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

• Nov 17th 2008, 07:44 PM
hatsoff
Probability of the next entry in a series of discrete trials.
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:

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

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 $\displaystyle n$, such that the probability of this occurring is:

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

So, how do we determine P(X) ?

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

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

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

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

Thanks!
• Nov 17th 2008, 09:46 PM
hatsoff
We could use the confidence interval to find a minimum probability. Let's say that the error is given as $\displaystyle E$. So:

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

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

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

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

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

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

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

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

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

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

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

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

And substitute for E:

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

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

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

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