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\heading{Fluctuations in the number of particles} 
\auname{Amir Bernat}


{\bf The problem:}

\Dn

A closed tank of volume ${V_{0}}$ has ${N_{0}}$ particles. "The system" is a subvolume ${V}$. The number of particles in $V$ is the random
variable ${N}$. Define the random variables ${\hat{X}_{n}}$ which determines if a certain particle is in the system. 
\[
X_n=\left\{\begin{matrix} 1, & \mbox{the particle is in }V \\
													0, & \mbox{the particle is not in }V
													\end{matrix}\right.
													\]

\begin {itemize}

\item[(a)]
  Express $\hat{N}$ using $\hat{X}_n$. By fundamental theorems on adding independent random variables find $\langle N \rangle$, $Var(N)$. 

\item[(b)]
Find the probability function ${f\left(N\right)}$. Using combinatorial considerations and calculate $\langle \hat{L}\rangle$, $Var(L)$.

\item[(c)]Assume ${|V/V_{0}-\frac{1}{2}|<<1}$ and treat ${N}$ as a continuous
random variable. Find the probability function ${f\left(N\right)}$.

\end {itemize}

\Dn\Dn


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{\bf The solution:}

\Dn

(a) The probability for finding the particle in the subvolume is $p = \frac{V}{V_0}$. As each particle is in $(X_n=1)$ or out $(X_n=0)$ of the subvolume the number of particles is the sum $N=\sum_{n=1}^N X_n$. The random variable $\langle \hat{X_n} \rangle = 1\cdot p + 0\cdot q$ can also be sumed, as the expectation operator $\operatorname{E}$ is linear in the sense that $\operatorname{E}(X + Y)=  \operatorname{E}(X) + \operatorname{E}(Y)$, then we can denote:

\[\tag{1}
\langle N \rangle =\sum_{n=1}^{N_0}\langle X_n \rangle = N_0 p
\]

To find $Var(N)$ let us find the variance of $X_n$ and use the linearity property $\operatorname{Var}(X+Y)=\operatorname{Var}(X)+\operatorname{Var}(Y)+2 \operatorname{Cov}(X,Y)$. The random variables $X_n$ are not dependant, meaning they have a Covariance of $0$. The variance operator of $N$ is then:

\[\tag{2}
Var(N)=\sum_{n=1}^{N_0} Var(X_n) = N_0\left( (p) - (p)^2 \right) = N_0p(1-p) = N_0pq
\]

\Dn
\Dn
(b) The number of ways to choose $N$ out of $N_0$ objects with no importance to the order is $C_{N_0}^N= \frac{N_0!}{N!(N_0-N)!}$. So the probability for the number of particles to be $N$ takes the form of binomial distribution:
\[\tag{3}
f(N) = \sum_{n=1}^{N_0} C_N^{N_0} p^N  q^{N_0-N}
\]

The expectation value and variance can be found by use of the moment generating function of the binomial distribution $M(v) = \left(pe^{v} + (1-p)\right)$ and the linearity properties.

\[\tag{4}
\langle N \rangle= \sum_{n=1}^{N_0} \langle X_n \rangle = \sum_{n=1}^{N_0} \frac{\partial M(v)}{\partial v}_{|v=0} =N_0p
\]
 In a similar way using the second moment (second derivative) we can find the variance:
 \[\tag{5}
 Var(N) = \sum_{n=1}^{N_0} Var(X_n) = \sum_{n=1}^{N_0} \left[ \frac{\partial^2 M(v)}{\partial v ^2} - \left(\frac{\partial M(v)}{\partial v} \right)^2 \right] = N_0p(1-p)
 \]

\Dn
\Dn

(c) The case when ${|V/V_{0}-\frac{1}{2}|<<1}$ is of particular interest, it is the extremum of $Var(N)=np(1-p)$ as a function of $p$.

Considering $N$ as a continues random variable is analogues to having a very large number of particles, thus allowing us to make use of the central limit theorem. 
The probability function of a random variable with $Var(X) = 	\sigma^2(x)$ and $\langle X \rangle = \bar{X}$ is:
\[\tag{6}
f(X) = \frac{1}{\sqrt{2\pi}\sigma}exp\left[-\frac{(X-\langle X \rangle)^2}{2\sigma^2}\right]
\]

It is possible to approximate the probability function found in (b) by making use of Stirling's formula $n! \simeq n^n e^{-n}\sqrt{2 \pi n}$, which holds for large $n$. The full derivation is known as the \textit{De Moivre} - \textit{Laplace theorem} which states:
\[ \tag{7}
f(N) = \sum_{n=1}^{N_0} C^N_{N_0} p^N q^{N_0}\simeq \frac{1}{\sqrt{2{\mathbf \pi }N_0pq}}exp\left[\frac{-{\left(N-N_0p\right)}^2}{2N_0pq} \right]
\]

Full derivation can be found at http://mathworld.wolfram.com/BinomialDistribution.html (Eq 34 - 65). For the case where $p=V/V_0 = \frac{1}{2}$ we have:
\[ \tag{8}
f(N) = \frac{1}{\sqrt{{\mathbf \pi }\frac{N_0}{2}}}exp\left[\frac{-{\left(N-\frac{N_0}{2}\right)}^2}{\frac{N_0}{2}} \right]
\]

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