Question

Individuals $1$through n,n > $1$, are to be recruited into a firm in the following manner: Individual $1$starts the firm and recruits individual $2$. Individuals $1$and $2$will then compete to recruit individual $3$. Once individual $3$is recruited, individuals $1,2$, and $3$will compete to recruit individual $4$, and so on. Suppose that when individuals $1,2$,...,i compete to recruit individual i + $1$, each of them is equally likely to be the successful recruiter.

(a) Find the expected number of the individuals $1$,...,n who did not recruit anyone else.

(b) Derive an expression for the variance of the number of individuals who did not recruit anyone else, and evaluate it for n=$5$.

Short answer

From the information,

a) The expected number of the individuals $1$,...,n who did not recruit anyone else is $=\frac{(i-1)(n-1)}{(n-1{)}^2}$

b) An expression for the variance of the number of individuals who did not recruit anyone else is

$\Rightarrow \mathrm{Var}\left(\sum _{i=1}^n{X}_i\right)=\frac{1}{(n-1{)}^2}\sum _{i=1}^n(i-1)(n-i)-\frac{1}{(n-2)(n-1{)}^2}\sum _{i=1}^{n-1}(i-1)(n-i)(n-i-1)$

Step-by-step solution

GIven Information (part a)

Find the expected number of the individuals $1$...,n who did not recruit anyone else.

Explanation (part a)

Let Xi=$1$If individual i don't recruit anyone

=$0$otherwise

$\mathrm{E}\left[{\mathrm{X}}_{\mathrm{i}}\right]=P\{\mathrm{i}\text{doesn't recruit any of}\mathrm{i}+1,\mathrm{i}+2,\dots \mathrm{n}\}$

$=\left(\frac{i-1}{i}\right)\left(\frac{i}{i+1}\right)\dots \left(\frac{n-2}{n-1}\right)$

$\Rightarrow \mathrm{E}\left[{\mathrm{X}}_i\right]=\frac{i-1}{n-1}$

$\Rightarrow \mathrm{E}\left[\sum _{i=1}^{\mathrm{n}}{\mathrm{X}}_{\mathrm{i}}\right]=\frac{\sum _{i=1}^n(i-1)}{(n-1)}$

$=\frac{n}{2}$

$\mathrm{Var}\left({\mathrm{X}}_i\right)=\frac{(i-1)}{(n-1)}[1-\frac{(i-1)}{(n-1)}]$

$=\frac{(i-1)(n-1)}{(n-1{)}^2}$

Final Answer (part a)

Find the expected number of the individuals $1$...,n who did not recruit anyone else is$=\frac{(i-1)(n-1)}{(n-1{)}^2}$

Given Information (part b)

Derive an expression for the variance of the number of individuals who did not recruit anyone else, and evaluate it for n=$5$.

Explanation (part b)

$\text{For}i<j,\mathrm{E}\left[{\mathrm{X}}_i{\mathrm{X}}_j\right]=\frac{i-1}{i}\dots \dots \frac{j-2}{j-1}\frac{j-2}{j}\frac{j-1}{j+1},\dots +\frac{n-3}{n-1}$

$=\frac{(i-1)(j-2)}{(n-2)(n-1)}$

$\Rightarrow \mathrm{Cov}[X_i,X_j]=\frac{(i-1)(j-2)}{(n-2)(n-1)}-\frac{(i-1)(j-1)}{(n-1)(n-1)}$

$=\frac{(i-1)(j-n)}{(n-2)(n-1{)}^2}$

$\Rightarrow \mathrm{Var}\left(\sum _{i=1}^n{X}_i\right)=\sum _{i=1}^n\mathrm{Var}\left(X_i\right)+2\sum _{i=1}^{n-1}\sum _{j=i+1}^n\mathrm{Cov}(X_i,X_j)$

$=\frac{\sum _{i=1}^n(i-1)(n-i)}{(n-1{)}^2}+2\sum _{i=1}^{n-1}\sum _{j=i+1}^n\frac{(\mathrm{i}-1)(j-n)}{(n-2)(n-1{)}^2}$

$\Rightarrow \mathrm{Var}\left(\sum _{i=1}^n{X}_i\right)=\frac{1}{(n-1{)}^2}\sum _{i=1}^n(i-1)(n-i)-\frac{1}{(n-2)(n-1{)}^2}\sum _{i=1}^{n-1}(i-1)(n-i)(n-i-1)$

Final Answer(part b)

Derive an expression for the variance of the numberof individuals who did not recruit anyone else is

$\Rightarrow \mathrm{Var}\left(\sum _{i=1}^n{X}_i\right)=\frac{1}{(n-1{)}^2}\sum _{i=1}^n(i-1)(n-i)-\frac{1}{(n-2)(n-1{)}^2}\sum _{i=1}^{n-1}(i-1)(n-i)(n-i-1)$