91Ó°ÊÓ

Here \(Y\) be discrete random variable. We have to show \(X\) has continuous random variable.

From given information to find that \(X = \left| Y \right|\) .

Firstly from random variable\(Y\)and \(f\left( y \right) = \frac{1}{{2\left| Y \right|\left( {\left| Y \right| + 1} \right)}}\). Then mean of\(Y\)is given by

\(\begin{array}{c}E\left( Y \right) = \sum\limits_{i = 1}^\infty {Y \times \frac{1}{{2\left| Y \right|\left( {\left| Y \right| + 1} \right)}}} \\ = \infty \end{array}\)

The random variable\(X\)is given by

\(f\left( X \right) = \frac{1}{{X\left( {X + 1} \right)}}\)

If we find the mean of\(X\) then it is given by

\(\begin{array}{c}E\left( x \right) = \sum\limits_{i = 1}^\infty {X \times \frac{1}{{2\left| X \right|\left( {\left| X \right| + 1} \right)}}} \\ = \infty \end{array}\)

If we define this random variable by pdf then it is defined by

\(f\left( x \right) = \frac{1}{{X\left( {X + 1} \right)}}\) where\(1 \le X \le \infty \)

We know that sum of the all pdf is 1.Then integrating to get

\(f\left( X \right) = \int_1^\infty {\frac{1}{{X\left( {X + 1} \right)}}} dx = 1\)

Hence proof that \(X\) is continuous random variable.