Q.4.26 - A First Course in Probability

Question

Let α be the probability that a geometric random variable $X$ with parameter $p$ is an even number.

(a) Find $\alpha$ by using the identity $\alpha =\sum _{i=1}^{\infty }\mathrm{P}\{\mathrm{X}=2\mathrm{i}\}$.

(b) Find α by conditioning on whether $X = 1$ or $X > 1$. 

Step-by-Step Solution

Answer

$a)$$\alpha$ is found to be$\frac{1-p}{2-p}$

$b)$$\alpha$ is found to be$\frac{1-p}{2-p}$

Step 1: Given Information ( Part a)

Let $\alpha$ be the probability that a geometric random variable $X$ with parameter $p$ is an even number. 

Step 2: Calculation ( Part a)

Define $X~Geom\left(p\right)$

We have that,

$\alpha =\sum _{i=1}^{\infty }P(X=2i)$

$=\sum _{i=1}^{\infty }(1-p{)}^{2i-1}p$

Substitute the given expression,

$=\frac{p}{1-p}\sum _{i=1}^{\infty }(1-p{)}^{2i}$

$=\frac{p}{1-p}\sum _{i=1}^{\infty }{\left((1-p{)}^2\right)}^i$

Substitute he given expression,

$=\frac{p}{1-p}·\frac{(1-p{)}^2}{1-(1-p{)}^2}$

We get,

$=\frac{1-p}{2-p}$.

Step 3: Final answer ( Part a)

The $\alpha$ is found to be $\frac{1-p}{2-p}$.

Step 4: Given Information (Part b)

Let α be the probability that a geometric random variable $X$ with parameter $p$ is an even number. 

Step 5: Calculation ( Part b)

For mathematically accurate writing, indicate with $E$an event where $X$ is an even number. Using LOTP, we have that

$P\left(E\right)=P(E\mid X=1)P(X=1)+P(E\mid X>1)P(X>1)=0·p+P(E\mid X>1)·q$

Now, the trick here is to utilize the memoryless property of Geometric distribution. If we are given that $X>1$, we can begin regarding random variable $X$ from the second trial and so on.

Hence, the necessary event, in that case, is that the unique variable ($X$shifted for one trial) is odd. The probability for that event is $1-\alpha$. Thus, we end up with an equation.

$\alpha =q·(1-\alpha )\Rightarrow \alpha =\frac{1-p}{2-p}$

Step 6: Final answer (Part b)

The $\alpha$ is found to be $\frac{1-p}{2-p}$.

 It remembers the memoryless property of Geometric distribution