Chapter 3, Problem 115SE

There are two Certified Public Accountants in a particular office who prepare tax returns for clients. Suppose that for a particular type of complex form, the number of errors made by the first preparer has a Poisson distribution with mean value μ₁, the number of errors made by the second preparer has a Poisson distribution with mean value μ₂, and that each CPA prepares the same number of forms of this type. Then if a form of this type is randomly selected, the function

$p(x;{\mu }_1,{\mu }_2)\text{ = }\text{.5}\frac{e^{-{\mu }_1}{\mu }_1^x}{x!}\text{ + }\text{.5}\frac{e^{-{\mu }_2}{\mu }_2^x}{x!}x\text{ = 0, 1, 2,}\mathrm{...}$

gives the pmf of X = the number of errors on the selected form.

1. a. Verify that p(x; μ₁, μ₂) is in fact a legitimate pmf (≥ 0 and sums to 1).
2. b. What is the expected number of errors on the selected form?
3. c. What is the variance of the number of errors on the selected form?
4. d. How does the pmf change if the first CPA prepares 60% of all such forms and the second prepares 40%?

To determine

Verify whether $p\left(x;{\mu }_1,{\mu }_2\right)$ is a legitimate pmf.

Answer to Problem 115SE

The pmf $p\left(x;{\mu }_1,{\mu }_2\right)$ is a legitimate pmf.

Explanation of Solution

Given info:

Between two Certified Public Accountants in a particular office, the number of errors made by the first preparer has a Poisson distribution with mean value ${\mu }_1$ and the number of errors made by the second preparer has a Poisson distribution with mean value ${\mu }_2$. Each prepares the same number of forms of this type. The function is

$p\left(x;{\mu }_1,{\mu }_2\right)=0.5\frac{e^{-{\mu }_1}{\mu }_1{}^x}{x!}+0.5\frac{e^{-{\mu }_2}{\mu }_2{}^x}{x!},x=0,1,2,\mathrm{3...}$

Where, X be the number of errors in the selected form.

Calculation:

The properties of legitimate pmf are,

1. 1. $p\left(x\right)\ge 0$
2. 2. ${\displaystyle \sum _{allx}p\left(x\right)}=1$

The pmf is $p\left(x;{\mu }_1,{\mu }_2\right)=0.5\frac{e^{-{\mu }_1}{\mu }_1{}^x}{x!}+0.5\frac{e^{-{\mu }_2}{\mu }_2{}^x}{x!},x=0,1,2,\mathrm{3...}$a

In the expression $\frac{e^{-{\mu }_1}{\mu }_1{}^x}{x!}$ and $\frac{e^{-{\mu }_2}{\mu }_2{}^x}{x!}$ are two Poisson pmfs with parameter ${\mu }_{1}and{\mu }_2$.

As, the both pmfs are greater than equal to 0, the function $0.5\frac{e^{-{\mu }_1}{\mu }_1{}^x}{x!}+0.5\frac{e^{-{\mu }_2}{\mu }_2{}^x}{x!}\ge 0$.

Thus, 1^st property satisfied.

Now, ${\displaystyle \sum _{x=0}^{\infty }p\left(x;{\mu }_1,{\mu }_2\right)}$ is,

$\begin{array}{c}{\displaystyle \sum _{x=0}^{\infty }p\left(x;{\mu }_1,{\mu }_2\right)}={\displaystyle \sum _{x=0}^{\infty }0.5\frac{e^{-{\mu }_1}{\mu }_1{}^x}{x!}}+{\displaystyle \sum _{x=0}^{\infty }0.5\frac{e^{-{\mu }_2}{\mu }_2{}^x}{x!}}\\ =0.5{\displaystyle \sum _{x=0}^{\infty }\frac{e^{-{\mu }_1}{\mu }_1{}^x}{x!}}+0.5{\displaystyle \sum _{x=0}^{\infty }\frac{e^{-{\mu }_2}{\mu }_2{}^x}{x!}}\end{array}$

In the expression $\frac{e^{-{\mu }_1}{\mu }_1{}^x}{x!}$ and $\frac{e^{-{\mu }_2}{\mu }_2{}^x}{x!}$ are two Poisson pmfs with parameter ${\mu }_1$ and ${\mu }_2$.

