Question

(a) Suppose \(X\) is distributed according to a Poisson distribution with parameter \(\hat{2} .\) The parameter \(\lambda\) is itself a random variable whose distribution law is exponential with mean \(=1 / c .\) Find the distribution of \(X\). (b) What if \(\lambda\) follows a gamma distribution of order \(\alpha\) with scale parameter \(c\), i.e., the density of \(\lambda\) is \(e^{\alpha+1} \frac{\lambda^{a}}{\Gamma(\alpha+1)} e^{-\lambda c}\) for \(\lambda>0 ; 0\) for \(\lambda \leq 0\)

Short answer

In summary: (a) When \(\lambda\) follows an exponential distribution with mean \(=1/c\), the distribution of X is given by: \(P(X=k) = \frac{c}{(1 + c)^{k+1}}\) (b) When \(\lambda\) follows a Gamma distribution of order \(\alpha\) with scale parameter \(c\), the distribution of X is given by: \(P(X=k) = \int_{0}^{\infty} \frac{\lambda^k e^{-\lambda}}{k!} e^{\alpha+1} \frac{\lambda^{a}}{\Gamma(\alpha+1)} e^{-\lambda c} d\lambda\) In this case, we may need to rely on numerical methods and simulations to find the distribution of X.

Step-by-step solution

(a) Find the conditional distribution of X given \(\lambda\)#

Since X follows a Poisson distribution with parameter \(\hat{2}\) and \(\lambda\) follows an exponential distribution with mean \(=1/c\), the conditional probability of X is: \(P(X=k|\lambda) = \frac{\lambda^k e^{-\lambda}}{k!}\)

(a) Use the law of total probability to find the distribution of X#

Now, we need to integrate over all possible values of \(\lambda\) to find the overall probability of X. \(P(X=k) = \int_{0}^{\infty} P(X=k|\lambda) f_{\lambda}(\lambda) d\lambda\) Where, f_{\lambda}(\(\lambda\)) is the probability density function of the exponential distribution with mean \(1/c\): \(f_{\lambda}(\lambda)=ce^{-c\lambda}\) for \(\lambda > 0\).

(a) Evaluate the integral to find the distribution of X#

We can simplify the integral by plugging the conditional probability and the pdf of \(\lambda\): \(P(X=k) = \int_{0}^{\infty} \frac{\lambda^k e^{-\lambda}}{k!} ce^{-c\lambda} d\lambda\) \(P(X=k) = \frac{c}{k!} \int_{0}^{\infty} \lambda^k e^{-(1+c)\lambda} d\lambda\) This integral is in the form of a Gamma function, \(\Gamma(\alpha, \beta)\). To solve this integral, let's consider the general form of the Gamma function: \(\Gamma(\alpha,\beta) = \int_0^\infty x^{\alpha - 1} e^{-\beta x} dx\) In our case, \(\alpha = k + 1\) and \(\beta = 1 + c\). Then we can write: \(P(X=k) = \frac{c}{k!} \times \frac{\Gamma(k+1, 1+c)}{(1 + c)^{k+1}}\) Since, \(\Gamma(k + 1) = k!\), we can simplify the expression for the distribution of X.

(a) Final result for the distribution of X#

The distribution of X is given by: \(P(X=k) = \frac{c}{(1 + c)^{k+1}}\)

(b) Find the conditional distribution of X given \(\lambda\) when \(\lambda\) follows a Gamma distribution#

Since X follows a Poisson distribution with parameter \(\hat{2}\) and \(\lambda\) follows a Gamma distribution of order \(\alpha\) with scale parameter \(c\): \(P(X=k| \lambda) = \frac{\lambda^k e^{-\lambda}}{k!}\)

(b) Use the law of total probability to find the distribution of X#

Now, we need to integrate over all possible values of \(\lambda\) to find the overall probability of X. \(P(X=k) = \int_{0}^{\infty} P(X=k|\lambda) f_{\lambda}(\lambda) d\lambda\) Where, f_{\lambda}(\(\lambda\)) is the probability density function of the Gamma distribution: \(f_{\lambda}(\lambda)=e^{\alpha+1} \frac{\lambda^{a}}{\Gamma(\alpha+1)} e^{-\lambda c}\) for \(\lambda>0 ; 0\) for \(\lambda \leq 0\)

(b) Evaluate the integral to find the distribution of X#

We can simplify the integral by plugging the conditional probability and the pdf of \(\lambda\): \(P(X=k) = \int_{0}^{\infty} \frac{\lambda^k e^{-\lambda}}{k!} e^{\alpha+1} \frac{\lambda^{a}}{\Gamma(\alpha+1)} e^{-\lambda c} d\lambda\) Since this integral is challenging to evaluate in a closed form, we may need to rely on numerical methods and simulations to find the distribution of X.

Key concepts

Conditional Probability

Understanding conditional probability is essential when dealing with statistics and probability calculations. It represents the likelihood of an event occurring, assuming that another event has already happened. In formal terms, the conditional probability of an event A given event B is denoted as P(A|B), and is calculated using the formula \[ P(A|B) = \frac{P(A \cap B)}{P(B)} \]if P(B) > 0.
In the context of a Poisson distribution, we often look at finding the probability of a certain number of events happening, given a particular rate parameter (lambda). For example, in the exercise, we needed to calculate the conditional distribution of X, a random variable following a Poisson distribution, given the rate \(\lambda\) which itself is a random variable with a known distribution. Here, conditional probability helps us integrate over all possible values of \(\lambda\) to determine the overall probability distribution of X.

Gamma Distribution

The gamma distribution is a two-parameter family of continuous probability distributions. It's a generalization of the exponential distribution and is commonly used to model waiting times for multi-phase processes. The probability density function (pdf) for the gamma distribution, when the shape parameter is an integer, resembles that of a Poisson distribution with parameters \(\alpha\) and \(c\).
The pdf of a gamma distribution with shape \(\alpha\) and scale \(c\) is given by: \[ f_{\lambda}(\lambda) = \frac{c^{\alpha} \lambda^{\alpha-1} e^{-c\lambda}}{\Gamma(\alpha)} \] for \(\lambda > 0\), and 0 otherwise, where \(\Gamma(\alpha)\) is the gamma function evaluated at \(\alpha\).
In the exercise, when \(\lambda\) follows a gamma distribution, we compute the conditional probability of X using a formula that incorporates the gamma distribution's properties. This allows for the examination of how the Poisson process behaves when its rate parameter is itself subject to random variation.

Law of Total Probability

The law of total probability is a fundamental rule that provides a way to break down complex probability problems into simpler parts. It is often used when we have a set of mutually exclusive and exhaustive events, B1, B2, ..., Bn, and we are interested in the probability of event A. The law of total probability states that:
\[ P(A) = \sum_{i=1}^{n} P(A|B_i)P(B_i) \]
This rule allows for the calculation of the overall probability of A by summing the probability of A occurring in conjunction with each B event, weighted by the probability of each B event.
In the case of the exercise, by using the law of total probability, we integrate the product of the conditional probability of X given \(\lambda\) and the probability density function of \(\lambda\) over all possible values of \(\lambda\). This effectively 'averages out' the Poisson distribution over the uncertainty in the rate parameter \(\lambda\), which may itself follow an exponential or gamma distribution. Thus, it merges the multiple possible scenarios into a single comprehensive distribution for X.