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\begin{document} 
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\heading{Average length of chain molecule} 
\auname{Amir Bernat}


{\bf The problem:}

\Dn
A molecule can be described as a chain of ${N}$ monomers, each monomer has the probability ${p}$ to be positioned horizontally, adding length ${a}$ to the molecule, otherwise the monomer is of length ${b}$. Let ${L}$ be the molecule's length, define a random variable $\hat{X}_n$ so that:
\[
X_n=\left\{\begin{matrix} a, & \mbox{the monomer is horizontal} \\
													b, & \mbox{the monomer is vertical}
													\end{matrix}\right.
													\]

\Dn

We assume that  the ratio $\frac{a}{b}$ is an Irrational number. 

\begin {itemize}

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


\item[(b)]
 Find  $f\left(L\right)\equiv {P}\left(L=na+\left(N-n\right)b\right)$. Using combinatorial considerations and calculate $\langle \hat{L}\rangle$, $Var(L)$.


\item[(c)]
 Define $\sigma_{L} = \sqrt{Var(L)}$ What is the behavior of ${\sigma_{L}/\langle L\rangle}$ as a function of ${N}$?

\end {itemize}
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\Dn


\Dn\Dn


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{\bf The solution:}

\Dn

(a) As each monomer is of lenth $a$ or $b$, the lengh of the molecule is the sum $L=\sum_{n=1}^N X_n$.  The random variable $\langle \hat{X_n} \rangle = a\cdot p + b\cdot q$ where $q=(1-p)$ can also be sumed. 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 :

\[\tag{1}
\langle L \rangle = \sum_{n=1}^N  \langle X_n \rangle = N \langle X_n \rangle = N \left( ap + bq \right)
\]

To find $Var(L)$ 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 of $L$ is the sum 

\[\tag{2}
Var(L)=\sum_{n=1}^N Var(X_n) = N\left( (p a^2 + q b^2) - (pa + qb)^2 \right)
\]

\Dn
\Dn

(b) The number of ways to choose $n$ out of $N$ objects with no importance to the order in which they were chosen is $C_N^n= \frac{N!}{n!(N-n)!}$. So the probablety for a molecule to be of lengh $L$ takes the form of binomial distribution 
\[\tag{3}
f(L) = {P}\left(L=na+\left(N-n\right)b\right) = \sum_{n=1}^N C_N^n p^n  q^{N-n} \left( na + (N-n)b \right)
\]

The expectation value and variance can be found by use of the moment generating function of the binomial distribution $M(v) = \left(pe^{av} + qe^{bv}\right)$ and the linearity properties.
\[\tag{4}
\langle L \rangle = \sum_{n=1}^N \langle X_n \rangle = N \frac{\partial M(v)}{\partial v} \left(pe^{av} + q e^{bv} \right)_{|v=0}=N\left(pa + qb \right)
\]
 In a similar way using the second moment (second derivative) we can find the variance:
 \[\tag{5}
 Var(L) = \sum_{n=1}^N Var(X_n) =\sum_{n=1}^N \left[ \frac{\partial^2 M(v)}{\partial v ^2} - \left(\frac{\partial M(v)}{\partial v} \right)^2 \right] = N\left( (p a^2 + q b^2) - (pa + qb)^2 \right)
 \]

\Dn
\Dn

(c) 
by inspecting the behavior of the function with respect to $N$ one can see that:
\[\tag{6}
\frac{\sigma}{\langle L\rangle} = \frac{\sqrt{N(pa^2 + qb^2 - (pa+qb)^2)}}{N(pa+qb)} \sim \frac{1}{\sqrt{N}}
\]
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\end{document}
