Question

What is the formula for the expected value of a geometric random variable?

Answer 1

Hint: To derive the formula for the expected value of a geometric random variable, we have to consider a random variable X that has the values ${{x}_{1}},{{x}_{2}},...$ with corresponding probability $P\left( {{x}_{1}} \right),P\left( {{x}_{2}} \right),...$ . We know that the expected value of the random variable (also known as mean) is given by $E\left( X \right)=\sum{XP\left( X \right)}$ . Now, we have to expand the sum and substitute $P\left( X=x \right)={{\left( 1-p \right)}^{x-1}}p...$ in it. Then, we have to simplify and multiply $\left( 1-p \right)$ to the result. Then, we have to subtract this result from the previous result and simplify. Finally, we will get a geometric series and we have to simplify it using the sum to infinite formula of a GP.

Complete step-by-step answer:
Let us first see what is meant by expected value. Let us consider a random variable X that has the values ${{x}_{1}},{{x}_{2}},...$ with corresponding probability $P\left( {{x}_{1}} \right),P\left( {{x}_{2}} \right),...$ , then the expected value of the random variable (mean) is given by:
$E\left( X \right)=\sum{XP\left( X \right)}$
Let us expand the sum.
$\Rightarrow E\left( X \right)=1\cdot P\left( X=1 \right)+2\cdot P\left( X=2 \right)+3\cdot P\left( X=3 \right)+....\text{ }...\left( i \right)$
We know that geometric distribution states that the probability that first success appears after x number of trials. Let us consider p to be the probability of success or failure of each trial. Then, the probability that success appears on the ${{x}^{th}}$ trial is given by
$P\left( X=x \right)={{\left( 1-p \right)}^{x-1}}p...$
Let us substitute the above result equation (i).
$\begin{align}
  & \Rightarrow E\left( X \right)=1\cdot {{\left( 1-p \right)}^{1-1}}p+2\cdot {{\left( 1-p \right)}^{2-1}}p+3\cdot {{\left( 1-p \right)}^{3-1}}p+.... \\
 & \Rightarrow E\left( X \right)=1\cdot p+2\cdot \left( 1-p \right)p+3\cdot {{\left( 1-p \right)}^{2}}p+.... \\
 & \Rightarrow E\left( X \right)=p+2p\left( 1-p \right)+3p{{\left( 1-p \right)}^{2}}+....\text{ }...\left( ii \right) \\
\end{align}$
Let us multiply equation (ii) $\left( 1-p \right)$ times.
$\Rightarrow \left( 1-p \right)E\left( X \right)=p\left( 1-p \right)+2p{{\left( 1-p \right)}^{2}}+3p{{\left( 1-p \right)}^{3}}+....\text{ }...\left( iii \right)$
Now, we have to subtract equation (iii) from (ii).
$\Rightarrow E\left( X \right)-\left( 1-p \right)E\left( X \right)=p+2p\left( 1-p \right)+3p{{\left( 1-p \right)}^{2}}+....-\left[ p\left( 1-p \right)+2p{{\left( 1-p \right)}^{2}}+3p{{\left( 1-p \right)}^{3}}+.... \right]$
We have to rearrange the terms.
$\Rightarrow E\left( X \right)-\left( 1-p \right)E\left( X \right)=p+2p\left( 1-p \right)-p\left( 1-p \right)+3p{{\left( 1-p \right)}^{2}}-2p{{\left( 1-p \right)}^{2}}+....$
Now, let us perform the subtraction.
$\Rightarrow E\left( X \right)-\left( 1-p \right)E\left( X \right)=p+p\left( 1-p \right)+p{{\left( 1-p \right)}^{2}}+....$
We have to simplify the LHS by using distributive property.
$\begin{align}
  & \Rightarrow E\left( X \right)-E\left( X \right)+pE\left( X \right)=p+p\left( 1-p \right)+p{{\left( 1-p \right)}^{2}}+.... \\
 & \Rightarrow pE\left( X \right)=p+p\left( 1-p \right)+p{{\left( 1-p \right)}^{2}}+.... \\
\end{align}$
Let us take p outside from the RHS.
$\Rightarrow pE\left( X \right)=p\left[ 1+\left( 1-p \right)+{{\left( 1-p \right)}^{2}}+.... \right]$
We can now cancel p from both the sides.
$\Rightarrow E\left( X \right)=1+\left( 1-p \right)+{{\left( 1-p \right)}^{2}}+....\text{ }...\left( iv \right)$
Now, we can see that the RHS is a geometric progression (GP). We have to find the common ratio, r. For this, we have to divide the second term by the first term or divide the third term by the second term. Either way, we have get the same value of r.
$r=\dfrac{\left( 1-p \right)}{1}=\left( 1-p \right)$
We know that sum to infinity of a geometric progression is given by the formula
${{S}_{\infty }}=\dfrac{a}{1-r};-1 < r < 1$
Where a is the first term of the GP. From the RHS of equation (iv), we can see that $a=1$ . Therefore, we can write the RHS of equation (iv) as
$\Rightarrow E\left( X \right)=\dfrac{1}{1-\left( 1-p \right)}$
Let us simplify the denominator of the RHS.
$\begin{align}
  & \Rightarrow E\left( X \right)=\dfrac{1}{1-1+p} \\
 & \Rightarrow E\left( X \right)=\dfrac{1}{p} \\
\end{align}$
Hence, the formula for the expected value of a geometric random variable is $E\left( X \right)=\dfrac{1}{p}$ .

Note: Students must know the general formula of probability to derive the expectation. Expectation of a geometric random variable is also known as the mean $\mu $ . Students must know to find the common ratio of a GP and also must be very thorough with the formulas related to the GP. We have obtained a geometric progression during the derivation. This is the reason we call geometric random variables.