Problem 4 - A Modern Introduction to Probability and Statistics: Understanding Why and How

Question

Let \(X\) have a Pois \((2)\) distribution. What is \(\mathrm{P}(X \leq 1) ?\)

Step-by-Step Solution

Answer

\( \mathrm{P}(X \leq 1) \approx 0.4059 \)

Step 1: Understand the Poisson Distribution

The Poisson distribution is a discrete probability distribution that expresses the probability of a given number of events occurring in a fixed interval of time or space, given that these events happen with a known constant mean rate and independently of the time since the last event. Here, the parameter of the distribution, denoted as \( \lambda \), is 2.

Step 2: Identify the Relevant Probability

We need to find the probability that the random variable \( X \) takes on a value less than or equal to 1, which can be expressed as \( \mathrm{P}(X \leq 1) \). This is equivalent to finding the sum of the probabilities that \( X \) is 0 and 1, i.e., \( \mathrm{P}(X = 0) + \mathrm{P}(X = 1) \).

Step 3: Use the Poisson Probability Formula

The probability mass function for a Poisson distribution is given by: \[ \mathrm{P}(X = k) = \frac{e^{-\lambda} \cdot \lambda^k}{k!} \]where \( \lambda = 2 \) for this problem, and \( k \) takes on integer values 0, 1, 2, ... .

Step 4: Calculate P(X = 0)

Plug \( k = 0 \) and \( \lambda = 2 \) into the formula:\[ \mathrm{P}(X = 0) = \frac{e^{-2} \cdot 2^0}{0!} = e^{-2} \cdot 1 \]Since \( 0! = 1 \), this simplifies to \( e^{-2} \).

Step 5: Calculate P(X = 1)

Plug \( k = 1 \) and \( \lambda = 2 \) into the formula:\[ \mathrm{P}(X = 1) = \frac{e^{-2} \cdot 2^1}{1!} = \frac{2e^{-2}}{1} = 2e^{-2} \]

Step 6: Add Probabilities Together

Now we add \( \mathrm{P}(X = 0) \) and \( \mathrm{P}(X = 1) \) together:\[ \mathrm{P}(X \leq 1) = e^{-2} + 2e^{-2} \]Simplifying gives:\[ \mathrm{P}(X \leq 1) = 3e^{-2} \]

Step 7: Compute the Numerical Value

Use the approximate value of \( e \), which is roughly 2.718, to compute \( e^{-2} \):\[e^{-2} \approx 0.1353 \]Thus, \[ 3e^{-2} \approx 3 \times 0.1353 = 0.4059 \].

Key Concepts

Probability Mass Function

In the realm of statistics, a probability mass function (PMF) is a fundamental concept when dealing with discrete probability distributions. The PMF gives the probability that a discrete random variable is precisely equal to some value. Mathematically, for a random variable \(X\), it is written as \(\mathrm{P}(X=k)\).

The PMF provides valuable insights into how likely each outcome is within a distribution. For a Poisson distribution, the PMF is defined by the formula:
\[ \mathrm{P}(X = k) = \frac{e^{-\lambda} \cdot \lambda^k}{k!} \]
Where:

• \(e\) is Euler's number, approximately equal to 2.718.
• \(\lambda\) is the average number of occurrences (the rate) of the event.
• \(k\) is the number of occurrences.
• \(!\) denotes factorial, which means multiplying the number by all of its integers less than itself.

In the given exercise, the PMF helps us calculate \(\mathrm{P}(X=0)\) and \(\mathrm{P}(X=1)\) by plugging in these specific values and computing the results. The PMF essentially "maps out" the exact likelihood of each potential outcome within the distribution.

Discrete Probability Distribution

A discrete probability distribution is all about probabilities of countable events, meaning events that have specific, separate values, unlike continuous distributions where outcomes fall within a range.

One common discrete distribution is the Poisson distribution, which is characterized by a probability of a given number of events occurring in a fixed interval of time or space. The key characteristics of discrete distributions include:

• They consist of a finite or countably infinite set of values.
• The sum of all probabilities in the distribution equals 1, meaning something must happen within the set of possible outcomes.
• Each individual outcome has a specific probability assigned by the PMF.

The Poisson distribution, with a parameter \(\lambda\), is a type of discrete distribution particularly useful in real-world scenarios where we count occurrences of an event, like phone calls to a call center per hour, or, in our exercise, where events happen at an average rate of 2 per time interval described by \(X\). Understanding these distributions helps in making predictions and decisions based on probabilistic models.

Expected Value

Expected value, often symbolized as \(E(X)\), provides a measure of the central tendency of a probability distribution, essentially representing "the mean" or average value we can expect from repeated trials of a random process.

For a discrete random variable, the expected value is calculated by summing the products of each value the random variable can take and its corresponding probability:
\[ E(X) = \sum_{k=0}^{\infty} k \cdot \mathrm{P}(X=k) \]
This formula essentially weighs each possible outcome \(k\) by its probability \(\mathrm{P}(X=k)\) and adds them up.

In the context of a Poisson distribution, the expected value of a Poisson random variable is equal to the parameter \(\lambda\), which in this exercise is 2. This means, on average, there are 2 events expected to occur in the given interval. This feature of the Poisson distribution not only activates models of real-world scenarios but also helps verify the performance and applicability of stochastic models in various fields.