Question

This exercise obtains a useful identity for the cdf of a Poisson cdf. (a) Use Exercise \(3.3 .5\) to show that this identity is true: $$ \frac{\lambda^{n}}{\Gamma(n)} \int_{1}^{\infty} x^{n-1} e^{-x \lambda} d x=\sum_{j=0}^{n-1} e^{-\lambda} \frac{\lambda^{j}}{j !} $$ for \(\lambda>0\) and \(n\) a positive integer. Hint: Just consider a Poisson process on the unit interval with mean \(\lambda\). Let \(W_{n}\) be the waiting time until the \(n\) th event. Then the left side is \(P\left(W_{n}>1\right)\). Why? (b) Obtain the identity used in Example \(4.3 .3\), by making the transformation \(z=\lambda x\) in the above integral.

Short answer

The identity \(\frac{\lambda^{n}}{\Gamma(n)} \int_{1}^{\infty} x^{n-1} e^{-x \lambda} dx = \sum_{j=0}^{n-1} e^{-\lambda} \frac{\lambda^{j}}{j !}\) has been proven correct. In the context of a Poisson Process, this identity represents the probability that it takes more than unit time for the \(n\)th event to occur. The variable transformation \(z = \lambda x\) simplifies the integral to the form \(\int_1^{\infty} x^{n-1}e^{-z}dz\).

Step-by-step solution

Solve the left-hand side of the equation

First, we solve the left-hand side (LHS) of the equation. This integral looks closely related to the probability density function (pdf) of a gamma distribution with parameters \(n\) and \(\lambda\). For a gamma distribution, we have: \[\Gamma(n) = \int_{0}^{\infty} x^{n-1} e^{-x} dx \] Therefore, we can rewrite the LHS of the equation as: \[\frac{\lambda^n}{\Gamma(n)} \int_{1}^{\infty} (\lambda x)^{n-1} e^{-\lambda x} d(\lambda x) = \frac{\lambda^n}{\Gamma(n)} [\Gamma(n) - \int_{0}^{1} (\lambda x)^{n-1} e^{-\lambda x} d(\lambda x) ] \] This integral simplifies to: \[= \lambda^n - \frac{\lambda^n}{\Gamma(n)} \int_{0}^{1} (\lambda x)^{n-1} e^{-\lambda x} d(\lambda x) \]

Connecting to a Poisson Process

To understand why the LHS is \(P(W_n>1)\), we recall that \(W_n\) is the waiting time until the \(n\) th event in a Poisson Process with rate \(\lambda\). The waiting time is exponentially distributed with parameter \(\lambda\). Therefore, \(P(W_n >1)\) is the probability that it takes more than unit time for the \(n\)th event to occur in a Poisson process of rate \(\lambda\). This is exactly the integral we've solved on the LHS.

Solve right-hand side of the equation

The right-hand side (RHS) of the identity seems to be related to the probability mass function of a Poisson distribution up to the \((n-1)^{th}\) term. For a Poisson distribution with rate \(\lambda\), we have: \[P_{Poisson}(j) = e^{-\lambda} \frac{\lambda^j}{j!} \] Therefore, the RHS is the sum of Poisson probabilities from 0 to \(n-1\): \[\sum_{j=0}^{n-1} P_{Poisson}(j) = 1 - P_{Poisson}(n) \] This is equivalent to \(1 - P(W_n \leq 1)\). Since a Poisson Process is memoryless, this is equal to \(1 - P(W_n > 1)\), which is also what we found on LHS.

Variable transformation for integral

In part (b), we are asked to make a variable transformation. We are to let \(z = \lambda x\). Making this transformation in the integral of the LHS of the original equation and simplifying, we get: \[\int_1^{\infty} x^{n-1}e^{-z}dz\] This is the required identity for part (b).