Mean and variance of the Poisson distribution. Find the MGF of the Poisson distribution and use it to compute the mean and variance. Use simulations to check the MGF by plotting the empirical and theoretical MGF on the same graph.
The correct answer and explanation is :
Poisson Distribution: Mean, Variance, and MGF
The Poisson distribution is often used to model the number of events occurring within a fixed interval of time or space. The random variable ( X ) that follows a Poisson distribution with parameter ( \lambda ) (rate or average number of occurrences in the interval) is denoted as ( X \sim \text{Poisson}(\lambda) ).
Mean and Variance of the Poisson Distribution
- Mean:
The mean of a Poisson distribution is given by:
[
\mu = E[X] = \lambda
]
This means that the expected number of events occurring in a fixed interval is equal to the rate ( \lambda ). - Variance:
The variance of a Poisson distribution is also equal to ( \lambda ):
[
\text{Var}(X) = \lambda
]
This property reflects the fact that in a Poisson process, the variance is directly proportional to the mean.
Moment Generating Function (MGF)
The moment generating function (MGF) of a random variable ( X ) is given by:
[
M_X(t) = E[e^{tX}]
]
For the Poisson distribution, the MGF is derived as follows:
- The probability mass function (PMF) of the Poisson distribution is:
[
P(X = k) = \frac{\lambda^k e^{-\lambda}}{k!}, \quad k = 0, 1, 2, \dots
] - The MGF is:
[
M_X(t) = E[e^{tX}] = \sum_{k=0}^{\infty} e^{tk} \frac{\lambda^k e^{-\lambda}}{k!}
]
This simplifies to:
[
M_X(t) = e^{-\lambda} \sum_{k=0}^{\infty} \frac{(\lambda e^t)^k}{k!} = e^{-\lambda} e^{\lambda e^t}
]
Therefore, the MGF of the Poisson distribution is:
[
M_X(t) = e^{\lambda (e^t – 1)}
]
Using the MGF to Compute the Mean and Variance
- The mean can be computed by taking the first derivative of the MGF at ( t = 0 ):
[
\mu = M_X'(0) = \lambda
] - The variance can be computed by taking the second derivative of the MGF at ( t = 0 ) and using the formula ( \text{Var}(X) = M_X”(0) – (M_X'(0))^2 ):
[
\text{Var}(X) = \lambda
]
Thus, the mean and variance of the Poisson distribution are both equal to ( \lambda ), confirming the earlier results.
Simulations
To verify the theoretical MGF, we can perform simulations of the Poisson distribution, plot the empirical MGF, and compare it with the theoretical MGF.
Here’s a general procedure for performing the simulation and plotting the graphs:
- Simulate Poisson data: Generate a large number of random samples from a Poisson distribution with a given ( \lambda ).
- Compute empirical MGF: For each simulated value ( X_i ), compute the empirical MGF ( e^{t X_i} ) for different values of ( t ).
- Plot: Plot the empirical MGF against the theoretical MGF ( e^{\lambda (e^t – 1)} ) for various values of ( t ).
This comparison should show that the empirical MGF closely matches the theoretical MGF, verifying that the MGF correctly captures the mean and variance of the Poisson distribution.
Conclusion
The Poisson distribution has a mean and variance both equal to ( \lambda ), and the MGF is given by ( M_X(t) = e^{\lambda (e^t – 1)} ). By performing simulations, we can verify the accuracy of the MGF, which should match the empirical MGF in practice. This approach provides an excellent way to confirm theoretical results with data.