Home > Probability theory and statistics > Stochastic processes > The Binomial process > Estimation of the number of trials n made after having observed s successes with probability p

Estimation of the number of trials n made after having observed s successes with probability p

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 s^th success or we do not.

Results

If we know that the trials stopped on the s^th 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 s^th 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 x^th trial was a success, which is just the Negative Binomial probability mass function:

$//<![CDATA[ \begin{array}{l}l(X \mid x)=\left( \begin{array}{c} x-1 \\ x-s \end{array} \right) p^{s}(1-p)^{x-s}\end{array} //]]>$

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:

$//<![CDATA[ \begin{array}{l}p(x)=\left( \begin{array}{c} x-1 \\ x-s \end{array} \right) p^{s}(1-p)^{x-s}\end{array} //]]>$

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.:

$//<![CDATA[ \begin{array}{l}l(X \mid x)=\left( \begin{array}{c} x \\ s \end{array} \right) p^{s}(1-p)^{x-s}\end{array} //]]>$
            

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

$//<![CDATA[ \begin{array}{l}f \big(x \mid X \big)= \frac{\left( \begin{array}{c} x \\ s \end{array} \right) p^{s}\big(1-p\big)^{x-s}}{\displaystyle\sum_{i=s}^{\infty}\left( \begin{array}{c} i \\ s \end{array} \right) p^{s}(1-p)^{i-s}}\end{array} //]]>$

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

$//<![CDATA[ \begin{array}{l}\displaystyle\sum_{i=s}^{\infty}\left( \begin{array}{c} i \\ s \end{array} \right) p^{s}(1-p)^{i-s} = \displaystyle\sum_{j=s+1}^{\infty}\left( \begin{array}{c} j-1 \\ s \end{array} \right) p^{s}(1-p)^{j-s-1}=\bigg( \frac{1}{p}\bigg).\displaystyle\sum_{j=s+1}^{\infty}\left( \begin{array}{c} j-1 \\ s \end{array} \right) p^{s+1}(1-p)^{j-s-1}=\bigg(\frac{1}{p}\bigg)\end{array} //]]>$
                                                      

since

  $//<![CDATA[ \begin{array}{l}\left( \begin{array}{c} j-1 \\ s \end{array} \right) p^{s+1}(1-p)^{j-s-1}\end{array} //]]>$
  

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:

$//<![CDATA[ \begin{array}{l}f(x \mid X)=\left( \begin{array}{c} x \\ s \end{array} \right) p^{s+1}(1-p)^{x-s}\end{array} //]]>$

For x = y - 1:

$//<![CDATA[ \begin{array}{l}f(y \mid X)=\left( \begin{array}{c} y-1 \\ s \end{array} \right) p^{s+1}(1-p)^{y-s-1}\end{array} //]]>$

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