91Ó°ÊÓ

For a discrete distribution, it is given that \(f\left( x \right) = 0\) for \(x \in \left[ {0,1} \right]\).

Let f be the probability function for a discrete distribution.

Suppose \(f\left( x \right) = 0\) for\(x \notin \left[ {0,1} \right]\).

Let X have the probability function as f. Assume that \(Var\left( X \right) > 0\) (otherwise, the value for the variance is surely less than\(\frac{1}{4}\)).

Let’s suppose that X has only two possible values, 0 and 1.

Let \(p = \Pr \left( {X = 1} \right)\)

Then,

\(\begin{array}{c}E\left( X \right) = E\left( {{X^2}} \right)\\ = p\end{array}\)

And the variance of X is \({\mathop{\rm var}} \left( X \right) = p - {p^2}\)

The largest possible value of \(p - {p^2}\) occurs when\(p = \frac{1}{2}\)and the value is\(\frac{1}{4}\).

Therefore, \(Var\left( X \right) \le \frac{1}{4}\) if X has any possible values 0 and 1, there is a random variable Y taking values 0 and 1\(Var\left( Y \right) \ge Var\left( X \right)\).

So assume that X takes at least one value strictly between 0 and 1.

Assume that one of those possible values is between 0 and \(\mu \)(replacing X by 1-X will give the same value for the variance)

Let \(\mu = E\left( X \right)\), and let \(x{ _1},{x_2},...\) be the values such that \({x_1} \le \mu ,f\left( {{x_i}} \right) > 0\)

Define the new random variable as shown below:

\(\begin{array}{c}{X^*} = 0\;\;\;\;\;if\,X \le \mu \\ = X\;\;\;\;if\;X > \mu \end{array}\)

The probability function of \({X^*}\) is \({\mu ^*} = \mu - \sum\limits_i {{x_i}f\left( {{x_i}} \right)} \).

The mean of \({X^{*2}}\) is \(E\left( {{X^2}} \right) - \sum\limits_i {x_{_i}^2} f\left( {{x_i}} \right)\) , so the variance of \({X^*}\) is

\(\begin{array}{c}Var\left( {{X^*}} \right) = E\left( {{X^2}} \right) - \sum\limits_i {x_i^2{f_i}{{\left[ {\mu - \sum\limits_i {{x_i}f\left( {{x_i}} \right)} } \right]}^2}} \\E\left( {{X^2}} \right) = Var\left( {{X^*}} \right) - \sum\limits_i {x_i^2f\left( {{x_i}} \right) - {{\left[ {\sum\limits_i {{x_i}f\left( {{x_i}} \right)} } \right]}^2}} \end{array}\)……………………………. (1)

Since, \({x_i} \le \mu \) for each i, then,

\( - \sum\limits_i {x_i^2} f\left( {{x_i}} \right) + 2\mu \sum\limits_i {{x_i}} f\left( {{x_i}} \right) \ge \sum\limits_i {f\left( {{x_i}} \right)} \) …………………………….. (2)

Let,

\(t = \sum\limits_i {f\left( {{x_i}} \right)} > 0\)

Then

\(\begin{array}{c}g\left( x \right) = \frac{{f\left( x \right)}}{t}\;\;\;for\,x \in \left\{ {{x_1},{x_2},...} \right\}\\ = 0\;\;\;otherwise\end{array}\)

is a probability function.

Let Z be a random variable whose probability function is g. Then

\(\begin{array}{l}E\left( Z \right) = \frac{1}{t}\sum\limits_i {f\left( {{x_i}} \right)} \\Var\left( Z \right) = \frac{1}{t}\sum\limits_i {x_i^2f\left( {{x_i}} \right)} \end{array}\)

Since \(Var\left( Z \right) \ge 0\) and \(t \le 1\) then

\(\begin{array}{c}\sum\limits_i {x_i^2f\left( {{x_i}} \right)} \ge \frac{1}{t}{\left[ {\sum\limits_i {{x_i}f\left( {{x_i}} \right)} } \right]^2}\\ \ge {\left[ {\sum\limits_i {{x_i}f\left( {{x_i}} \right)} } \right]^2}\end{array}\)

From equations (1) and (2), \(Var\left( {{X^*}} \right) \ge Var\left( X \right)\)

If \(f\left( x \right) > 0\) , for some x strictly between 0 and 1, replace X by \(1 - {X^*}\) and repeat the process to produce the desired random variable Y.

Hence, it is proved that the variance of this distribution is at the most \(\frac{1}{4}\) .