0
$\begingroup$

Let's consider the Pólya urn model. Inside an urn there are two kinds of balls in two different colours - black and white (b - the amount of black ones, w - the amount of white ones.)
We draw one ball, then put it back with and add $d$ more balls in the same colour.
For example:
the possibility of drawing a white ball is equal to:
1) $P_1^w =\frac{w}{b+w}$ in the first draw,
2) $P_2^w = \frac{b+d}{b+w+d}$ in the second draw.
And so on.
I am to find out the possibility of extraction exactly $k$ black balls in $n$ trials.
I thought of using Bernoullie's scheme: $$P(S_n = k) = {n\choose k}p^k(1-p)^{n-k}$$ where the success is pulling out the black ball of course.
There is a problem however. It is easy to say how the denominator changes after each extraction, the k-th denomitor will look like this: $$p_k = \frac{\alpha}{b+w+(k-1)d}$$ The counter is the problem. I think that: $$p_k^b = \frac{b + md}{b+w+(k-1)d}$$ where $m$ can be any number from the set $\{ 0, 1, 2, 3, 4,...k-1\}$
I would appreciate any tips.

$\endgroup$
1
$\begingroup$

Note that Polya’s urn give rise to exchangeable sequences--- conceptually the probability measure of the outcome is invariant under permutation. For example, let $W_t$ and $B_t$ be the event that the $t$-th draw is white and black respectively, then observing $2$ white balls in $3$ draws is same regardless of the sequence it is observed: \begin{align} \Pr[W_1W_2B_3] = \frac{w}{w+b}\frac{w+d}{w+b+d}\frac{b}{w+b+2d}\\ =\Pr[W_1B_2W_3] = \frac{w}{w+b}\frac{b}{w+b+d}\frac{w+d}{w+b+2d}\\ =\Pr[B_1W_2W_3] = \frac{b}{w+b}\frac{w}{w+b+d}\frac{w+d}{w+b+2d}. \end{align}

Therefore, observing $k$ black balls in $n$ draws is equal to $$\sum_{\text{possible sequence}}\frac{\prod_{i=0}^{k-1} (b+id)\cdot \prod_{j=0}^{n-k-1} (w+jd)}{\prod_{i=0}^{n-1} (w+b+id)} = {n\choose k}\frac{\prod_{i=0}^{k-1} (b+id)\cdot \prod_{j=0}^{n-k-1} (w+jd)}{\prod_{i=0}^{n-1} (w+b+id)}$$

$\endgroup$
  • $\begingroup$ Thank you very much! At the end I figured it out myself and our solutions are the same so it must be correct. Thanks once again! :) $\endgroup$ – Hendrra Mar 20 '18 at 14:20

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.