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A book has $p$ pages and contains $q$ errors. The random variable $X$ is defined as the number of errors in a given page.

  • What probability distribution law does $X$ follows and why?
  • What is its expected value, its variance and its standard deviation?
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    If the book has more than 30 pages, and the probability is small then there is pretty much no difference between Poisson and binomial, however in practice we would use Poisson.2012-10-27

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Generally the Poisson Distribution is used to model the number of occurrences of a random variable in a given time interval or in your case the mean number of errors in your book $\ (q/p) $ and in general for events that do not occur very frequently.

Now a Poisson Distribution is defined to be

$ \ f(X|\lambda) = \lambda^{x}e^{-\lambda}/x!\, $

Where $\lambda$ would be the mean number of errors in your book:

$ \ \lambda = q/p $

and your random variable X is then the number of errors on any given page. We can now read the function $f(X|\lambda)$ as the probability of X errors on a page given that the mean number of errors in the book is $\lambda$.

Finding the expected value of a probability distribution is just a fancy way of asking what is the mean; for the Poisson Distribution that is the same thing as the mean we found earlier.

The variance can also be found by working out the simple sum

$ E[X(X-1)] = E[X^2]- E[X] = \sum_0^\infty x(x-1)f(X|\lambda) $

*hint along the way do a change of variables $y=x-2$

And by definition the variance we conclude that

$ Var(X) = E[X^2]- (E[X])^2 = \lambda $

The standard deviation is taken as the square root of the variance.

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    thank you and welcome to MSE2012-10-26
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Sounds like Poisson distribution because it satisfies the properties of a Poisson experiment. I believe mean and variance are the same $\lambda$, and sd is $\sqrt\lambda$.

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    Such a small sample size? Never mind. Neglect one property and we can still use Poisson -- does not make any difference.2012-10-26
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You do have to make an assumption about how errors appear on a page.

If you assume that each of the $q$ errors has an equal probability of appearing on each page and that the page each error appears on is independent of the pages where other errors appear then given $p$ and $q$, for each page the probability distribution is binomial, so $\Pr(X=x)= {q \choose x}\left(\frac{1}{p}\right)^x \left(1-\frac{1}{p}\right)^{q-x} = {q \choose x}\frac{\left(p-1\right)^{q-x}}{p^q}$ with mean $\frac{q}{p}$, variance $\frac{q(p-1)}{p^2}$ and standard deviation $\sqrt{\frac{q(p-1)}{p^2}}.$

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    @glebovg: imagining say 1,200 pages and say 100 errors, my computer can find the whole distribution for any one page almost instantly in R using the code `dbinom(0:100, 100, 1/1200)`. Meanwhile `dpois(0:100, 100/1200)` is also almost instant.2012-10-27
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Probability distribution function:

$f_X(n)=\frac{\binom{q+p-1-n}{p-2}}{\binom{q+p-1}{p-1}}$

for $n \le q$

The formula is equal to the numbers of ways to spread $q-n$ errors over $p-1$ pages divided by the number of ways to spread $q$ errors over $p$ pages.

http://en.wikipedia.org/wiki/Composition_(number_theory)

Expected value is easy:

$E(X)=\frac{q}{p}$

Variance and standard deviation will require some calculations...

$Var(X)=\sum_{n=0}^{q}\frac{\binom{q+p-1-n}{p-2}}{\binom{q+p-1}{p-1}}\cdot n^2-\frac{q^2}{p^2}$

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    Yes that is a special case. $f_X(q)$ should be one in case p=1.2012-10-26