This is my problem:
Consider an i.i.d sequence of random variables $ X_1, X_2,\ldots,X_n,$ and define $m_n = 1 - \min(X_1, \ldots, X_n)$. Find the distribution of $ m_n $ in terms of $ F_X $ the CDF of X .
For context, I have solved a similar problem which is shown below:
Consider an i.i.d sequence of random variables $ X_1, X_2,\ldots, X_n,$ and define $M_n = \max(X_1, \ldots, X_n)$.
Find the distribution of $ M_n $ in terms of $ F_X $ the CDF of X .
Answer:
\begin{align} P(M_n \leq x) &= P(\max(X_1, \ldots, X_n) \leq x) \\&= P(X_1 \leq X_1,\ldots, X_n \leq x) \\&= P(X_1 \leq x)\ldots P(X_n \leq x) \\&= F_x(x) \ldots F_x(n) \\&=(F_x(x))^n \end{align}
I am just wondering if i am missing some rule between max() and min() that I could use to solve this problem? or could i use the following
$$P(\min(X,Y) \leq x) = F_x(x)+ F_y(x) - F_x(x) F_y(x)$$
, but I am not sure how to include that with the 1 - term in the probability function.
Any advice would be very helpful. Thank you.
Answering your original problem directly:
$$\{m_n\leq x\}=\{1-\min(X_1,\dots,X_n)\leq x\}=\{\min(X_1,\dots,X_n)\geq1-x\}=$$$$\{X_1\geq 1-x\}\cap\dots\cap\{X_n\geq1-x\}$$
On base of independepence we find:$$P(\{X_1\geq 1-x\}\cap\dots\cap\{X_n\geq1-x\})=P(X\geq1-x)^n=(1-P(X<1-x))^n\tag1$$
Here: $$P(X<1-x)=\lim_{z\downarrow x}F_X(1-z)$$ and equals $F_X(1-x)$ if $P(X=1-x)=0$.
In that case you can write the RHS of $(1)$ as: $$(1-F_X(1-x))^n$$