Q. 3.3 - A First Course in Probability

Question

Consider a school community of $m$ families, with $n_i$ of them having $i$ children, $i=1,\dots ,k,\sum _{i=1}^k{n}_i=m$ Consider the following two methods for choosing a child:

$1$. Choose one of the $m$ families at random and then randomly choose a child from that family.

$2$. Choose one of the $\sum _{i=1}^{k}i{n}_i$ children at random.

Show that method $1$ is more likely than method $2$ to result

in the choice of a firstborn child.

Hint: In solving this problem, you will need to show that

$\sum _{i=1}^{k}i{n}_i\sum _{j=1}^k\frac{n_j}{j}\ge \sum _{i=1}^k{n}_i\sum _{j=1}^k{n}_j$

To do so, multiply the sums and show that for all pairs $i, j$, the coefficient of the term $n_i{n}_j$ is greater in the expression on the left than in the one on the right.

Step-by-Step Solution

Answer

We have proven by following the hint.

$P_1\left(O\right)\ge P_2\left(O\right)$

Step 1: Given Information

Given

$m$ families

$n_i$ of them have $i=1,2,\dots ,k$ children

A child is selected.

Event:

$O$ - the selected child is the oldest in the family

$F_i$ - the child is from a family with $i$ children

Method $1$, probability of being chosen is equally distributed among $m$ families, and in a family, among $i$ children, only one of which is the oldest

$P_1\left(F_i\right)=\frac{n_i}{m}$

$P_1\left(O\mid F_i\right)=\frac{1}{i}$

$\text{ for } i=1,2,\dots ,k$

Method $2$ , since there are $m$ families, there are $m$ oldest children. Every child is equally likely to be chosen, therefore:

$P_2\left(O\right)=\frac{m}{{\sum }_{i=1}^{k}i{n}_i}$

Prove: $P_1\left(O\right)\ge P_2\left(O\right)$

Step 2: Explanation

Bayes formula using $C_i\text{ for }i=1,2,\dots ,k$ as hypothesis yields

$P\left(O\right)=\sum _{i=1}^{k}P\left(O\mid F_i\right)P\left(F_i\right)=\sum _{i=1}^k\frac{1}{i}·\frac{n_i}{m}$

The wanted inequality is then: 

$\sum _{i=1}^k\frac{1}{i}·\frac{n_i}{m}\ge \frac{m}{{\sum }_{i=1}^{k}i{n}_i} /·m·\sum _{j=1}^{k}j{n}_j$

$\left(\sum _{i=1}^k\frac{n_i}{i}\right)·\left(\sum _{j=1}^{k}j{n}_j\right)\ge m^2$

$\text{ Since }m=\sum _{i=1}^k{n}_i\Rightarrow$

$\left(\sum _{i=1}^k\frac{n_i}{i}\right)·\left(\sum _{j=1}^{k}j{n}_j\right)\ge {\left(\sum _{i=1}^k{n}_i\right)}^2$

The hint states that both sides have to be transformed into sums. Organizing by $n_i,n_j$ the inequality is:

$\sum _{(i,j)=(1,1)}^{(n,n)}{n}_i{n}_j·\left(\frac{1}{i}·j\right)=\sum _{(i,j)=(1,1)}^{(n,n)}{n}_i{n}_j·1$

$\text{ With }(i,j)\nsim (j,i)$

Step 3: Final Answer

If $i=j$

Left side coefficients $\frac{1}{i}·i=1$, right side coefficients $1$

Else $i\ne j$, there are two elements of the sum on both sides that correspond to that two numbers $-n_i{n}_j$ and $n_j{n}_i$, if we add those two together on both sides we get

Left side coefficients $\frac{1}{i}·j+\frac{1}{j}·i$, right side coefficients $2$

$\frac{i}{j}+\frac{j}{i}\ge 2$

$\frac{i^2+j^2}{ij}\ge 2 /·ij, i,j\ge 0$

$i^2+j^2\ge 2ij$

$(i-j{)}^2\ge 0$

These are all tautologies, this means the coefficient on the left-hand side is greater than on the right-hand side. Since we are adding corresponding positive numbers, greater coefficients mean greater sum, therefore:

$P_1\left(O\right)\ge P_2\left(O\right)$.