Problem:
Let $X$ and $Y$ be independent random variables having geometric
densities with parameters $p_1$ and $p_2$ respectively. Find the density of
$min(X,Y)$.
Answer:
Let $Z = \min(X,Y)$.
\begin{eqnarray*}
P( Z = 0 ) &=& P( X = 0 ) P(Y = 0) \\
P( Z = 1 ) &=& P( X = 1 ) P(Y < 1) + P(Y = 1)P(X < 1) + P(x = 1)P(Y = 1) \\
P( Z = 2 ) &=& P( X = 2 ) P(Y < 2) + P(Y = 2)P(X < 2) + P(x = 2)P(Y = 2) \\
P( Z = i ) &=& P( X = i ) P(Y < i) + P(Y = i)P(X < i) + P(X = i)P(Y = i) \\
P( X = i ) &=& p_1(1-p_1)^{i} \\
P( Y = i ) &=& p_2(1-p_2)^{i} \\
P(X < i) &=& \sum_{i = 0}^{i-1} P( X = i ) = \sum_{i = 0}^{i-1} p_1(1-p_1)^{i} \\
\end{eqnarray*}
We have a finite geometric series with $a = p_1$ and $r = 1 - p_1$.
\begin{eqnarray*}
P(X < i) &=& \frac{a(1-r^i)}{1 - r} = \frac{p_1(1-(1 - p_1)^i)}{1 - (1 - p_1)} \\
P(X < i) &=& \frac{p_1(1-(1 - p_1)^i)}{p_1} = 1 - (1 - p_1)^i \\
P(Y < i) &=& 1 - (1 - p_2)^i \\
P( Z = i ) &=& p_1(1-p_1)^i(1 - (1 - p_1)^i) +
p_2(1 - p_2)^i(1 - (1 - p_1)^i) + p_1 p_2 (1 - p_1)^i(1 - p_2)^i \\
\end{eqnarray*}
The book's answer is: geometric with parameter $p_1 + p_2 - p_1 p_2$.
That means, to me, that:
\begin{eqnarray*}
P( Z = i ) &=& p_1 + p_2 - p_1 p_2( 1 - (p1 + p_2 - p_1 p_2))^i \\
\end{eqnarray*}
which I believe is different than my answer. So, where did I go wrong?
Based upon the comments I got, I now have what I believe to be the correct answer:
Let $Z = min(X,Y)$. \begin{eqnarray*} P( Z = i ) &=& P ( X = i )P(Y > i) + P( Y = i)P(X > i) + P(X = i)P(Y = i)\\ P( X = i ) &=& p_1(1-p_1)^{i} \\ P( Y = i ) &=& p_2(1-p_2)^{i} \\ P( Y > i) &=& \sum_{j = i+1}^{\infty} P( Y = j ) = \sum_{j = i+1}^{\infty} p_2(1-p_2)^{j} \\ \end{eqnarray*} Now we have a geometric series with $a = p_2(1 - p_2)^{i+1}$ and $r = 1 - p_2$. \begin{eqnarray*} P( Y > i) &=& \frac{p_2(1 - p_2)^{i+1}}{1 - (1 - p_2)} = (1 - p_2)^{i+1} \\ P( X > i) &=& \frac{p_1(1 - p_1)^{i+1}}{1 - (1 - p_1)} = (1 - p_1)^{i+1} \\ P( Z = i ) &=& p_1(1-p_1)^{i}( (1 - p_2)^{i+1} ) + p_2(1-p_2)^{i}((1 - p_1)^{i+1} ) \\ &+& p_1(1-p_1)^{i}p_2(1-p_2)^{i} \\ P( Z = i ) &=& (1 - p_1)^i ( 1 - p_2)^i ( p_1(1 - p_2) + p_2(1 - p_2) + p_1 p_2 ) \\ P( Z = i ) &=& (1 - p_1)^i ( 1 - p_2)^i ( p_1 - p_1 p_2 + p_2 - p_2 p_1 + p_1 p_2 ) \\ P( Z = i ) &=& (1 - p_1)^i ( 1 - p_2)^i ( p_1 + p_2 - p_2 p_1 ) \\ P( Z = i ) &=& ((1 - p_1) ( 1 - p_2))^i ( p_1 + p_2 - p_2 p_1 ) \\ P( Z = i ) &=& ( p_1 + p_2 - p_2 p_1 ) ( 1 - (p_1 + p_2 - p_1 p_2 )^i\\ \end{eqnarray*} Therefore $Z$ is a geometric random variable with parameter $p_1 + p_2 - p_1 p_2$.
Bob