On a statistical mechanics problem set this week, we were asked to ``backwards'' derive the Poisson distribution from a statement like this:
"You notice that, as you're riding high above the ground in airplane, the probability for there to be no cars on some stretch of road -- say, the length x -- to be $e^{-ax}$. From this observation, can you express the likelihood of observing exactly $n$ cars in a length of highway $z$?"
This is a very tricky question. But we should start by looking at our first piece of information in view of the Poisson distribution
\begin{eqnarray}
P(n) &=& \frac{\lambda^n}{n!}e^{-\lambda}
\end{eqnarray}
$\lambda$ is the mean number of occurrences of some event, for some fixed interval. We see that if we set $n=0$ we get the original part of our statement:
\begin{eqnarray}
P(0) &=&e^{-\lambda}
\end{eqnarray}
This is the probability of observing zero ``events'' within some fixed interval. (Which, in our case is length. But this could easily be time or energy or whatnot.) It's worth noting that, if we write $\lambda=ax$, we can see that this is just the cumulative Probability of an exponential distribution:
\begin{eqnarray}
P(x \ge X) &=& \int_X^\infty ae^{-ax}dx=e^{-aX}
\end{eqnarray}
Which is a ``memory-less'', distribution. What does that mean? I like to think about the exponential distribution in terms of waiting for a train at the train station. If you've already been waiting 10 minutes, and you ask yourself, ``what's the probability that I won't see a train for another 5 minutes''? Mathematically this means
\begin{eqnarray}
P(t \ge X+Y \vert t \ge Y) &=& P(t \ge X)
\end{eqnarray}
Where $X$ is five minutes and $Y$ is the original 10 minutes. For $X$ and $Y$ treated as random variables, this means that we have conditional dependence between $X$ and $Y$, or, the doesn't care that you've already been waiting 10 minutes, the expectation of arrival is ubiquitous in time.
Now that's all very entertaining, but how does this relate to Poisson? Not seeing a train arrive in time interval $t$ is the same as not seeing a car on the highway over some interval $x$, and, to be exhaustingly pedantic, is the same as not seeing a quantum energy level in some band $\Delta E$. What about the probability of seeing exactly $n$ such events in said ``width''?
The first thing we need to do is define, a new variable, which I'll call $t_n$, which is the sum of $n$ people's waiting time for a train:
\begin{eqnarray}
t_n &=& t_1+t_2+\dots t_n
\end{eqnarray}
The probability distribution on $t_n$ is just a convolution of the independently, identically distributed variables,
\begin{eqnarray}
P(t_n) &=& \frac{a^n t_n^{n-1}}{(n-1)!}e^{-at_n}
\end{eqnarray}
Now, this almost looks Poisson, but not quite. What if we looked at the cumulative distribution of such a variable; say, of the probability that the sum of $n+1$-people's waiting times adds up to something greater than $x$?
\begin{eqnarray}
P(t_{n+1} \ge x) &=& \int_x^\infty \frac{a^{n+1} t_n^{n}}{(n)!}e^{-at_n} dt_n \\
&=& \frac{(ax)^n}{n!}e^{-ax}+\int_x^\infty \frac{a^n t_n^{n-1}}{(n-1)!}e^{-at_n}
\end{eqnarray}
We see that the second term looks like the cumulative probability of $n$-peoples waiting times add up to something greater than $x$. And so we take the difference:
\begin{eqnarray}
P(t_{n+1} \ge x) &=& \frac{(ax)^n}{n!}e^{-ax} +P(t_{n} \ge x) \\
P(t_{n+1} \ge x) -P(t_{n} \ge x) &=& \frac{(ax)^n}{n!}e^{-ax}
\end{eqnarray}
and get Poisson! So the Poisson distribution can be thought of as the probability that the sum of exactly $n$ waiting times at a train station adds up to some total $x$. In this case, $x$ is our width, be it along a highway, energy spectrum, or time, and $n$ is the number of successes, or events, or train arrivals.
Or, perhaps, in a common -- and quite close to the heart example -- baby births.
----------------------------------------------------------------------------------------------------
Note, it is important to remember that the mean of the exponential distribution is $\langle x \rangle=a^{-1}$, so $\lambda$ should really be written:
\begin{eqnarray}
P(n) &=& \frac{(t_n/\mu)^n}{n!}e^{-\frac{t_n}{\mu}}
\end{eqnarray}
Where $\mu$ is the average``waiting time'' for each independently distributed member, $t$.
