Question

Show that there does not exist any numbercsuch that the following function would be a p.f.:

\(f\left( x \right) = \left\{ \begin{array}{l}\frac{c}{x}\;\;\;\;for\;x = 1,2,...\\0\;\;\;\;otherwise\end{array} \right.\)

Short answer

There does not exist any number c such that \(f\left( x \right) = \left\{ \begin{array}{l}\frac{c}{x}\;\;for\;x = 1,2,...\\0\;\;\;\;otherwise\end{array} \right.\) is a probability function.

Step-by-step solution

Given information

The function is given as,

\(f\left( x \right) = \left\{ \begin{array}{l}\frac{c}{x}\;\;for\;x = 1,2,...\\0\;\;\;\;otherwise\end{array} \right.\)

Calculate the value for c

For a function to be a probability function, the sum of the probabilities in the distribution (over the support ) should be equal to 1.

From the provided function, the sum of the probabilities is computed as,

\(\begin{aligned}{c}\sum\limits_i {f\left( x \right)} &= \sum\limits_{x = 1}^\infty {\frac{c}{x}} \\ &= \left( {c + \frac{c}{2} + \frac{c}{3} + \frac{c}{4} + ...} \right)\\ &= c\left( {1 + \frac{1}{2} + \frac{1}{3} + ...} \right)\\ &= c\left( {\sum\limits_{n = 1}^\infty {\frac{1}{n}} } \right)\end{aligned}\)

It is known that the series diverges.

Hence, there does not exist any number c such that the provided function is a probability function.