Problem with Two Geometric Random Variables

Question

Please consider the following problem and my attempt to solve it.

Let $X$ and $Y$ be independent random variables having geometric densities with parameters $p_1$ and $p_2$ respectively. Find $P(X >= Y)$.
Answer:
\begin{eqnarray*} P( X >= Y ) &=& P( X = 0)P(Y <= 0) + P(x = 1)P(Y <= 1) \\ &+& P(x = 2)P(Y <= 2) + P(X = 3)P(Y <= 3) + P(X = 4)P(Y <= 4) + ... \\ P( X = 0) &=& p_1 \\ P( X = 1) &=& p_1(1-p_1) \\ P( X = 2) &=& p_1(1-p_1)^2 \\ P( X = 3 ) &=& p_1(1-p_1)^3 \\ P( X = 4 ) &=& p_1(1-p_1)^4 \\ P(Y <= 0) &=& p_2 \\ P(Y <= 1) &=& P(Y = 0 ) + P( Y = 1) = p_2 + p_2(1- p_2) \\ P(Y <= 2) &=& P(Y = 0 ) + P(Y = 1 )+ P( Y = 2) \\ P(Y <= 2) &=& p_2 + p_2(1- p_2) + p_2(1- p_2)^2 \\ P(Y <= 3) &=& P( Y <= 2) + P(Y = 3) \\ P(Y <= 3) &=& p_2 + p_2(1- p_2) + p_2(1- p_2)^2 + p_2(1- p_2)^3 \\ P(Y <= 4) &=& P( Y <= 3) + P(Y = 4)\\ P(Y <= 4) &=& p_2 + p_2(1- p_2) + p_2(1- p_2)^2 + p_2(1- p_2)^3 + p_2(1- p_2)^4 \\ P( X >= Y ) &=& p_1( p_2 ) + p_1(1-p_1)( p_2 + p_2(1- p_2) ) \\ &+& p_1(1-p_1)^2 ( p_2 + p_2(1- p_2) + p_2(1- p_2)^2 ) \\ &+& p_1(1-p_1)^3 ( p_2 + p_2(1- p_2) + p_2(1- p_2)^2 + p_2(1- p_2)^3 ) \\ &+& P(X = 4)P(Y <= 4) + ... \\ P( X >= Y ) &=& p_1( p_2 ) + p_1(1-p_1)( p_2 + p_2(1- p_2) ) \\ &+& p_1(1-p_1)^2 ( p_2 + p_2(1- p_2) + p_2(1- p_2)^2 ) \\ &+& p_1(1-p_1)^3 ( p_2 + p_2(1- p_2) + p_2(1- p_2)^2 + p_2(1- p_2)^3 ) \\ &+& p_1(1-p_1)^4 ( p_2 + p_2(1- p_2) + p_2(1- p_2)^2 + p_2(1- p_2)^3 \\ && +p_2(1- p_2)^4 ) + ... \\ \end{eqnarray*}

Am I on the right track? Do I have a geometric series here? How do I complete the problem?

Bob

Answer 1

Recall the law of total probability, i.e., $$ \mathbb{P}(A) = \sum_i\mathbb{P}(A|B_i)\mathbb{P}(B_i), $$ where $\cup_{i=1}^nB_i=\Omega$ and $B_i\cap B_j = \emptyset$ for $i\neq j$. Hence, \begin{align} \mathbb{P}(X\ge Y) &= \sum_y\mathbb{P}(X\ge Y|Y=y)\mathbb{P}(Y=y)\\ & = \sum_{y=1}(1-p_1)^{y-1}(1-p_2)^{y-1}p_2\\ & = \frac{p_2}{ 1 - (1-p_1)(1-p_2)} \\ & = \frac{p_2}{p_1 + p_2 - p_1p_2}, \end{align} where $\mathbb{P}(Y=y) = (1-p_2)^{y-1}p_2 $ and $$ \mathbb{P}(X\ge y | Y = y) = \sum_{y}(1-p_1)^{y-1}p_1 = (1-p_1)^{y-1}. $$