# Page History

## Key

• This line was added.
• This line was removed.
• Formatting was changed.

#### The problem

Consider the situation where we have observed s successes and know the probability of success p, but would like to know how many trials were actually done to have observed those successes. We wish to estimate a value that is fixed, so we require a distribution that represents our uncertainty about what the true value is. There are two possible situations: we either know that the trials stopped on the sth success or we do not.

#### Results

If we know that the trials stopped on the sth success, we can model our uncertainty about the true value of n as:

n = NegBinomial(p,s)

If, on the other hand, we do not know that the last trial was a success (though it could have been), then our uncertainty about n is modeled as:

n = NegBinomial(p, s+1) - 1

Both of these formulae result from a Bayesian analysis with Uniform priors for n.

#### Derivations

Let x be the number of trials that were needed to obtain the sth success. We will use a uniform prior for x, i.e. p(x) = c, and, from the binomial distribution, the likelihood function is the probability that at the (x-1)th trial there had been (s-1) successes and then the xth trial was a success, which is just the Negative Binomial probability mass function:

LaTeX Math Block
alignment left
 $$l(X \mid x)=\left( \begin{array}{c} x-1 \\ x-s \end{array} \right) p^{s}(1-p)^{x-s}$$

...

Since we are using a uniform prior (assuming no prior knowledge), and the equation for l(X|x) comes directly from a distribution (so it must sum to unity) we can dispense with the formality of normalizing the posterior distribution to one, and observe:

...

LaTeX Math Block
alignment left
 $$p(x)=\left( \begin{array}{c} x-1 \\ x-s \end{array} \right) p^{s}(1-p)^{x-s}$$

i.e. that x = NegBinomial(p,s).

In the second case, we do not know that the last trial was a success, only that in however many trials were completed, there were just s successes. We have the same Uniform prior for the number of trials, but our likelihood function is just the binomial probability mass function, i.e.:

...

LaTeX Math Block
alignment left
 $$l(X \mid x)=\left( \begin{array}{c} x \\ s \end{array} \right) p^{s}(1-p)^{x-s}$$

...

Since this does not have the form of a probability mass function of a known distribution, we need to complete the Bayesian analysis, so:

LaTeX Math Block
alignment left
 f \big(x \mid X \big)= \frac{\left( \begin{array}{c} x \\ s \end{array} \right) p^{s}\big(1-p\big)^{x-s}}{\displaystyle\sum_{i=s}^{\infty}\left( \begin{array}{c} i \\ s \end{array} \right) p^{s}(1-p)^{i-s}}

...

Look at the denominator, and substituting j = i+1 gives:

LaTeX Math Block
alignment left
\displaystyle\sum_{i=s}^{\infty}\left( \begin{array}{c} i \\ s \end{array} \right) p^{s}(1-p)^{i-s} = \displaystyle\sum_{j=s+1}^{\infty}\left( \begin{array}{c} j-1 \\ s \end{array} \right) p^{s}(1-p)^{j-s-1}=\bigg( \frac{1}{p}\bigg).\displaystyle\sum_{j=s+1}^{\infty}\left( \begin{array}{c} j-1 \\ s \end{array} \right) p^{s+1}(1-p)^{j-s-1}=\bigg(\frac{1}{p}\bigg)

since

LaTeX Math Block
alignment left
\left( \begin{array}{c} j-1 \\ s \end{array} \right) p^{s+1}(1-p)^{j-s-1}

is the probability mass function for the NegBinomial(s+1,p) distribution for j and therefore sums to 1. The posterior distribution then reduces to:

...

LaTeX Math Block
alignment left
f(x \mid X)=\left( \begin{array}{c} x \\ s \end{array} \right) p^{s+1}(1-p)^{x-s}

...

i.e. y = NegBinomial(p,s+1), and therefore x = NegBinomial(p,s+1) -1 distribution.

...