2 \) to get \( \lambda \) once again.
This result highlights that the mean and variance of the Poisson distribution are equal, a unique and defining characteristic.
Mathematical Derivation
Mathematical derivation is the process of logically deducing properties or outcomes from known principles. For the Poisson distribution, it involves utilizing series expansions and calculus.
Let's take the Poisson's MGF as an example: \( M_X(t) = e^{\lambda (e^t - 1)} \). To derive it, initiate with the definition \( M_X(t) = E[e^{tX}] \).
The sum \( \sum_{x=0}^{\infty} e^{tx} \frac{e^{-\lambda} \lambda^x}{x!} \) requires recognizing the famous series expansion of the exponential function \( e^{\lambda e^t} \). This step-by-step approach of simplifying complex functions forms the basis of mathematical derivation.
Let's take the Poisson's MGF as an example: \( M_X(t) = e^{\lambda (e^t - 1)} \). To derive it, initiate with the definition \( M_X(t) = E[e^{tX}] \).
The sum \( \sum_{x=0}^{\infty} e^{tx} \frac{e^{-\lambda} \lambda^x}{x!} \) requires recognizing the famous series expansion of the exponential function \( e^{\lambda e^t} \). This step-by-step approach of simplifying complex functions forms the basis of mathematical derivation.
Probability Theory
Probability theory is the mathematical framework for measuring and quantifying uncertainty. It defines concepts such as random variables, probability distributions, and expected values, all crucial in analyzing random phenomena.
A Poisson distribution model is valuable in probability theory as it represents the likelihood of a given number of events happening in a fixed interval of time or space, provided the events occur with a constant mean rate \( \lambda \) and independently of the time since the last event.
Understanding distributions like the Poisson helps build intuition for random processes and their outcomes, which mathematicians and statisticians use extensively to make informed predictions.
A Poisson distribution model is valuable in probability theory as it represents the likelihood of a given number of events happening in a fixed interval of time or space, provided the events occur with a constant mean rate \( \lambda \) and independently of the time since the last event.
Understanding distributions like the Poisson helps build intuition for random processes and their outcomes, which mathematicians and statisticians use extensively to make informed predictions.
