91Ó°ÊÓ

The Poisson distribution is a probability distribution that is used to model the number of events that occur within a fixed interval of time or space. These events are assumed to occur with a known constant mean rate and independently of the time since the last event.
In the context of our exercise, we are dealing with typo errors made by two typists. Errors occur randomly and can be considered to follow a Poisson distribution because of their random and independent nature. Each typist has a different error rate, denoted by \( \lambda \) and \( \mu \), which represent the mean number of errors per page made by each typist. The Poisson PMF (probability mass function) is given by:
\[ p(x; \lambda) = e^{-\lambda} \frac{\lambda^x}{x!} \]
where \( x \) is a non-negative integer representing the number of errors, \( \lambda \) is the average number of errors, and \( e \) is the base of the natural logarithm. This formula helps calculate the probability of observing exactly \( x \) errors in a page.

The expected value, often denoted as \( E(X) \), is a measure of the central tendency of a random variable, indicating the average outcome one would expect from the random process. For a Poisson distribution, the expected value is equal to the parameter \( \lambda \).
In the exercise, when both typists are equally involved in typing, the expected number of errors on a page is calculated as the average of the two error rates. Thus, the expected value for the distribution \( p(x; \lambda, \mu) \) is:
\[ E(X) = \frac{1}{2}(\lambda + \mu) \]
If the typists are involved in different proportions, such as the first typist typing 60% of the pages, the expected value changes accordingly:
\[ E(X) = 0.6\lambda + 0.4\mu \]
This formula reflects the average errors expected per page considering the different contributions of each typist.

Variance measures how much the values of the random variable \( X \) are spread out from the expected value. For a Poisson-distributed random variable, the variance is equal to its expected value, \( \sigma^2 = \lambda \).
In our scenario, when both typists contribute equally, the variance of the number of errors made in a page is:
\[ \sigma^2(X) = \frac{1}{2}(\lambda + \mu) \]
This indicates the spread from the average number of errors considering an equal share of typing. When the typists contribute unequally, such as the first typist typing 60% of the pages, the variance adjusts proportionately:
\[ \sigma^2(X) = 0.6\lambda + 0.4\mu \]
This revised formula showcases how the distribution of errors' variance changes based on the varying weighting from each typist.

A vital property of any probability mass function (PMF) is the non-negativity condition. This condition requires all probabilities to be non-negative for any possible value \( x \) of the random variable. In simpler terms, probabilities can't be negative because they represent the likelihood of an event occurring.
Our function \( p(x; \lambda, \mu) \) satisfies this condition because:

• The exponential component \( e^{-\text{rate}} \) is always positive, no matter the rate.
• The factorial part, \( x! \), is positive for all non-negative integers \( x \).
• The prefactor \( \frac{1}{2} \) is positive, ensuring that \( p(x; \lambda, \mu) \geq 0 \) for \( x=0,1,2,\ldots \).

Since the PMF is explicitly set to zero for values not listed (such as negative x), it inherently adheres to non-negativity across its domain. This is a crucial step in validating that a function is indeed a PMF.