Good morning, I have one small question about an exercise:
Let $X_1, . . . , X_n$ independent, real-valued random variables with distribution functions $F_1, . . . , F_n$ If each $X_i$ is uniform distributed on $[0, 1]$, find the density of the minimum of the set.
I first calculated $F_{X_i}(x)$
$F_{i}(x)=\left\{\begin{matrix} & 0 & otherwise\\ & x & a \leq x \leq b\\ & 1 & x \geq b \end{matrix}\right.$
And with the formula
$F_{minX_i}(x)=\left\{\begin{matrix} & 0 & otherwise\\ & 1-\prod_{i=1}^{n}(1-F_{i})=1-(1-F_{i})^n=1-(1-x)^n & a \leq x \leq b\\ & 1 & x \geq b \end{matrix}\right.$
Knowing $f_{minX_i}(x)=F'_{minX_i}(x)$
$f_{minX_i}(x)=\left\{\begin{matrix} & 0 & otherwise\\ & -n(1-x)^{n-1} & a\leq x \leq b \end{matrix}\right.$
However what I can't explain is that in the solutions the right answer is:
$f_{minX_i}(x)=\left\{\begin{matrix} & 0 & otherwise\\ & 1-nx(1-x)^{n-1} & a\leq x \leq b \end{matrix}\right.$
Where does this result come from?
Am I doing something wrong?
It's $n(1-x)^{n-1}$, with no minus sign before, otherwise you would have a negative density. So up to a change of sign you are correct. I do not know what the other function refers to, it does not integrate to 1 so it is not even a density function.