Question

Denote by [x] the greatest integer not exceeding x. For the pdf in Example 2.2.1, show

that the CDF can be represented as F(x) = ([x]/12)2 for O < z < 13, zero if x O, and

one ifx13.

Answer 1

Let [x] be the greatest integer not exceeding x.

The probability density function in Example 2.2.1 is given as follows:

$f(x) = \frac{2x-1}{12^2}$

for x=1,2,…,12

The cumulative distribution function (CDF) of a random variable X is defined for any real x is,

F(x) =P(X≤x)

Calculate the cumulative distribution function by using the following table:

$X=1 \\ F(1) = \frac{(2 \times 1)-1}{(12)^2} = \frac{1}{(12)^2} = (\frac{[1]}{12})^2 \\ X=2 \\ F(2) = \frac{(2 \times 1)-1}{(12)^2} + \frac{(2 \times 2)-1}{(12)^2}= \frac{1}{(12)^2} + \frac{3}{(12)^2}= (\frac{[2]}{12})^2 \\ X=3 \\ F(3) = \frac{(2 \times 1)-1}{(12)^2} + \frac{(2 \times 2)-1}{(12)^2} + \frac{(2 \times 3)-1}{(12)^2}= \frac{1}{(12)^2} + \frac{3}{(12)^2} + \frac{5}{(12)^2}= (\frac{[3]}{12})^2$

Similarly, for the remaining terms, the cumulative distribution function is obtained as,

$X= 4 \\ F(4) = (\frac{[4]}{12})^2 \\ X= 5 \\ F(5) = (\frac{[5]}{12})^2 \\ X= 6 \\ F(6) = (\frac{[6]}{12})^2 \\ … \\ X= 12 \\ F(12) = (\frac{[12]}{12})^2$

By the above equations, the cumulative distribution function is represented as,

$F(x) = (\frac{[x]}{12})^2$

Therefore, the cumulative distribution function is,

$F(x)= \left\{\begin{matrix} 0\;;x\leqslant 0 & \\ ([x]/12)^2 \; ;0<x<13 & \\ 1 \; ; x \geqslant 13 & \end{matrix}\right.$