91Ó°ÊÓ

Define a function \(p(x ; \lambda, \mu)\) by $$ \begin{aligned} &p(x ; \lambda, \mu) \\ &=\left\\{\begin{array}{cc} \frac{1}{2} e^{-\lambda} \frac{\lambda^{x}}{x !}+\frac{1}{2} e^{-\mu} \frac{\mu^{x}}{x !} & x=0,1,2, \ldots \\ 0 & \text { otherwise } \end{array}\right. \end{aligned} $$ a. Show that \(p(x, \lambda, \mu)\) satisfies the two conditions necessary for specifying a pmf. [Note: If a firm employs two typists, one of whom makes typographical errors at the rate of ¿ per page and the other at rate \(\mu\) per page and they each do half the firm's typing, then \(p(x ; \lambda, \mu)\) is the pmf of \(X=\) the number of errors on a randomly chosen page.] b. If the first typist (rate \(\lambda\) ) types \(60 \%\) of all pages, what is the penf of \(X\) of part (a)? c. What is \(E(X)\) for \(p(x, \lambda, \mu)\) given by the displayed expression? d. What is \(\sigma^{2}\) for \(p(x ; \lambda, \mu)\) given by that expression?

Step by step solution

01

Understand the Expression

The given function is a mixture of two Poisson distributions, each with parameters \( \lambda \) and \( \mu \), and weights 0.5 for each. The function is defined for non-negative integers \( x = 0, 1, 2, \ldots \), and 0 otherwise.

02

Check PMF Conditions

A probability mass function must satisfy two conditions: 1) \( p(x; \lambda, \mu) \geq 0 \) for all \( x \). Here, \( p(x; \lambda, \mu) \) is a sum of two non-negative terms, so it is always non-negative. 2) \( \sum_{x=0}^{\infty} p(x; \lambda, \mu) = 1 \). This sum can be split into two sums of Poisson probabilities, each summing to 1, multiplied by 0.5, confirming that the total is 1.

03

Modify for Different Typing Rates

When the first typist handles 60% of pages and the second 40%, the PMF becomes: \( p(x; \lambda, \mu) = 0.6 e^{-\lambda} \frac{\lambda^x}{x!} + 0.4 e^{-\mu} \frac{\mu^x}{x!} \). Adjust the weights according to the proportions of pages typed by each typist.

04

Calculate Expected Value, \( E(X) \)

For the original expression, \( E(X) = 0.5 \times \lambda + 0.5 \times \mu \), using the properties of the Poisson distribution. For the modified rates, adjust to: \( E(X) = 0.6 \lambda + 0.4 \mu \).

05

Calculate Variance, \( \sigma^2(X) \)

For the original distribution: \( \sigma^2 = 0.5 \times \lambda + 0.5 \times \mu \) using the properties of the Poisson distribution variance. For modified rates, it becomes: \( \sigma^2 = 0.6 \lambda + 0.4 \mu \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Probability Mass Function

A Probability Mass Function (PMF) is crucial for understanding discrete random variables. It enables us to determine the likelihood of a random variable taking on a specific value. Let's delve into the two essential conditions a PMF must fulfill:

• Firstly, a PMF must be non-negative for all possible values of the random variable, which means that for every integer value of our random variable, the probability assigned must be ≥ 0.


• Secondly, the total \( \sum_{x=0}^{\infty} p(x; \lambda, \mu) = 1 \). This requirement ensures that we account for all possible outcomes, reflecting the certainty that something will happen.

In our context with the function \( p(x; \lambda, \mu) \), it is a blend of two Poisson distributions, equally weighted at 0.5 each. It satisfies the PMF conditions because both components are always non-negative, and their total sum equals 1. This model applies to real-world situations, such as predicting errors made by typists working at different rates.

Expected Value

The expected value is like the average or mean of a random variable, providing insight into its long-term behavior when the process is repeated many times. For a Poisson Mixture Model, calculating \(E(X)\) involves understanding the contribution of each component of the mixture. In mathematical terms, the expected value for our function is:

• For equal typing rates: \(E(X) = 0.5 \times \lambda + 0.5 \times \mu\).
• If the typing rates are adjusted, such as one typist handling 60% of pages, then: \(E(X) = 0.6 \lambda + 0.4 \mu\).

The expected value calculation reflects the weighted average based on the likelihood of each typist's contribution, informing us about the general number of errors.

Variance

Variance measures the spread or dispersion around the expected value, capturing the variability of the random variable. Knowing how much the data can vary is essential for understanding the reliability of predictions.
In a Poisson distribution, the variance is equal to the mean. Here, for the Poisson Mixture Model, calculating variance involves a similar approach to the expected value:

• For equal weights: \(\sigma^2 = 0.5 \times \lambda + 0.5 \times \mu\).
• For adjusted weights: \(\sigma^2 = 0.6 \lambda + 0.4 \mu\).

Variance provides insights into how much the actual number of errors can differ from the expected number, playing a vital role in risk assessment and reliability.

Poisson Distribution

A Poisson Distribution is used to model the number of events occurring in a fixed interval of time or space, assuming these events happen with a constant mean rate and are independent of each other. It's especially helpful for discrete events like counting accident occurrences, or as in our exercise, typographical errors.
A Poisson Distribution is characterized by its parameter \( \lambda \), the average rate of occurrence, and its Probability Mass Function (PMF) is given by:
\[ P(X = x) = e^{-\lambda} \frac{\lambda^x}{x!} \] In the case of the mixture model with two Poisson distributions (\(\lambda\) and \(\mu\)), we leverage the properties of the Poisson PMF to address situations where two different processes might influence the outcome, reflecting more complex real-life scenarios. By understanding the Poisson distribution, we can design better strategies for managing and predicting various types of events.