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Say there are three jars, $j_1, j_2, j_3$ filled with different binary sequences of length two.

The distribution of the binary sequences in each of the jars is given by the $p_i^k(1-p_i)^{n-k}$, where $p_i = \frac{i}{m + 1}$ where $m$ is the number of jars, $i$ is the jar index, $k $is number of 1$$'s and $n$ is the length of the string.

So for three jars we have $p_1 = 0.25, p_2 = 0.5$, and $p_3 = 0.75$ for $j_1, j_2, j_3$ respectively.

Here are the sequences and their probabilities for $j_1$ with $p_1 = 0.25$:

\begin{align*} P(00) = 9 / 16 \\ P(10) = 3 / 16 \\ P(01) = 3 / 16 \\ P(11) = 1 / 16. \end{align*}

If I tell you that I have selected a binary sequence and the first element is $1$ what is the E($p_i$)?

Well, this can be calculated by looking at each of the jars and adding up the probability of candidate sequences times the value of $p_i$.

Edit: I wasn't normalizing this conditionally space properly. I'm skipping a step which I'll explain, someone wants.

\begin{equation*} E(p_i) = (4/24 * 1/4) + (8/24 * 1/2) + (12/24 * 3/4) = 14 / 24 = 0.58. \end{equation*}

So the question is ... what is $E(p_i)$ when the numbers of jars goes to infinity (or alternatively, when $p$ can take on values between $0$ and $1$)? Also what happens when the size of the binary strings goes to infinity? Does it have an effect on the outcome? If it does, does the order we take the limits change the answer?

And most importantly what is the general case for when I have $s$ 1's and $r$ $0$'s?, with a continuous $p$ from $0$ to $1$ and infinite sequences?

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    `If I tell you that I have selected a binary sequence and the first element is 1 what is the expected value of X...` What is X? I don't see where you've defined it.2010-07-27
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    E(value of $p_i$). It might be confusing where the probabilities in the expected value computation came from which I can explain. The other point is that when we are calculating the expected $p_i$ it is just a value associated with the jars, it is not the probability that you picked a binary sequence from the jar.2010-07-27
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    That was confusing, my bad2010-07-27
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    In retrospect, after finally catching my error in the calculation of P(1...), since we only care about the probability that the first digit is 1 and the rest of the digits (however many there may be) are free, P(1...) should be the probability of a single digit being 1, which is p_i, regardless of n.2010-07-27
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    Yes, that's what I would expect, since this question was attempt at rephrasing this one: http://math.stackexchange.com/questions/695/probability-of-a-bent-coin-closed. I'm not positive they are equivalent, but I think they are...2010-07-27
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    It seems likely to me that they are equivalent. Assuming they are, I wonder if there's a reason why the answer is the harmonic mean of 1/2 and 1 (as in, does the harmonic mean have something to do with it?).2010-07-29
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    I had never heard of the harmonic mean before so I did not notice that. On first blush the formulas show some similarity with the expansion you worked out, so maybe? What I find more interesting is that this is a combinatorial argument for the rule of succession (or at least a simple version of it) that does not require the assumption of a uniform prior.2010-07-29

1 Answers 1

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First, sticking to strings of length n=2:

$P(10)=p_i (1-p_i)=\frac{i}{m+1}\left (1-\frac{i}{m+1} \right )=\frac{i(m+1-i)}{(m+1)^2}$ and $P(11)=p_i^2=\left (\frac{i}{m+1} \right )^2=\frac{i^2}{(m+1)^2}$;

$E(p_i)=\sum_{i=1}^{m}p_iP(p_i)=\sum_{i=1}^{m}\frac{i}{m+1}\left(\frac{P(10)+P(11)}{\sum_{i=1}^{m}(P(10)+P(11))} \right )$ $=\frac{\sum_{i=1}^{m}\frac{i}{m+1}\left(\frac{i(m+1-i)}{(m+1)^2}+\frac{i^2}{(m+1)^2} \right )}{\sum_{i=1}^{m}\left(\frac{i(m+1-i)}{(m+1)^2}+\frac{i^2}{(m+1)^2} \right )}$ $=\frac{\sum_{i=1}^{m}\frac{i}{m+1}\left(i(m+1-i)+i^2\right )}{\sum_{i=1}^{m}\left(i(m+1-i)+i^2\right )}$ $=\frac{\sum_{i=1}^{m}i^2}{(m+1)\sum_{i=1}^{m}i}$ $=\frac{\left(\frac{m(m+1)(2m+1)}{6} \right )}{(m+1)\frac{m(m+1)}{2}}$ $=\frac{2m+1}{3(m+1)}$;

$\lim_{m\to\infty}E(p_i)=\lim_{m\to\infty}\frac{2m+1}{3(m+1)}=\frac{2}{3}$

For general n:

$P(1...)=p_i\sum_{k=0}^{n-1}\left({n-1 \choose k}p_i^k(1-p_i)^{n-1-k} \right )=p_i$ (yes, P(1...) does not depend on n at all!);

$E(p_i)=\sum_{i=1}^{m}p_iP(p_i)=\sum_{i=1}^{m}p_i\left(\frac{P(1...)}{\sum_{i=1}^{m}P(1...)} \right )=\sum_{i=1}^{m}p_i\left(\frac{p_i}{\sum_{i=1}^{m}p_i} \right )$ $=\frac{\sum_{i=1}^{m}p_i^2}{\sum_{i=1}^{m}p_i}=\frac{\sum_{i=1}^{m}\left(\frac{i}{m+1} \right )^2}{\sum_{i=1}^{m}\frac{i}{m+1}}$ $=\frac{\sum_{i=1}^{m}i^2}{(m+1)\sum_{i=1}^{m}i}=\frac{2m+1}{3(m+1)}$ (no surprise there, given that $P(1...)$ doesn't depend on n);

$\lim_{m\to\infty}E(p_i)=\lim_{m\to\infty}\frac{2m+1}{3(m+1)}=\frac{2}{3}$