Q. 4.27 - A First Course in Probability

Question

If $X$ is a geometric random variable, show analytically that

$P\{X=n+k\mid X>n\}=P\{X=k\}$

Using the interpretation of a geometric random variable, give a verbal argument as to why the preceding equation is true. 

Step-by-Step Solution

Answer

We proved that

$P\{X=n+k\mid X>n\}=P\{X=k\}$

Step 1 Definition

A random variable that takes the value k, a non-negative integer with probability pk(1-p).

Step 2 Calculation

$\mathrm{P}\{\mathrm{X}=\mathrm{n}+\mathrm{k}\mid \mathrm{X}>\mathrm{n}\}$ $=\frac{\mathrm{P}\{\mathrm{X}=\mathrm{n}+\mathrm{k}\cap \mathrm{X}>\mathrm{n}\}}{P(X>n)}$

$P\{X=n+k\cap X>n\}$ $=(1-p{)}^{n+k-1}\mathrm{p}$

$\mathrm{P}(\mathrm{X}>\mathrm{n})$ $=$First $n$ are failures

$=(1-p{)}^n$

$\Rightarrow \mathrm{P}\{\mathrm{X}=\mathrm{n}+\mathrm{k}\mid \mathrm{X}>\mathrm{n}\}=\frac{(1-\mathrm{p}{)}^{n+k-1}}{(1-p{)}^n}.\mathrm{p}$

$=(1-p{)}^{k-1}p$

$=\mathrm{P}\{\mathrm{X}=\mathrm{k}\}$

Hence proved

Step 3 continue Calculation

Also because trials are independent

Given that first $\mathrm{n}$ trials does'nt result in success

Next $k^{\text{th }}$ will result in success and next $(\mathrm{k}-1)$ in failures $=(1-p{)}^{k-1}p$

$=P\{X=k\}$

$\Rightarrow \mathrm{P}\{\mathrm{X}=\mathrm{n}+\mathrm{k}\mid \mathrm{X}>\mathrm{n}\}=\mathrm{P}\{\mathrm{X}=\mathrm{k}\}$