Question

Question: Show that the sum of the probabilities of a random variable with geometric distribution with parameter\(p\), where\(0 < p < 1\), equals \(1\).

Short answer

Answer:

The sum of the probabilities of a random variable with geometric distribution with parameter, where\(0{\rm{ }} < {\rm{ }}p{\rm{ }} \le {\rm{ }}1\), equals \(1\).

Step-by-step solution

Given information

Random variable with geometric distribution with parameter \(p\), where \(0{\rm{ }} < {\rm{ }}p{\rm{ }} \le {\rm{ }}1\)

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 sum of the probabilities

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

Thus, the sum of the probabilities of a random variable with geometric distribution with parameter, where\(0{\rm{ }} < {\rm{ }}p{\rm{ }} \le {\rm{ }}1\), equals \(1\).