Hence, ${\displaystyle \sum _{x=0}^{\infty }\frac{e^{-{\mu }_1}{\mu }_1{}^x}{x!}=1\text{and}}{\displaystyle \sum _{x=0}^{\infty }\frac{e^{-{\mu }_2}{\mu }_2{}^x}{x!}}=1$

Substituting the values,

$\begin{array}{c}{\displaystyle \sum _{x=0}^{\infty }p\left(x;{\mu }_1,{\mu }_2\right)}=0.5{\displaystyle \sum _{x=0}^{\infty }\frac{e^{-{\mu }_1}{\mu }_1{}^x}{x!}}+0.5{\displaystyle \sum _{x=0}^{\infty }\frac{e^{-{\mu }_2}{\mu }_2{}^x}{x!}}\\ =0.5\times 1+0.5\times 1\\ =1\end{array}$

Thus, 2^nd property has satisfied.

Hence, the pmf is a legitimate pmf.

To determine

Find the expected number of errors on the selected form.

Answer to Problem 115SE

The expected number of errors on the selected form is $E\left(X\right)=0.5\times {\mu }_1+0.5\times {\mu }_2$

Explanation of Solution

Calculation:

The expected number of errors on the selected form:

Using the formula, $E\left(X\right)={\displaystyle \sum _{x=0}^{\infty }x\times p\left(x\right)}$

$\begin{array}{c}E\left(X\right)={\displaystyle \sum _{x=0}^{\infty }x\times p\left(x;{\mu }_1,{\mu }_2\right)}\\ =0.5{\displaystyle \sum _{x=0}^{\infty }x\times \frac{e^{-{\mu }_1}{\mu }_1{}^x}{x!}}+0.5{\displaystyle \sum _{x=0}^{\infty }x\times \frac{e^{-{\mu }_2}{\mu }_2{}^x}{x!}}\\ =0.5\times E\left(X,X\sim Poisson\left({\mu }_1\right)\right)+0.5\times E\left(X,X\sim Poisson\left({\mu }_2\right)\right)\end{array}$

Here, $\frac{e^{-{\mu }_1}{\mu }_1{}^x}{x!}$ and $\frac{e^{-{\mu }_2}{\mu }_2{}^x}{x!}$ are two Poisson pmfs with parameter ${\mu }_{1}and{\mu }_2$.

It is known that for Poisson distribution, $E\left(X\right)=\mu$, If $X\sim Poisson\left(\mu \right)$

Hence, above expression will be

$\begin{array}{c}E\left(X\right)=0.5\times \left[E\left(X,X\sim Poisson\left({\mu }_1\right)\right)\right]+\left[0.5\times E\left(X,X\sim Poisson\left({\mu }_2\right)\right)\right]\\ =0.5\times \left({\mu }_1\right)+\left(0.5\times {\mu }_2\right)\end{array}$

Hence, the expected number of errors on the selected form is $E\left(X\right)=0.5\times {\mu }_1+0.5\times {\mu }_2$.

To determine

Find the variance of the number of errors on the selected form.

Answer to Problem 115SE

The variance of the number of errors on the selected form is $V\left(X\right)=0.5{\mu }_1+0.5{\mu }_2+0.25{\left({\mu }_1-{\mu }_2\right)}^2$.

Explanation of Solution

Calculation:

It is known that for a random variable X ,$V\left(X\right)=E\left(X^2\right)-{\left\{E\left(X\right)\right\}}^2$

Using the formula of $E\left(X^2\right)={\displaystyle \sum _{x=0}^{\infty }{x}^2\times p\left(x\right)}$

$\begin{array}{c}E\left(X^2\right)={\displaystyle \sum _{x=0}^{\infty }{x}^2\times p\left(x;{\mu }_1,{\mu }_2\right)}\\ =0.5{\displaystyle \sum _{x=0}^{\infty }{x}^2\times \frac{e^{-{\mu }_1}{\mu }_1{}^x}{x!}}+0.5{\displaystyle \sum _{x=0}^{\infty }{x}^2\times \frac{e^{-{\mu }_2}{\mu }_2{}^x}{x!}}\\ =0.5\times E\left(X^2,X\sim Poisson\left({\mu }_1\right)\right)+0.5\times E\left(X^2,X\sim Poisson\left({\mu }_2\right)\right)\end{array}$

Here, $\frac{e^{-{\mu }_1}{\mu }_1{}^x}{x!}$ and $\frac{e^{-{\mu }_2}{\mu }_2{}^x}{x!}$ are two Poisson pmfs with parameter ${\mu }_{1}and{\mu }_2$.

