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Exponential Random Variable 11/4


\begin{displaymath}f(x)=\lambda e^{-\lambda x},\mbox{ for } x \geq 0\end{displaymath}

This is used to modelize waiting times, at telephone booths, at postoffices and in science for time until decay of an atom in radioactive decay. Call T the time to decay for an atom, $T \sim Exp(\lambda)$,

\begin{displaymath}P(T>t)=e^{-\lambda t}\qquad P(a \leq T < b)= e^{-\lambda a} - e^{-\lambda b}\end{displaymath}

The physicists do not use this parametrization, they use the notion of half life h, this is the median of the exponential:

\begin{displaymath}P(T<h)=P(T\leq h)= \frac{1}{2}=e^{-\lambda h}\end{displaymath}

So that $h=\frac{2}{\lambda}$

Example: Strontium 90 is a dangerous component of nuclear fallout. It has a half life of 28 years.

\begin{displaymath}\lambda_{strontium}= \frac{log(2)}{28}=\frac{.693147}{28}=.0248 \;\end{displaymath}

How long does it take for $99\%$ of the strontium to disappear?

\begin{displaymath}e^{-\lambda t}= \frac{1}{100}\qquad t=\frac{log(100)}{.0248}\equiv
186\; years\end{displaymath}

Physics and Exponential Decay

Susan Holmes