91影视

Given

$m$families

$n_i$of them have $i=1,2,鈥�,k$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

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

$P_1\left(O鈭�F_i\right)=\frac{1}{i}$

$\text{for}i=1,2,鈥�,k$

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

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

Prove:$P_1\left(O\right)鈮�P_2\left(O\right)$

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

$P\left(O\right)=\underset{i=1}{\overset{k}{鈭�}}P\left(O鈭�F_i\right)P\left(F_i\right)=\underset{i=1}{\overset{k}{鈭�}}\frac{1}{i}路\frac{n_i}{m}$

The wanted inequality is then:

$\underset{i=1}{\overset{k}{鈭�}}\frac{1}{i}路\frac{n_i}{m}鈮�\frac{m}{{鈭�}_{i=1}^{k}i{n}_i}/路m路\underset{j=1}{\overset{k}{鈭�}}j{n}_j$

$\left(\underset{i=1}{\overset{k}{鈭�}}\frac{n_i}{i}\right)路\left(\underset{j=1}{\overset{k}{鈭�}}j{n}_j\right)鈮�m^2$

$\text{Since}m=\underset{i=1}{\overset{k}{鈭�}}{n}_i鈬�$

$\left(\underset{i=1}{\overset{k}{鈭�}}\frac{n_i}{i}\right)路\left(\underset{j=1}{\overset{k}{鈭�}}j{n}_j\right)鈮�{\left(\underset{i=1}{\overset{k}{鈭�}}{n}_i\right)}^2$

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

$\underset{(i,j)=(1,1)}{\overset{(n,n)}{鈭�}}{n}_i{n}_j路\left(\frac{1}{i}路j\right)=\underset{(i,j)=(1,1)}{\overset{(n,n)}{鈭�}}{n}_i{n}_j路1$

$\text{With}(i,j)鈮�(j,i)$

If $i=j$

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

Else $i鈮�j$, 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 $\frac{1}{i}路j+\frac{1}{j}路i$, right side coefficients $2$

$\frac{i}{j}+\frac{j}{i}鈮�2$

$\frac{i^2+j^2}{ij}鈮�2/路ij,i,j鈮�0$

$i^2+j^2鈮�2ij$

$(i-j{)}^2鈮�0$

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:

$P_1\left(O\right)鈮�P_2\left(O\right)$.