It is known that for Poisson distribution, $E\left(X\right)=\mu ,V\left(X\right)=\mu$, If $X\sim Poisson\left(\mu \right)$.

Thus,

$\begin{array}{c}E\left(X^2\right)=V\left(X\right)+{\left\{E\left(X\right)\right\}}^2\\ =\mu +{\mu }^2\end{array}$

Using this formula,

$\begin{array}{c}E\left(X^2\right)=0.5\times \left[E\left(X^2,X\sim Poisson\left({\mu }_1\right)\right)\right]+\left[0.5\times E\left(X^2,X\sim Poisson\left({\mu }_2\right)\right)\right]\\ =0.5\times \left({\mu }_1+{\mu }_1^2\right)+0.5\times \left({\mu }_2+{\mu }_2^2\right)\end{array}$

The variance of the number of errors on the selected form:

$\begin{array}{c}V\left(X\right)=E\left(X^2\right)-{\left\{E\left(X\right)\right\}}^2\\ =0.5\times \left({\mu }_1+{\mu }_1^2\right)+0.5\times \left({\mu }_2+{\mu }_2^2\right)-{\left\{0.5\times {\mu }_1+0.5\times {\mu }_2\right\}}^2\\ =0.5{\mu }_1+0.5{\mu }_1^2+0.5{\mu }_2+0.5{\mu }_2^2-0.25{\left({\mu }_1+{\mu }_2\right)}^2\\ =0.5{\mu }_1+0.5{\mu }_1^2+0.5{\mu }_2+0.5{\mu }_2^2-0.25\left\{{\mu }_1^2+2\times {\mu }_1\times {\mu }_2+{\mu }_2^2\right\}\end{array}$

           $\begin{array}{c}=0.5{\mu }_1+0.5{\mu }_1^2+0.5{\mu }_2+0.5{\mu }_2^2-0.25{\mu }_1^2-0.5\times {\mu }_1\times {\mu }_2-0.25{\mu }_2^2\\ =0.5{\mu }_1+0.5{\mu }_2+0.25{\mu }_1^2+0.25{\mu }_2^2-0.5\times {\mu }_1\times {\mu }_2\\ =0.5{\mu }_1+0.5{\mu }_2+0.25{\left({\mu }_1-{\mu }_2\right)}^2\end{array}$

Hence, the variance of the number of errors on the selected form is $V\left(X\right)=0.5{\mu }_1+0.5{\mu }_2+0.25{\left({\mu }_1-{\mu }_2\right)}^2$.

To determine

Find the changes in the pmf if the first CPA prepares 60% of all such forms and the second prepares 40%.

Answer to Problem 115SE

The pmf will be

$p\left(x;{\mu }_1,{\mu }_2\right)=0.6\frac{e^{-{\mu }_1}{\mu }_1{}^x}{x!}+0.4\frac{e^{-{\mu }_2}{\mu }_2{}^x}{x!},x=0,1,2,\mathrm{3...}$.

Explanation of Solution

Calculation:

The first CPA prepares 60% of all such forms means the proportion is 0.6 for 1^st CPA prepares.

Thus, the  weights of the pmf corresponding to first CPA is 0.6.

The second CPA prepares 40% of all such forms means the proportion is 0.4 for 2^nd CPA prepares.

Thus, the weights of the pmf corresponding to second CPA is 0.4.

Hence, the pmf will be

$p\left(x;{\mu }_1,{\mu }_2\right)=0.6\frac{e^{-{\mu }_1}{\mu }_1{}^x}{x!}+0.4\frac{e^{-{\mu }_2}{\mu }_2{}^x}{x!},x=0,1,2,\mathrm{3...}$