Let $\ X_1, X_2, \dots X_n $ be independent variable with geometric distribution with parameter $\ p $ $\ (0 < p < 1 ) $ Find the distribution of $\ \min X_i , i = 1,2, \dots,n $
My attempt:
$\ P\{\min X_i = j \} = P\{\min X_i \ge j \} - P\{\min X_i \ge j+1 \} $
verbally, if I understand correctly it means I need at least $\ j $ number of fails and the probability of fail is $\ (1-p) $ so
$\ P\{ \min X_i \ge j \} - P\{ \min X_i \ge j+1 \} = (1-p)^{jn} - (1-p)^{jn+n} $
Yet in the solutions sheet the answer is $\ (1-p)^{nj-n} - (1-p)^{nj} $
What am I missing here ?
$ \min X_i \ge j $ means first $j-1$ trails fail instread of $j$, so $\ P\{ \min X_i \ge j \}=\ (1-p)^{nj-n}$.
The latter one is similar.
By the way, $ \min X_i$ is a geometric distribution with parameter $(1-(1-p)^n)$.