2
$\begingroup$

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?
$\endgroup$
2
  • 1
    $\begingroup$ Like I said use Poisson. For future reference anything along the lines of: the number of typing errors per page, number of telephone calls per hour, the number of customers during a time period, the number of employees who fill in a form, etc. usually means Poisson. $\endgroup$
    – glebovg
    Oct 25, 2012 at 23:32
  • $\begingroup$ 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. $\endgroup$
    – glebovg
    Oct 27, 2012 at 1:00

4 Answers 4

2
$\begingroup$

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 $\sqrt{\frac{q(p-1)}{p^2}}.$

$\endgroup$
5
  • $\begingroup$ and before wnvl chips in, for large $p$ the variance is close to the mean, but for small $p$ it is not. $\endgroup$
    – Henry
    Oct 25, 2012 at 23:56
  • $\begingroup$ Yes, it a better solution than mine. +1 My solution should give the same answer. $\endgroup$
    – wnvl
    Oct 26, 2012 at 0:08
  • $\begingroup$ If the sample size is large enough and probability is small then it does not make any difference. Poisson or binomial. Given that most books are longer than 30 pages, Poisson is a good approximation to binomial. Poisson should give the same answer and your calculator will not explode. $\endgroup$
    – glebovg
    Oct 26, 2012 at 0:16
  • $\begingroup$ Imagine using binomial on something like Les Miserables, your computer will take forever. Publishers use Poisson. $\endgroup$
    – glebovg
    Oct 27, 2012 at 1:00
  • 1
    $\begingroup$ @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. $\endgroup$
    – Henry
    Oct 27, 2012 at 19:11
2
$\begingroup$

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 $\sqrt\lambda$.

$\endgroup$
9
  • $\begingroup$ what properties? $\endgroup$
    – user31280
    Oct 25, 2012 at 23:16
  • $\begingroup$ Poisson experiment i.e successes or failures, the probability of more than one success occurring within a very short interval is small, etc. $\endgroup$
    – glebovg
    Oct 25, 2012 at 23:19
  • $\begingroup$ Poisson could be a good approximation, but an exact solution is possible. $\endgroup$
    – wnvl
    Oct 25, 2012 at 23:21
  • $\begingroup$ @wnvl Observe that the number of successes in any interval is independent of the number of successes in any other interval, the probability of success in an interval is the same for all equal-sized intervals, the probability of more than one success in an interval approaches 0 as the interval becomes smaller, etc. It is Poisson! $\endgroup$
    – glebovg
    Oct 25, 2012 at 23:25
  • 1
    $\begingroup$ The number of successes in any interval is not independent of the number of successes in any other interval, as the total sum of all errors should be q. $\endgroup$
    – wnvl
    Oct 25, 2012 at 23:32
1
$\begingroup$

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...

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

$\endgroup$
4
  • $\begingroup$ This sound utterly unnecessary and misleading. Poisson distribution is good enough unless you are interested in mean, variance, probability etc. to more than 4 decimal places. $\endgroup$
    – glebovg
    Oct 25, 2012 at 23:44
  • $\begingroup$ I don't understand your remark, with Poisson you can also calculate mean and variance (you showed it in your answer). For practical application I would probably choose for Poisson as well, just wanted to show that an exact solution is possible. $\endgroup$
    – wnvl
    Oct 25, 2012 at 23:49
  • $\begingroup$ How does your first probability function cope with $p=1$? $\endgroup$
    – Henry
    Oct 25, 2012 at 23:54
  • $\begingroup$ Yes that is a special case. $f_X(q)$ should be one in case p=1. $\endgroup$
    – wnvl
    Oct 26, 2012 at 0:01
0
$\begingroup$

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.

$\endgroup$
3
  • $\begingroup$ Yes. That is my answer and my comment summarized. $\endgroup$
    – glebovg
    Oct 26, 2012 at 0:05
  • $\begingroup$ Hey sorry. This is my first answer so it took me some time to write out the equation so I hadn't noticed any new posts until I finished. $\endgroup$
    – kuantumbro
    Oct 26, 2012 at 0:09
  • $\begingroup$ thank you and welcome to MSE $\endgroup$
    – user31280
    Oct 26, 2012 at 23:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy