91Ó°ÊÓ

X is a random variable with c.d.f F. The numbers a and b are medians of X such that \(a < b\).

(a)

We know that a is a median of X. Hence,\(P\left( {X \le a} \right) \ge \frac{1}{2}{\rm{ and }}P\left( {X \ge a} \right) \ge \frac{1}{2}.\)

\(\begin{aligned}{}Now,\,\,P\left( {X \le a} \right) \ge \frac{1}{2}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \Rightarrow F\left( a \right) \ge \frac{1}{2} \ldots \ldots \ldots .....\left( 1 \right)\end{aligned}\)

Again,

\(\begin{aligned}{}Again,\,\,P\left( {X \ge a} \right) \ge \frac{1}{2}\\ \Rightarrow 1 - P\left( {X < a} \right) \ge \frac{1}{2}\\ \Rightarrow P\left( {X < a} \right) \le \frac{1}{2}\\ \Rightarrow F\left( a \right) \le \frac{1}{2} \ldots \ldots ......... \ldots \left( 2 \right)\end{aligned}\)

By solving equation (1) and (2), we get:

Hence, \(F\left( a \right) = \frac{1}{2}.\)

(b)

We know that a and b are the medians of X.

Let, c be the smallest value of a, and d be the smallest value of b.

Then, it is clear that c and d are also medians of X.

That means:

\(\begin{aligned}{}P\left( {X \le c} \right) \ge \frac{1}{2}{\rm{ and }}P\left( {X \ge c} \right) \ge \frac{1}{2}\\P\left( {X \le d} \right) \ge \frac{1}{2}{\rm{ and }}P\left( {X \ge d} \right) \ge \frac{1}{2}\end{aligned}\)

Now, let m be another number such that \(m \in \left( {c,d} \right)\)i.e., \(c \le m \le d\).

We know that the cumulative distribution function of X is a non-decreasing function.

\(\begin{aligned}{}Hence,\,\,P\left( {X \le m} \right) \ge P\left( {X \le c} \right) \ge \frac{1}{2}\\ \Rightarrow P\left( {X \le m} \right) \ge \frac{1}{2}\end{aligned}\)

\(\begin{aligned}{}Also,\,\,P\left( {X \ge m} \right) \ge P\left( {X \ge d} \right) \ge \frac{1}{2}\\ \Rightarrow P\left( {X \ge m} \right) \ge \frac{1}{2}\end{aligned}\)

Thus, m satisfies both the conditions of being a median. Hence, for any \(m \in \left( {c,d} \right)\), m is a median of X.

(c)

Since d is a median, \(P\left( {X \le d} \right) \ge \frac{1}{2}{\rm{ and }}P\left( {X \ge d} \right) \ge \frac{1}{2}.\)

\(\begin{aligned}{}Now,\,\,P\left( {X \le d} \right) \ge \frac{1}{2}\\ \Rightarrow F\left( d \right) \ge \frac{1}{2}.............\left( 3 \right)\end{aligned}\)

\(\begin{aligned}{}Again,\,\,P\left( {X \ge d} \right) \ge \frac{1}{2}\\ \Rightarrow 1 - P\left( {X < d} \right) \ge \frac{1}{2}\\ \Rightarrow F\left( {d - 1} \right) \le \frac{1}{2}...................\left( 4 \right)\end{aligned}\)

Since, X has a discrete distribution,\(F\left( d \right) > F\left( {d - 1} \right).\)

Also, since any number in the closed interval\(\left( {c,d} \right)\)is a median,\(d - 1\)is also a median. Thus,\(F\left( {d - 1} \right) = \frac{1}{2}{\rm{ }}\left( {{\rm{complying with equation }}\left( 4 \right)} \right).\)

So,

\(\begin{aligned}{}So,\,\,F\left( d \right) > F\left( {d - 1} \right) = \frac{1}{2}\\\,\,\, \Rightarrow F\left( d \right) > \frac{1}{2}\end{aligned}\)

Hence, \(F\left( d \right) > \frac{1}{2}.\)