He everyone! I got the following question in class today, and I was wondering if you can shed some light on the question. Just to clarify, I am not looking for a solution. I am more so looking for a explanation about the problem. Anyway, here is the problem I was given:

A certain typing agency employs 2 typists. The average number of errors per article is 3 when typed by the first typist and 4.2 when typed by the second. If the article is equally likely to be typed by either typist, approximate the probability that it will have no errors.

Here is my solution to the problem:

$E[X] = \sum_{x:p(x)>0} xp(x)$

$\implies E[X] = 3(.5) + 4.2(.5)$

$\implies E[X] = 3.6$

Assuming a Poisson Distribution, we can say the following:

$p(i) = \frac{e^{-\lambda}\lambda^i}{i!}$

$\implies p(0) = \frac{e^{-3.6}(3.6)^0}{0!}$

$\implies p(0) = e^{-3.6}$

However, here is the interesting part. Being quite the perfectionist, I look up all of the answers once I complete the homework. Doing so, I found a similar but different answer which proceeds like this:

$p(0) = p(0|T_1)P(T_1) + p(0|T_2)P(T_2)$

Assuming a Poisson Distribution:

$\implies p(0) = \frac{e^{-3}(3)^0}{0!}(.5) + \frac{e^{-4.2}(4.2)^0}{0!}(.5)$

$\implies p(0) = .5(e^{-3} + e^{-4.2})$

So, which of these are correct? Are they both correct? An explanation would be amazing.

  • $\begingroup$ You cannot commute expectation with non-linear function: $Eg(X)\ne g(E(X))$. $\endgroup$ – A.S. Feb 25 '16 at 2:13

Your second answer is correct. The issue is that an average of Poison processes does not have a rate equal to the average of individual processes. Imagine the limiting case where one typist has a 0 error rate, and other has an infinite error rate. Your first solution would give a zero probability of no errors, but the answer is obviously 50 percent.

  • $\begingroup$ But then wouldn't a Poisson Distribution not apply to the problem since there is a high probability of success? $\endgroup$ – Birdman2246 Feb 25 '16 at 4:24

As shown by the second answer, while indeed you obtained the expectation of $X$, that random variable does not have a Poison distribution.

$$\begin{align}\mathsf E(X) ~ = & ~ \mathsf E(X\mid T_1)~\mathsf P(T_1)+\mathsf E(X\mid T_2)~\mathsf P(T_2) \\[1ex] ~= & ~ \tfrac 12\big(\mathsf E(X\mid T_1)+\mathsf E(X\mid T_1)\big) \\[1ex] = & ~ \dfrac{\lambda_1+\lambda_2}2 \\[1ex] = 3.6\end{align}$$

However, the probability that $X=k$ errors are made is: $$\begin{align}\mathsf P(X=k) ~ = & ~ \mathsf P(X=k\mid T_1)~\mathsf P(T_1)+\mathsf P(X=k\mid T_2)~\mathsf P(T_2) \\[1ex] = & ~ \dfrac{\lambda_1^k\mathsf e^{-\lambda_1}+\lambda_2^k\mathsf e^{-\lambda_2}}{2\times k!} & \neq ~ \dfrac{\left(\frac{\lambda_1+\lambda_2}2\right)^k\mathsf e^{(\lambda_1+\lambda_2)/2}}{k!} \\[1ex] = & ~ \dfrac{3^k\mathsf e^{-3}+4.2^k\mathsf e^{-4.2}}{2\times k!} & \neq ~ \dfrac{3.6^k\mathsf e^{-3.6}}{k!} \end{align}$$


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