"You notice that, as you're riding high above the ground in airplane, the probability for there to be no cars on some stretch of road -- say, the length x -- to be $e^{-ax}$. From this observation, can you express the likelihood of observing exactly $n$ cars in a length of highway $z$?"
This is a very tricky question. But we should start by looking at our first piece of information in view of the Poisson distribution
\begin{eqnarray}
P(n) &=& \frac{\lambda^n}{n!}e^{-\lambda}
\end{eqnarray}
$\lambda$ is the mean number of occurrences of some event, for some fixed interval. We see that if we set $n=0$ we get the original part of our statement:
\begin{eqnarray}
P(0) &=&e^{-\lambda}
\end{eqnarray}
This is the probability of observing zero ``events'' within some fixed interval. (Which, in our case is length. But this could easily be time or energy or whatnot.) It's worth noting that, if we write $\lambda=ax$, we can see that this is just the cumulative Probability of an exponential distribution:
\begin{eqnarray}
P(x \ge X) &=& \int_X^\infty ae^{-ax}dx=e^{-aX}
\end{eqnarray}
Which is a ``memory-less'', distribution. What does that mean? I like to think about the exponential distribution in terms of waiting for a train at the train station. If you've already been waiting 10 minutes, and you ask yourself, ``what's the probability that I won't see a train for another 5 minutes''? Mathematically this means
\begin{eqnarray}
P(t \ge X+Y \vert t \ge Y) &=& P(t \ge X)
\end{eqnarray}
Where $X$ is five minutes and $Y$ is the original 10 minutes. For $X$ and $Y$ treated as random variables, this means that we have conditional dependence between $X$ and $Y$, or, the doesn't care that you've already been waiting 10 minutes, the expectation of arrival is ubiquitous in time.
Now that's all very entertaining, but how does this relate to Poisson? Not seeing a train arrive in time interval $t$ is the same as not seeing a car on the highway over some interval $x$, and, to be exhaustingly pedantic, is the same as not seeing a quantum energy level in some band $\Delta E$. What about the probability of seeing exactly $n$ such events in said ``width''?
The first thing we need to do is define, a new variable, which I'll call $t_n$, which is the sum of $n$ people's waiting time for a train:
\begin{eqnarray}
t_n &=& t_1+t_2+\dots t_n
\end{eqnarray}
The probability distribution on $t_n$ is just a convolution of the independently, identically distributed variables,
\begin{eqnarray}
P(t_n) &=& \frac{a^n t_n^{n-1}}{(n-1)!}e^{-at_n}
\end{eqnarray}
Now, this almost looks Poisson, but not quite. What if we looked at the cumulative distribution of such a variable; say, of the probability that the sum of $n+1$-people's waiting times adds up to something greater than $x$?
\begin{eqnarray}
P(t_{n+1} \ge x) &=& \int_x^\infty \frac{a^{n+1} t_n^{n}}{(n)!}e^{-at_n} dt_n \\
&=& \frac{(ax)^n}{n!}e^{-ax}+\int_x^\infty \frac{a^n t_n^{n-1}}{(n-1)!}e^{-at_n}
\end{eqnarray}
We see that the second term looks like the cumulative probability of $n$-peoples waiting times add up to something greater than $x$. And so we take the difference:
\begin{eqnarray}
P(t_{n+1} \ge x) &=& \frac{(ax)^n}{n!}e^{-ax} +P(t_{n} \ge x) \\
P(t_{n+1} \ge x) -P(t_{n} \ge x) &=& \frac{(ax)^n}{n!}e^{-ax}
\end{eqnarray}
and get Poisson! So the Poisson distribution can be thought of as the probability that the sum of exactly $n$ waiting times at a train station adds up to some total $x$. In this case, $x$ is our width, be it along a highway, energy spectrum, or time, and $n$ is the number of successes, or events, or train arrivals.
Or, perhaps, in a common -- and quite close to the heart example -- baby births.
----------------------------------------------------------------------------------------------------
Note, it is important to remember that the mean of the exponential distribution is $\langle x \rangle=a^{-1}$, so $\lambda$ should really be written:
\begin{eqnarray}
P(n) &=& \frac{(t_n/\mu)^n}{n!}e^{-\frac{t_n}{\mu}}
\end{eqnarray}
Where $\mu$ is the average``waiting time'' for each independently distributed member, $t$.
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