I have an urn containing $N$ balls, of which $n$ ~ Uniform(1, ..., $N$) are white and the rest black. After sampling $i$ balls without replacement (of which $m$ are white), how can I find the conditional distribution on the colour of the next ball, p($x_{i+1}$ | $m$)?

I've been reading about hypergeometric and beta-binomial distributions but haven't been able to work out an answer. Thanks.


Define the events $A=\{x_{i+1}=\text{white}\}$, $B=\{\text{in $i$ firstly sampled balls $m$ are white}\}$. By definition of conditional probability, $$ p(x_{i+1}\mid m)=\mathbb P(A\mid B) = \frac{\mathbb P(A\cap B)}{\mathbb P(B)} $$ Find numerator and denominator by Law of total probability. $$ \mathbb P(B) = \sum_{n=1}^N p(n)\mathbb P(B\mid \text{$n$ white balls in urn})=\sum_{n=m}^N \frac1N \frac{\binom{n}{m}\binom{N-n}{i-m}}{\binom{N}{i}} $$ It is assumed that the binomial coefficients vanish if the subscript exceeds the upper index. So we can rewrite this sum in more convinient form $$ \mathbb P(B) =\frac1N \frac{1}{{\binom{N}{i}}} \sum_{n=0}^N \binom{n}{m}\binom{N-n}{i-m} = \frac1N \frac{\binom{N+1}{i+1}}{{\binom{N}{i}}}. \tag{1}\label{1} $$ In last equality we applied the equation (8) from this wiki page.

Similar way, $$ \mathbb P(A\cap B) = \sum_{n=1}^N p(n)\,\mathbb P(B\mid n)\,\mathbb P(A\mid B \cap \{n \text{ white balls\}}) $$ $$=\sum_{n=m+1}^N \frac1N \frac{\binom{n}{m}\binom{N-n}{i-m}}{\binom{N}{i}}\,\frac{n-m}{N-i}. $$ Rewrite $$ \binom{n}{m}\frac{n-m}{N-i} = \frac{m+1}{N-i} \binom{n}{m+1} $$ (simply check the identity!). Then
$$ \mathbb P(A\cap B) = \frac1N \frac{m+1}{N-i} \frac{1}{\binom{N}{i}} \sum_{n=m+1}^N {\binom{n}{m+1}\binom{N-n}{i-m}} $$ $$=\frac1N \frac{m+1}{N-i} \frac{1}{\binom{N}{i}} \sum_{n=0}^N {\binom{n}{m+1}\binom{N-n}{(i+1)-(m+1)}} = \frac1N \frac{m+1}{N-i} \frac{\binom{N+1}{i+2}}{\binom{N}{i}} \tag{2}\label{2}. $$ In last equality we used equation (8) from wiki again.

Finally, put (\ref{1}) and (\ref{2}) into the conditional probability: $$ p(x_{i+1}\mid m)=\frac{ \frac1N \frac{m+1}{N-i} \frac{\binom{N+1}{i+2}}{\binom{N}{i}}}{ \frac1N \frac{\binom{N+1}{i+1}}{\binom{N}{i}}} = \frac{m+1}{i+2}. $$

Edit: As rightly noted by Luke Hewitt, this solution implicitly assumes that $m> 0$. For this case only the equality (\ref{1}) is valid. For $m=0$,

$$ \mathbb P(B) =\frac{\sum_{n=1}^N \binom{n}{m}\binom{N-n}{i-m}}{N{\binom{N}{i}}} =\frac{1}{N{\binom{N}{i}}} \sum_{n=0}^N \binom{n}{m}\binom{N-n}{i-m}-\frac1N = \frac1N \frac{\binom{N+1}{i+1}}{{\binom{N}{i}}}-\frac1N. \tag{3}\label{3} $$ Divide (\ref{2}) to (\ref{3}) and obtain the conditional probability $p(x_{i+1}\mid m)$ for $m=0$: $$ p(x_{i+1}\mid m) =\frac{m+1}{i+2}\cdot\frac{N+1}{N-i}. $$

So, the final answer looks as: $$ p(x_{i+1}\mid m) =\begin{cases}\frac{m+1}{i+2}\cdot\frac{N+1}{N-i}, & m=0 \cr \frac{m+1}{i+2} & m>0 \end{cases} $$

  • $\begingroup$ Hi, thanks so much for your help! :) This answer is almost entirely correct, but solves the problem for n ~ Uniform(0, ..., N) whereas I want n ~ Uniform(1, ..., N). To get n ~ Uniform(1, ..., N), I think we need to subtract 1/N from equation (1) when m = 0 (in order to correct for the index starting at 0). That gives p(x_{i+1} | m) = ( (m+1)*(N+1))/((i+2)*(i+1) ) / ( (N+1)/(i+1) - 1 ) for m = 0 and the current answer, (m+1)/(i+2) when m > 0. $\endgroup$ – Luke Hewitt Jun 5 '17 at 18:19
  • $\begingroup$ No, this answer solves the problem for $n\sim \text{Uniform}(1,\ldots, N)$! $\endgroup$ – NCh Jun 5 '17 at 18:28
  • $\begingroup$ Consider N = 1. In your solution, p(x_1 | m = 0) = 1/2. Therefore, n ~ Uniform({0,1}). Is there something I'm missing? $\endgroup$ – Luke Hewitt Jun 5 '17 at 21:51
  • $\begingroup$ I'm sorry, the first time I read your reply inattentively. Thank you for your perseverance. Indeed, the case $m=0$ crushes equality (1). Or, you are right again, this answer without corrections is valid for $U(0,1\ldots,N)$. I'll make corrections in few mnutes. $\endgroup$ – NCh Jun 6 '17 at 10:44

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.