Question

Suppose that the number of typographical errors in a new text is Poisson distributed with mean \(\lambda\). Two proofreaders independently read the text. Suppose that each error is independently found by proofreader \(i\) with probability \(p_{i}, i=1,2 .\) Let \(X_{1}\) denote the number of errors that are found by proofreader 1 but not by proofreader \(2 .\) Let \(X_{2}\) denote the number of errors that are found by proofreader 2 but not by proofreader \(1 .\) Let \(X_{3}\) denote the number of errors that are found by both proofreaders. Finally, let \(X_{4}\) denote the number of errors found by neither proofreader. (a) Describe the joint probability distribution of \(X_{1}, X_{2}, X_{3}, X_{4}\). (b) Show that $$ \frac{E\left[X_{1}\right]}{E\left[X_{3}\right]}=\frac{1-p_{2}}{p_{2}} \text { and } \frac{E\left[X_{2}\right]}{E\left[X_{3}\right]}=\frac{1-p_{1}}{p_{1}} $$ Suppose now that \(\lambda, p_{1}\), and \(p_{2}\) are all unknown. (c) By using \(X_{i}\) as an estimator of \(E\left[X_{i}\right], i=1,2,3\), present estimators of \(p_{1}, p_{2}\) and \(\lambda\). (d) Give an estimator of \(X_{4}\), the number of errors not found by either proofreader.

Short answer

The joint probability mass function (pmf) of \(X_1, X_2\), and \(X_3\) is: \[ P(X_1=k_1, X_2=k_2, X_3=k_3)=\binom{\lambda}{k_1, k_2, k_3}(p_1p_2)^{k_3}(p_1(1-p_2))^{k_1}((1-p_1)p_2)^{k_2} \] We have, \[ \frac{E\left[X_{1}\right]}{E\left[X_{3}\right]} = \frac{1-p_2}{p_2} \\ \frac{E\left[X_{2}\right]}{E\left[X_{3}\right]} = \frac{1-p_1}{p_1} \] Estimators for \(p_1, p_2,\) and \(\lambda\) are: \[ \hat{p_2}=\frac{X_1}{X_1+X_3} \] \[ \hat{p_1} = \frac{X_2}{X_2+X_3} \] \[ \hat{\lambda} = X_1 + X_2 + X_3 \] And the estimator of the number of errors not found by either proofreader is: \(\hat{X_4} = 0\).

Step-by-step solution

There are four possible outcomes for each error: 1. The error is found by proofreader 1 but not by proofreader 2. This happens with probability \(p_1(1-p_2)\). 2. The error is found by proofreader 2 but not by proofreader 1. This happens with probability \((1-p_1)p_2\). 3. The error is found by both proofreaders. This happens with probability \(p_1p_2\). 4. The error is not found by either proofreader. This happens with probability \((1-p_1)(1-p_2)\). #Step 2: Describe the joint probability distribution#

The given problem can be modeled as a trinomial distribution with parameters \(n, p_1p_2, p_1(1-p_2)\), and \((1-p_1)p_2\). The joint probability mass function (pmf) of \(X_1, X_2\), and \(X_3\) is: \[ P(X_1=k_1, X_2=k_2, X_3=k_3)=\binom{\lambda}{k_1, k_2, k_3}(p_1p_2)^{k_3}(p_1(1-p_2))^{k_1}((1-p_1)p_2)^{k_2} \] (b) #Step 3: Find the expectation of \(X_1, X_2, X_3\)#

Using the properties of expectation: \[ E\left[X_1\right]=\sum_{k_1}k_1P(X_1=k_1)=\lambda p_1 (1-p_2) \] \[ E\left[X_2\right]=\sum_{k_2}k_2P(X_2=k_2)=\lambda (1-p_1) p_2 \] \[ E\left[X_3\right]=\sum_{k_3}k_3P(X_3=k_3)=\lambda p_1 p_2 \] #Step 4: Show the required ratio of expectations#

We have, \[ \frac{E\left[X_{1}\right]}{E\left[X_{3}\right]} = \frac{\lambda p_1 (1-p_2)}{\lambda p_1 p_2} = \frac{1-p_2}{p_2} \\ \frac{E\left[X_{2}\right]}{E\left[X_{3}\right]} = \frac{\lambda (1-p_1) p_2}{\lambda p_1 p_2} = \frac{1-p_1}{p_1} \] (c) #Step 5: Present estimators for \(p_1, p_2,\) and \(\lambda\)#

Since \(X_i\) is an estimator of \(E\left[X_i\right]\) for \(i=1,2,3\), we have: \[ \hat{p_2}=\frac{X_1}{X_1+X_3} \] \[ \hat{p_1} = \frac{X_2}{X_2+X_3} \] \[ \hat{\lambda} = X_1 + X_2 + X_3 \] (d) #Step 6: Give an estimator of \(X_4\)#

We can find an estimator of \(X_4\) as follows: \[ \hat{X_4} = \hat{\lambda} - (X_1 + X_2 + X_3) = 0 \] Note that \(\hat{X_4}\) is the number of errors found by neither proofreader. Since the estimators for \(p_1\) and \(p_2\) are based on the errors found by each proofreader, we cannot estimate the number of errors that are missed by both. In this case, the estimator of \(X_4\) is zero.