91Ó°ÊÓ

A certain population of individuals is k different strata.

The proportion of individuals in the total population belong to the stratum i is\({p_i}\)

For the individuals in stratum i, this characteristic has mean \({\mu _i}\) and variance \(\sigma _i^2\)

Let X denote the value of the characteristic for a person chosen at random from the total population, and let\({A_i}\)denote the event that the person belongs to stratum i\(\left( {i = 1,...,k} \right)\)

Then,

\(\begin{aligned}{}\mu &= E\left( X \right)\\ &= \sum\limits_{i = 1}^k {E\left( {X\left| {{A_i}} \right.} \right)P\left( {{A_i}} \right)} \\ &= \sum\limits_{i - 1}^k {{\mu _i}{p_i}} \end{aligned}\)

Also,

\(\begin{aligned}{}E\left( {\hat \mu } \right) &= \sum\limits_{i = 1}^k {{p_i}E\left( {{{\bar X}_i}} \right)} \\ &= \sum\limits_{i = 1}^k {{p_i}{\mu _i}} \\ &= \mu \end{aligned}\)

Hence, \(\hat \mu = \sum\nolimits_{i = 1}^k {{p_i}{{\bar X}_i}} \) is an unbiased estimator of \(\mu \). [Proved]

The samples are taken independently of each other; the variables\({\bar X_1},...,{\bar X_k}\)are independent

Therefore,

\(\begin{aligned}{}Var\left( {\hat \mu } \right) &= \sum\limits_{i = 1}^k {p_i^2Var\left( {{{\bar X}_i}} \right)} \\ &= \sum\limits_{i = 1}^k {\frac{{p_i^2\sigma _i^2}}{{{n_i}}}} \end{aligned}\)

Hence, the values of\({n_1},...,{n_k}\)must be chosen to minimize,

\(v = \sum\limits_{i = 1}^k {\frac{{{{\left( {{p_i}{\sigma _i}} \right)}^2}}}{{{n_i}}}} \)

Subject to the constraint that\(\sum\limits_{i = 1}^k {{n_i} = n} \)

If let, \[{n_k} = n - \sum\limits_{i = 1}^{k - 1} {{n_i}} \]

Then,

\(\begin{aligned}{}\frac{{\partial v}}{{\partial {n_i}}} &= \frac{\partial }{{\partial {n_i}}}\left( {\sum\limits_{i = 1}^k {\frac{{{{\left( {{p_i}{\sigma _i}} \right)}^2}}}{{{n_i}}}} } \right)\\ &= \frac{{ - {{\left( {{p_i}{\sigma _i}} \right)}^2}}}{{n_i^2}} + \frac{{{{\left( {{p_k}{\sigma _k}} \right)}^2}}}{{n_k^2}}\,\,\,for\,\,\,i = 1,...,k - 1\end{aligned}\)

When each of the partial derivatives is set equal to 0, it is found that\(\frac{{{n_i}}}{{\left( {{p_i}{\sigma _i}} \right)}}\)has the same value for\(i = 1,...,k\)

Therefore,

\({n_i} = c{p_i}{\sigma _i}\)for some constant c.

It follows that

\(\begin{aligned}{}n &= \sum\limits_{j = 1}^k {{n_j}} \\ &= c\sum\limits_{j = 1}^k {{p_j}{\sigma _j}} \end{aligned}\)

Hence, \(c = \frac{n}{{\sum\limits_{j = 1}^k {{p_j}{\sigma _j}} }}\)

Then,

\({n_i} = \frac{{n{p_i}{\sigma _i}}}{{\sum\limits_{j = 1}^k {{p_j}{\sigma _j}} }}\)

The minimum integers for v would presumably be near the minimizing values of\({n_1},...,{n_k}\)which have just been found.

Therefore, \({n_i} = \frac{{n{p_i}{\sigma _i}}}{{\sum\limits_{j = 1}^k {{p_j}{\sigma _j}} }}\), the values \({n_1},...,{n_k}\) for which variance is minimum