Probability of the next entry in a series of discrete trials.

**hatsoff** · November 17th 2008, 07:44 PM

Hi, all.

Suppose you have a series of events, all the same, which can result in X or ~X. The probability of each is unknown. As you watch the series go forward, every entry (to start) is X. So:

$S = \{X, X, X, X, X... \}$

At any point, however, it is possible that the pattern will cease, and that we will get a ~X. But let's say it hasn't happened yet.

Let's also say that the number of X's we have observed so far is $n$ , such that the probability of this occurring is:

$P(S_n)=[P(X)]^n$

So, how do we determine P(X) ?

We could determine a confidence level, I think...

$[P(X)]^n>.05$

$P(X)>.05^{\frac{1}{n}}$

So, there's a 95% chance that $P(X)\in(.05^\frac{1}{n},1]$ , yes? But how do we actually find P(X)?

Thanks!

**hatsoff** · November 17th 2008, 09:46 PM

We could use the confidence interval to find a minimum probability. Let's say that the error is given as $E$ . So:

$E^{\frac{1}{n}}$ is the lower bound of $P(X)$ in the $1-E$ confidence level. Let's call Q(X) the probable expectation that the next entry in the sequence is X. So, $Q(X)\geq(1-E)E^{\frac{1}{n}}$ . We can take the derivative with respect to $E$ to find the value of $E$ which maximizes $Q(X)$ :

$\frac{d}{dE}(1-E)E^{\frac{1}{n}}=0$

$\frac{d}{dE}(E^{\frac{1}{n}}-E^{\frac{1}{n}+1})=0$

$\frac{1}{n}E^{\frac{1}{n}-1}-[\frac{1}{n}+1]E^{\frac{1}{n}})=0$

$\frac{1}{En}E^{\frac{1}{n}}-[\frac{1}{n}+1]E^{\frac{1}{n}})=0$

$[\frac{1}{En}-\frac{1}{n}-1]E^{\frac{1}{n}}=0$

$\frac{1}{En}-\frac{1}{n}-1=0$

$\frac{1}{En}=\frac{n+1}{n}$

$En=\frac{n}{n+1}$

$E=\frac{1}{n+1}$

Now, recall our $Q(X)$ relationship:

$Q(X)\geq(1-E)E^{\frac{1}{n}}$

And substitute for E:

$Q(X)\geq(1-\frac{1}{n+1})[\frac{1}{n+1}]^{\frac{1}{n}}$

$Q(X)\geq\frac{n}{(n+1)^{\frac{1}{n}}(n+1)}$

$Q(X)\geq\frac{n}{(n+1)^{\frac{n+1}{n}}}$

It seems like we should be able to do better than this, though.

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