Problem 75 - An Introduction to Mathematical Statistics and Its Applications

Question

Let \(Y\) be an exponential random variable, where \(f_{Y}(y)=\lambda e^{-\lambda y}, 0 \leq y\). For any positive integer \(n\), show that \(P(n \leq Y \leq n+1)=e^{-\alpha n}\left(1-e^{-\lambda}\right)\). Note that if \(p=1-e^{-\lambda}\), the "discrete" version of the exponential pdf is the geometric pdf.

Step-by-Step Solution

Answer

The probability that an exponential random variable \(Y\) falls between any positive integer \(n\) and \(n+1\) can indeed be represented as \( e^{-\lambda n}(1 - e^{-\lambda}). By integrating the pdf for \(Y\) over the interval [n, n+1], we obtained this expression, demonstrating the exercise's statement.

Step 1: Understand the Probability Density Function

The Probability Density Function for an exponential random variable \(Y\) is defined as \(f_{Y}(y)=\lambda e^{-\lambda y}\) for \(0 \leq y\). This function represents the probability that the random variable \(Y\) assumes the value \(y\). The parameter \(\lambda\) is the rate parameter.

Step 2: Integrate the exponential pdf

We need to find the probability that \(Y\) falls between \(n\) and \(n+1\). For continuous random variables, this is obtained by integrating the pdf over the interval [n, n+1]. i.e., \[P(n \leq Y \leq n+1) = \int_{n}^{n+1}\lambda e^{-\lambda y}dy\]

Step 3: Compute the integral

The integral of the exponential function is given by \(-e^{-\lambda y}\). Evaluating our integral at the limits \(n\) and \(n+1\) we get: \[-\lambda e^{-\lambda(n+1)} + \lambda e^{-\lambda n} = \lambda e^{-\lambda n} - \lambda e^{-\lambda n}e^{-\lambda}.\]

Step 4: Simplification

The expression simplifies to \( \lambda e^{-\lambda n}(1 - e^{-\lambda})\), which is our desired expression, considering that \(p = 1- e^{-\lambda}\).

Key Concepts

Probability Density Function

The probability density function (PDF) is a statistical expression that defines a continuous probability distribution. Often represented by the notation \( f_Y(y) \) for a random variable \(Y\), the PDF provides the likelihood that any value within a continuous set is equal to the random variable. The definition of the PDF for an exponential random variable is \( f_{Y}(y) = \boldsymbol{\bf{\boldsymbol\lambda}} e^{-\boldsymbol{\bf{\boldsymbol\lambda}} y} \), where \( \lambda \) is the rate at which events occur, also known as the rate parameter.

When dealing with exponential random variables, the PDF plays a crucial role in describing the time between events in a process where events occur continuously and independently at a constant average rate. For instance, it could represent the time until the next customer arrives at a service center. The exponential PDF is uniquely identified by its 'memoryless' property, meaning the probability of an event occurring in the next moment is always the same, irrespective of how much time has already elapsed.

Integration of Exponential PDF

To determine the probability that an exponential random variable \( Y \) takes on a value within a specific interval, we integrate the exponential PDF over that interval. Integration is the continuous analog to the summation in discrete distributions and gives us the total probability across a range of values.

For instance, if we want to find the probability that \( Y \) falls between \( n \) and \( n + 1 \), we would calculate \( P(n ≤ Y ≤ n+1) \) by integrating the PDF from \( n \) to \( n+1 \). Using the integral of the PDF, \( \int_{n}^{n+1}\text{\lambda} \exp{-\text{\lambda} y} \, \text{dy} \), we are essentially summing up all infinitesimal probabilities in that interval, which leads us to the cumulative distribution for that specific range. Through the process of integration, \exp{-\text{\lambda} y} becomes \( -e^{-\text{\lambda} y} \), and evaluating this antiderivative at the bounds \( n \) and \( n + 1 \) gives us the desired probability.

Geometric Probability Distribution Function

While exponential distribution describes the waiting time between continuous events, the geometric probability distribution function (PDF) deals with the number of trials until the first success in a sequence of independent and identically distributed Bernoulli trials. In these trials, there are only two possible outcomes: 'success' or 'failure'.

The geometric PDF is given by \( P(X = k) = (1-p)^{k-1}p \), where \( p \) is the probability of success on any single trial, and \( k \) is the number of trials until the first success. One interesting connection to note is that when we define \( p = 1 - e^{-\text{\lambda}} \), the geometric distribution becomes a discrete analog to the continuous exponential distribution, with \( \lambda \) playing the role of the success rate in continuous time. Thus, it is often used in similar contexts but for discrete quantities, like the number of clicks until a website visitor makes a purchase.