Math Problem Statement

If X is Poisson distribution with parameter µ, prove that\displaylines{\displaylines\\ }

Solution

To prove that XX follows a Poisson distribution with parameter μ\mu, we'll work through the definition and properties of the Poisson distribution.

Step 1: Define the Poisson distribution

A random variable XX follows a Poisson distribution with parameter μ\mu (where μ>0\mu > 0) if its probability mass function (PMF) is given by: P(X=k)=μkeμk!,k=0,1,2,P(X = k) = \frac{\mu^k e^{-\mu}}{k!}, \quad k = 0, 1, 2, \dots

Step 2: Derive the mean and variance of the Poisson distribution

For the Poisson distribution, the mean E(X)E(X) and variance Var(X)\text{Var}(X) are both equal to μ\mu. We will now show this.

Mean:

The mean of a Poisson-distributed random variable XX is given by: E(X)=k=0kP(X=k)E(X) = \sum_{k=0}^{\infty} k \cdot P(X = k) Substituting the PMF of the Poisson distribution: E(X)=k=0kμkeμk!E(X) = \sum_{k=0}^{\infty} k \cdot \frac{\mu^k e^{-\mu}}{k!}

We can simplify this using the fact that k×1k!=1(k1)!k \times \frac{1}{k!} = \frac{1}{(k-1)!}: E(X)=μeμk=1μk1(k1)!E(X) = \mu e^{-\mu} \sum_{k=1}^{\infty} \frac{\mu^{k-1}}{(k-1)!} Changing the index of summation j=k1j = k - 1, we get: E(X)=μeμj=0μjj!E(X) = \mu e^{-\mu} \sum_{j=0}^{\infty} \frac{\mu^j}{j!} Recognize that the sum is the series expansion of eμe^{\mu}: E(X)=μeμeμ=μE(X) = \mu e^{-\mu} \cdot e^{\mu} = \mu So, the mean of the Poisson distribution is μ\mu.

Variance:

The variance of XX is given by: Var(X)=E(X2)(E(X))2\text{Var}(X) = E(X^2) - (E(X))^2 First, calculate E(X2)E(X^2): E(X2)=k=0k2P(X=k)E(X^2) = \sum_{k=0}^{\infty} k^2 \cdot P(X = k) Expanding k2k^2 as k(k1)+kk(k-1) + k, we get: E(X2)=k=0(k(k1)+k)μkeμk!E(X^2) = \sum_{k=0}^{\infty} (k(k-1) + k) \cdot \frac{\mu^k e^{-\mu}}{k!} Splitting this into two sums, and simplifying, it can be shown that: E(X2)=μ2+μE(X^2) = \mu^2 + \mu Thus, the variance is: Var(X)=E(X2)(E(X))2=μ2+μμ2=μ\text{Var}(X) = E(X^2) - (E(X))^2 = \mu^2 + \mu - \mu^2 = \mu So, the variance of the Poisson distribution is also μ\mu.

Conclusion:

We have shown that for a Poisson-distributed random variable XX with parameter μ\mu, the mean and variance are both equal to μ\mu, which is a characteristic property of the Poisson distribution.

Would you like to dive into more details, or do you have any questions? Here are five related questions you might find helpful:

  1. How is the Poisson distribution related to the binomial distribution?
  2. Can you explain the relationship between the Poisson and exponential distributions?
  3. How can you derive the Poisson distribution as a limit of the binomial distribution?
  4. What are some real-world examples where the Poisson distribution is used?
  5. How do you calculate the cumulative distribution function (CDF) for a Poisson-distributed variable?

Tip: The Poisson distribution is particularly useful for modeling the number of occurrences of an event over a fixed interval of time or space when the events are independent and occur at a constant average rate.

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Probability Theory
Poisson Distribution

Formulas

Poisson PMF: P(X = k) = (mu^k * e^(-mu)) / k!
Mean: E(X) = mu
Variance: Var(X) = mu

Theorems

-

Suitable Grade Level

Advanced Undergraduate