## Friday, September 16, 2011

### Working out a neat probability rule

Working out a neat probability rule from divisbyzero.com.

For small $$x$$ (on the order of $$1/10000$$), what is $$1 - (1-x)^t$$, where $$t$$ is some large number.

Using the binomial theorem expansion

$$(1+x)^n = {n\choose 0}x^0 + {n\choose 1}x^1 + \ldots + {n\choose n}x^n$$

we have:

\begin{eqnarray} (1-x)^t &=& {t\choose 0}(-x)^0 + {t\choose 1}(-x)^1 + \ldots + {t\choose t}(-x)^t \\ &\approx& 1 - {t\choose 1}x + O(x^2) \\ &=& 1 - tx + O(x^2) \\ \end{eqnarray} Thus the original equation can be approximated by:

\begin{eqnarray} 1-(1-x)^t &\approx& 1 - (1 - tx + O(x^2)) \\ &=& tx - O(x^2) \end{eqnarray} The original post was trying to approximate $$1-(1-D/P)^L$$, and with $$x=D/P$$ and $$t=L$$, we arrive at:

\begin{eqnarray} 1-(1-\frac{D}{P})^L &\approx& 1 - (1 - \frac{LD}{P} + O(\left(\frac{D}{P}\right)^2)) \\ &=& \frac{LD}{P} - O(\left(\frac{D}{P}\right)^2) \end{eqnarray}