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There is this problem on the textbook. The derivation of the answer in the solution was not very clear, but this is how the problem goes:

Calls arrive at an office, following a Poission distribution, at 20 calls per hour. 20% of all calls result in sales. Given that $x$ sales were made in a certain hour, find the probability function for $Y$, the number of calls made in that hour.

My steps:

Let $X={\text{calls}}$, $Y={\text{sales}}$. Now $$P(Y=y\mid X=x)=\frac{P(X=x\mid Y=y)P(X=x)}{P(Y=y)}$$ from Bayes' Theorem. Expanding we get $$P(Y=y\mid X=x) = \frac{\binom{X}{y}0.2^{y}0.8^{x-y}\frac{e^{-20}20^{x}}{x!}}{\sum_{k=y}^{\infty}\binom{k}{y}0.2^{y}0.8^{k-y}\frac{e^{-20}20^{k}}{k!}} = \frac{\binom{x}{y}0.2^{y}0.8^{x-y}\frac{e^{-20}20^{x}}{x!}}{\sum_{k=y}^{\infty}\frac{k!}{y!(k-y)!}0.2^{y}0.8^{k-y}\frac{e^{-20}20^{k}}{k!}} = \frac{\binom{x}{y}0.8^{x-y}\frac{20^{x}}{x!}}{\frac{1}{20^{y}}\sum_{k=y}^{\infty}\frac{0.8^{k-y}20^{k-y}}{(k-y)!}} = \frac{\binom{x}{y}0.8^{x-y}\frac{20^{x}}{x!}}{\frac{1}{20^{y}}\sum_{k=y}^{\infty}\frac{16^{k-y}}{(k-y)!}} = \frac{\binom{x}{y}0.8^{x-y}\frac{20^{x}}{x!}}{\frac{1}{20^{y}}e^{16}} = \frac{\binom{x}{y}0.8^{x-y}\frac{20^{x}20^{y}y!}{x!}}{e^{16}} = e^{-16}\binom{x}{y}0.8^{x-y}\frac{20^{x+y}}{(y-x)!} $$

However, the textbook gives a solution of $$\frac{16^{y-x}e^{-16}}{(y-x)!}$$

I am not very experienced at manipulating stuff with lots of factorials and sigmas in them. Are the two expressions equivalent? Or did I do some step wrong?

(p.s. sorry for the ugly latex. Feel free to edit it; I copied it from LyX and all the equarray formatting was lost)

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You are starting from the wrong formula. –  André Nicolas Mar 16 '13 at 2:41

2 Answers 2

up vote 1 down vote accepted

We want $\Pr(Y=y|X=x)$. By the usual definition of conditional probability, we have $$\Pr(Y=y|X=x)=\frac{\Pr((Y=y)\cap (X=x))}{\Pr(X=x)}.$$

By a calculation, or by a standard fact, $X$ has Poisson distribution with parameter $4$. So $\Pr(X=x)=e^{-4}\frac{4^x}{x!}$.

To calculate the probability that $Y=y$ and $X=x$, note that the probability that $Y=y$ is $e^{-20}\frac{20^y}{y!}$. And given that $Y=y$, the probability that $X=x$ is $\binom{y}{x}\left(\frac{1}{5}\right)^x \left(\frac{4}{5}\right)^{y-x}$.

Substitute, simplify a bit. We get the textbook answer.

Remark: We more or less used the Bayes Formula, except a little indirectly, starting from the defining formula for conditional probability. I sometimes deliberately have avoided mentioning the Bayes Formula, because students so often get it wrong. But students are always looking for formulas that make life easier, and will find the Bayes formula in the text. And then in a test get it wrong.

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Wow much better than my textbook explanation ^_^ –  user54609 Mar 16 '13 at 17:18

The computation has the following problems:

  • As André Nicolas says, the statement of the Bayes formula is not correct.
  • In the body of the computation, the starting formula is correct, except that the roles of $x$ and $y$ have been reversed.
  • Later, $20^k$ was incorrectly changed to $\frac{20^{k-y}}{20^y}$ in the denominator.
  • At the same time, a factor of $\frac{1}{y!}$ was dropped in the denominator.
  • Near the end of the computation, the $\frac{1}{y!}$ that was dropped earlier was correctly put back as $y!$ in the numerator (so, these two errors had no net effect.)
  • In the last step, $\frac{y!}{x!}$ was incorrectly simplified to $\frac{1}{(y-x)!}$.

Apart from these errors, the computation is correct.

This is a special case of the following result: if $Y$ is Poisson distributed with parameter $\lambda$, and $X$ is binomial with $Y$ trials and success probability $p$, then, considering only $X$ and $Y-X$, these two random variables are independent and both Poisson distributed, with parameters $p\lambda$ and $(1-p)\lambda$, respectively. In the problem given above, then, the number of calls which did not result in sales, $Y-X$, is independent of the number $X$ of sales, and the distribution of $Y-X$ is always Poisson with parameter $20\cdot (1-0.2)=16$.

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