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 $X>1$.

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

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

We have that,

role="math" localid="1646740231382" $\alpha =\sum _{i=1}^{\infty }P(X=2i)$

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

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,

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

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

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

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.

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

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

It remembers the memoryless property of Geometric distribution