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I'm trying to prove urns version of Laplace's law of succession my professor suggested. Laplace's law states that the chance that the next trial is a success given $j$ successes out of the first $n$ is $\frac{(j+1)}{(n+2)}$. Here is how the problem states:

"If we have $n+k+1$ urns and urn $i$ has $i$ balls labeled $1$ and $n+k-i$ labeled zero. We pick an urn at random and draw $n$ balls from it without replacement say $j$ of them are ones. Show that the conclusion of Laplace's law holds for this setup. In other word, the chance that the next ball is a one is $\frac{(j+1)}{(n+2)}$."

I've proved one version of the law, which close to this version An elementary version of Laplace's Method of Succession. I tried to use similar approach but somehow my answer always in form of $k$ and I can't get rid of it. What is the intuition behind this $k$? Is it just to increase the complexity of the problem, or it has some meaning behind it?

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Michael K
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  • That can't be right; for $j=n=0$ it should be $1/2$, not $1$. I seem to remember the denominator should be $n+2$. – joriki May 20 '13 at 22:52
  • The $i$th urn always has $i$ balls labeled $1$? That somehow sounds odd. – dfeuer May 20 '13 at 23:19

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Let $I$ be the random variable indicating the label of the urn picked, $J$ the number of ones picked (after picking $n$ balls), and $X$ the event that the next ball is one. We know that $$ P(J = j \mid I = i) = \frac{\binom{i}{j}\binom{n+k-i}{n-j}}{\binom{n + k}{n}} = \frac{\binom{j+(i-j)}{i-j}\binom{n+k-j-(i-j)}{k-(i-j)}}{\binom{n+k}{n}} $$ and $$ P(X \mid I = i, J = j) = \begin{cases} 0 & ; i \le j \\ \frac{i-j}k & ; j \le i \le k + j \end{cases} $$ Compute $P(I = i \mid J = j)$ from $$ P(I = i \mid J = j) = \frac{P(I = i, J = j)}{\sum_{\tilde i}P(I = \tilde i, J = j)} = \frac{P(J = j \mid I = i)}{\sum_{\tilde i} P(J = j \mid I = \tilde i)} $$ where we use $P(I = i) = \frac{1}{n + k + 1}$ in the last equation. Then, \begin{align*} P(X \mid J = j) & = \sum_{i} P(X \mid I = i, J = j) P(I = i \mid J = j) \\ & = \sum_{i=j}^{k+j} \frac{i-j}k \frac{P(J = j \mid I = i)}{\sum_{\tilde i} P(J = j \mid I = \tilde i)} \\ & = \frac{\sum_{i=j}^{k+j}(i-j)P(J = j \mid I = i)}{k \sum_{i=j}^{k+j} P(J = j \mid I = i)} \\ & = \frac{\sum_{i=j}^{k+j}(i-j)\binom{i}{j}\binom{n+k-i}{n-j}}{k \sum_{i=j}^{k+j} \binom{i}{j}\binom{n+k-i}{n-j}} \\ & = \frac{\sum_{i=0}^k i \binom{j+i}{i}\binom{n-j+(k-i)}{k-i}}{k\sum_{i=0}^k \binom{j+i}{i}\binom{n-j+(k-i)}{k-i}} \end{align*} I don't know how to make this into the desired expression $\frac{j+1}{n+2}$. Putting $k = 1$ seems to work though.

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