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I wonder a technique to extend the discrete probability space.

Here's an example from Concrete Mathematics EXERCISE 8.17:

Let $X_{n,p}$ and $Y_{n,p}$ have the binomial and negative binomial distributions, respectively, with parameters $(n,p)$. Prove that $\Pr(Y_{n,p}\le m) = \Pr(X_{m+n,p}\ge n)$.

The answer to the problem is also from Concrete Mathematics:

\begin{align} \Pr(Y_{n,p}\le m) &= \Pr(Y_{n,p}+n \le m+n) \\ &= \hbox{probability that we need $\le m+n$} \tag{1}\\ &= \hbox{probability that $m+n$ tosses yield $\ge n$ heads} \tag{2} \\ &= \Pr(X_{m+n,p}\ge n) \end{align}

Well, (1) and (2) are describing the same thing, but they're in different probability spaces, so we should extend these two probability spaces into a unique probability space, ensuring that the probability of each event doesn't change. How can we do it? I haven't a clear idea.

And the more general problem arises: How to extend a probability space? Is there any technique to do it, at least, treat part of problems?

Thanks for your help!

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1 Answer 1

up vote 2 down vote accepted

In general $X_{m+n,p}$ and $Y_{n,p}$ are all in different probability spaces, but we can define versions of them on the same probability space (technically this is known as coupling), by considering an infinite sequence of independent coin-flips with probability $p$ of heads in each one.

BTW, there seems to be some confusion here, due to different conventions for the negative binomial. The definition as given in the text has $Y_{n,p}$ as the number of flips until the $n$'th head, while the question appears to use a different definition where $Y_{n,p}$ is the number of tails until the $n$'th head. With the definition as given, the correct statement is that $\text{Pr}(Y_{n,p} \le m) = \text{Pr}(X_{m,p} \ge n)$.

EDIT: Oh, I see the problem. I was looking at (8.59), and they're taking (8.60) as the definition of the negative binomial. So it is the number of tails.

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The generating function of negative binomial distribution is $\displaystyle \left( \frac p {1-qz} \right)^n$, so it must be the number of tosses decreasing $n$. –  Frank Science Jun 6 '12 at 8:09
    
Coupling, yes, en.wikipedia.org/wiki/Coupling_(probability) I'm educated. –  Frank Science Jun 6 '12 at 8:12

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