Q. 4.4 - A First Course in Probability

Question

A certain community is composed of$m$ families, $n_i$ of which have $i$children, $\sum _{i=1}^r{n}_i=m$. If one of the families is randomly chosen, let$X$ denote the number of children in that family. If one of the $\sum _{i=1}^r{in}_i$ children is randomly chosen, let$Y$ denote the total number of children in that child's family. Show that $E\left[Y\right]\ge E\left[X\right]$

Step-by-Step Solution

Answer

The probability of choosing a child from a family with $i$ children is$\frac{i{n}_i}{{\sum }_{i=1}^{r}i{n}_i}$. To show, $\mathrm{E}\left[\mathrm{Y}\right]\ge \mathrm{E}\left[\mathrm{X}\right]$$\Rightarrow (i-j{)}^2\ge 0$. So proved as square can never be a negative term.

Step 1: Computation of E [ X ]

The probability that a randomly chosen family will have $i$children is $\frac{ni}{m}$

Hence,$\mathrm{E}\left[\mathrm{X}\right]=\frac{{\sum }_{\mathrm{i}=1}^{\mathrm{r}}{i}_i}{m}$

We get,

$=\frac{{\sum }_{i=1}^{r}i{n}_i}{{\sum }_{i=1}^r{n}_i}.$

Step 2: Computation of E [ Y ]

The probability of choosing a child from a family with $i$children is $\frac{i{n}_i}{{\sum }_{i=1}^{p}i{n}_i}$.

$E\left[Y\right]=\frac{{\sum }_{i=1}^t{i}^2{n}_i}{{\sum }_{i=1}^{r}i{n}_i}$

To show, $\mathrm{E}\left[\mathrm{Y}\right]\ge \mathrm{E}\left[\mathrm{X}\right]$

$\frac{{\sum }_{i=1}^r{\mathrm{i}}^2{n}_i}{{\sum }_{i=1}^t{i}_i}\ge \frac{{\sum }_{i=1}^r{i}_i}{{\sum }_{i=1}^t{n}_i}.$

Step 3: Prove E [ Y ] ≥ E [ X ]

Proved as square can never be a negative term

$\Rightarrow \sum _{i=1}^{r}i{n}_i\sum _{i=1}^r{i}^2{n}_i\ge \sum _{i=1}^{r}i{n}_i\sum _{i=1}^{r}i{n}_i$

$\Rightarrow \sum _{j=1}^t\sum _{i=1}^r{i}^2{n}_i{n}_j{{}^3\sum }_{i=1}^t\sum _{j=1}^{t}ij{n}_i{n}_j$

$\Rightarrow {\mathrm{i}}^2+{\mathrm{j}}^2\ge 2\mathrm{ij}\text{ [coefficients] }$

We get,

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

Step 4: Final Answer

$\mathrm{E}\left[\mathrm{Y}\right]\ge \mathrm{E}\left[\mathrm{X}\right]$$\Rightarrow (i-j{)}^2\ge 0$. Hence proved as square can never be a negative term.