# Poisson as Binomial

• January 3rd 2009, 02:06 AM
James0502
Poisson as Binomial
A textbook has n pages. The number of mistakes on a page is a poisson random variable with parameter a, and is independent of the number of mistakes on all other pages.

Question 1: What is the expected number of pages with no mistakes?

I get this as n*e^-a

Question 2: When reading the book you detect each mistake with probability p, independently of other mistakes. Let M denote the number of mistakes on a particular page and let D denote the number of mistakes detected. Write down P[D = k|M = m]

I modelled on binomial, parameters m and p

so = m choose k * p^k(1-p)^(m-k)

Question 3: Hence find, for each k >/= 0 P[D=k]

I'm not sure how to start this part

many thanks
• January 3rd 2009, 02:23 AM
mr fantastic
Quote:

Originally Posted by James0502
A textbook has n pages. The number of mistakes on a page is a poisson random variable with parameter a, and is independent of the number of mistakes on all other pages. What is the expected number of pages with no mistakes?

I get this as n*e^-a

When reading the book you detect each mistake with probability p, independently of other mistakes. Let M denote the number of mistakes on a particular page and let D denote the number of mistakes detected. Write down P[D = k|M = m]

I modelled on binomial, parameters m and p

so = m choose k * p^k(1-p)^(m-k)

Hence find, for each k >/= 0 P[D=k]

I'm not sure how to start this part

many thanks

Calculate the probability that a single page has no mistake: $p = \Pr(X = 0) = e^{-a}$.

Let Y be the random variable number of pages with no mistake.

Y ~ Binomial $\left(n, \, p = e^{-a}\right)$.

$E(Y) = np = n e^{-a}$.
• January 3rd 2009, 02:29 AM
James0502
yeh.. I got this part.. it's the next bit that got me..

sorry - my question was poorly written out..

any help with question 2 would be much appreciated!
• January 3rd 2009, 02:45 AM
mr fantastic
Quote:

Originally Posted by James0502
A textbook has n pages. The number of mistakes on a page is a poisson random variable with parameter a, and is independent of the number of mistakes on all other pages.

Question 1: What is the expected number of pages with no mistakes?

I get this as n*e^-a

Question 2: When reading the book you detect each mistake with probability p, independently of other mistakes. Let M denote the number of mistakes on a particular page and let D denote the number of mistakes detected. Write down P[D = k|M = m]

I modelled on binomial, parameters m and p

so = m choose k * p^k(1-p)^(m-k)

Hence find, for each k >/= 0 P[D=k]

I'm not sure how to start this part

many thanks

Question 2:

$\Pr(D = k \, | \, M = m) = ^mC_k \, p^k (1 - p)^{m-k}$.
• January 3rd 2009, 03:26 AM
mr fantastic
Quote:

Originally Posted by James0502
A textbook has n pages. The number of mistakes on a page is a poisson random variable with parameter a, and is independent of the number of mistakes on all other pages.

Question 1: What is the expected number of pages with no mistakes?

I get this as n*e^-a

Question 2: When reading the book you detect each mistake with probability p, independently of other mistakes. Let M denote the number of mistakes on a particular page and let D denote the number of mistakes detected. Write down P[D = k|M = m]

I modelled on binomial, parameters m and p

so = m choose k * p^k(1-p)^(m-k)

Question 3: Hence find, for each k >/= 0 P[D=k]

I'm not sure how to start this part

many thanks

Question 3:

If you detect k mistakes on a page then obviously $m \geq k$. So:

$\Pr(D = k) = \Pr(D = k \, | \, M = k) \cdot \Pr(M = k) +$ $\Pr(D = k \, | \, M = k + 1) \cdot \Pr(M = k + 1) + \, .....$

$= ^kC_k \, p^k (1 - p)^{k-k} \cdot \frac{e^{-a} a^k}{k!} +$ $^{k+1}C_k \, p^k (1 - p)^{k + 1 -k} \cdot \frac{e^{-a} a^{k+1}}{(k+1)!} + \, ....$

I suggest you add a couple more terms, simplify each one and take out the common factor. You're left with an infinite sum which doesn't look too bad to find I think.
• January 3rd 2009, 03:44 AM
mr fantastic
Quote:

Originally Posted by mr fantastic
Question 3:

If you detect k mistakes on a page then obviously $m \geq k$. So:

$\Pr(D = k) = \Pr(D = k \, | \, M = k) \cdot \Pr(M = k) +$ $\Pr(D = k \, | \, M = k + 1) \cdot \Pr(M = k + 1) + \, .....$

$= ^kC_k \, p^k (1 - p)^{k-k} \cdot \frac{e^{-a} a^k}{k!} +$ $^{k+1}C_k \, p^k (1 - p)^{k + 1 -k} \cdot \frac{e^{-a} a^{k+1}}{(k+1)!} + \, ....$

I suggest you add a couple more terms, simplify each one and take out the common factor. You're left with an infinite sum which doesn't look too bad to find I think.

I will leave it to you to show that the general term is

$\Pr(D = k \, | \, M = k + n) \cdot \Pr(M = k + n) =$ $^{k+n}C_k \, p^k (1 - p)^{k + n -k} \cdot \frac{e^{-a} a^{k+n}}{(k+n)!} = \frac{p^k a^k e^{-a}}{k!} \cdot \frac{[a(1 - p)]^n}{n!}$

where n = 0, 1, 2, 3, .....

and so your sum is $~ \frac{p^k a^k e^{-a}}{k!} \sum_{n=0}^{\infty} \frac{[a(1 - p)]^n}{n!}$.

Now recall the Maclaurin Series for $e^x$ ......