Question

Question: Show that if the random variable \(X\) has the geometric distribution with parameter \(p\), and \(j\) is a positive integer, then \({\bf{p}}\left( {{\bf{X}}{\rm{ }} \ge {\rm{ }}{\bf{j}}} \right){\rm{ }} = {\rm{ }}\left( {{\bf{1}}{\rm{ }} - {\rm{ }}{\bf{p}}} \right){\bf{j}} - {\bf{1}}\).

Short answer

Answer:

The random variable \(X\) has the geometric distribution with parameter\(p\), and \(j\) is a positive integer \({\bf{p}}\left( {{\bf{X}}{\rm{ }} \ge {\rm{ }}{\bf{j}}} \right){\rm{ }} = {\rm{ }}\left( {{\bf{1}}{\rm{ }} - {\rm{ }}{\bf{p}}} \right){\bf{j}} - {\bf{1}}\).

Step-by-step solution

Given information

The random variable \(X\) has the geometric distribution with parameter\(p\), and \(j\) is a positive integer

Step 2: Definition

A random variable\(X\)has a geometric distribution with parameter p if \(E(x) = \frac{1}{p}\)

\(p(X = k) = {(1 - p)^{k - 1}}p\)where p is a real number with\(0 \le p \le 1\)

Theorem: If the random variable \(X\) has the geometric distribution with parameter \(p\), then \(E(x) = \frac{1}{p}\)

Step 3: Calculate the geometric distribution

\(\begin{array}{l}\sum\limits_{k = 1}^\infty {P(X \ge j)} \\ = \sum\limits_{k = j}^\infty {p(X = k)} \\ = \sum\limits_{k = j}^\infty {{{(1 - p)}^{k - 1}}p} \\ = p\sum\limits_{k = j}^\infty {{{(1 - p)}^{k - 1}}} \\ = p{(1 - p)^{j - 1}}\sum\limits_{k = 0}^\infty {{{(1 - p)}^k}} \\ = p{(1 - p)^{j - 1}}.\frac{1}{{1 - (1 - p)}}\\ = p{(1 - p)^{j - 1}}.\frac{1}{p}\\ = {(1 - p)^{j - 1}}\end{array}\)

Thus, it can be concluded thatthe random variable \(X\) has the geometric distribution with parameter\(p\), and \(j\) is a positive integer\({\bf{p}}\left( {{\bf{X}}{\rm{ }} \ge {\rm{ }}{\bf{j}}} \right){\rm{ }} = {\rm{ }}\left( {{\bf{1}}{\rm{ }} - {\rm{ }}{\bf{p}}} \right){\bf{j}} - {\bf{1}}\).