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$(X_n)$ sequence of iid random variables with uniform distribution $U([0,1])$.
$m=\min(X_1,...X_n), M=\max(X_1,...X_n)$.

I want to find $f_{m,M}(s,t)$.

$$ \begin{split} P(m<s,M<t) &= P(m<s)P(M<t)1_{m\ne M}+P(X_1<\min(s,t))1_{m=M} \\ &= (1-(1-s))^nt^n1_{m \ne M}+\min(s,t)1_{m=M} \\ &=((st)^n+s)1_{s<t}+((st)^n+t)1_{s \ge t} \end{split} $$

When I differentiate it, I get $f_{m,M}(s,t)=n^2t^{n-1}s^{n-1}$.

Is this okay? And does it mean that $M$ and $m$ are independent and $f_{m,M}(s,t)=f_m(s)f_M(t)$?

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For a sequence of $n$ iid continuous samples, $(X_i)_{n=1}^n$, the minimum is less than $s$ and maximum less than $t$ iff all samples are less than $t$ and at least one is less than $s$.

$$\begin{split} \mathsf P(m\leqslant s, M\leqslant t) &=\mathsf P\Big(\big(\bigcup_{i=1}^n \{X_i\leqslant s\}\big)\cap\big(\bigcap_{i=1}^n\{X_i\leqslant t\}\big)\Big) \\&= \mathsf P\Big(\big(\bigcap_{i=1}^n\{X_i\leqslant t\}\big)\setminus\big(\bigcap_{i=1}^n\{s<X_i\}\big)\Big) \\ &= \mathsf P\big(\bigcap_{i=1}^n\{X_i\leqslant t\}\big)-\mathsf P\big(\bigcap_{i=1}^n\{s< X_i\leq t\}\big) \\ &= \prod_{i=1}^n\mathsf P\{X_i\leqslant t\}-\prod_{i=1}^n\mathsf P\{s<X_i\leqslant t\} \\ &= \big(\mathsf P\{X_i\leqslant t\}\big)^n-\big(\mathsf P\{s<X_i\leqslant t\}\big)^n \\ & =\begin{cases} 0 &:& s<0 ~\vee~ t<0 \\ t^n-(t-s)^n & :& 0\leqslant s\leqslant t< 1 \\ t^n &:& 0\leqslant t < \min (s,1) \\ 1-(1-s)^n &:& 0\leqslant s< 1\leqslant t \\ 1 &:& 1\leqslant s ~\wedge~ 1\leqslant t \end{cases} \\[2ex]f_{n,M}(s,t) &=\begin{cases} n(n-1)(t-s)^{n-2} & :& 0\leqslant s\leqslant t< 1 \\ 0 &:& \textsf{elsewhere} \end{cases} \end{split}$$

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  • $\begingroup$ How can I do these when $X_i$ have continuous distributions? $P(x \le t)=P(x < t)$, right? $\endgroup$
    – Michalina
    Feb 2 '18 at 1:12
  • $\begingroup$ I can use only the first part you wrote, right? And the density will be $nt^{n-1}-n(n-1)(t-s)^{n-2}$? $\endgroup$
    – Michalina
    Feb 2 '18 at 1:42
  • $\begingroup$ Ah, right. Then indeed, you just need the first and differentiate. $\endgroup$ Feb 2 '18 at 5:58
  • $\begingroup$ @GrahamKemp the last case $1\leq s$ and $1 \leq t$ should be $1$, right? $\endgroup$ Apr 5 '19 at 17:12
  • 1
    $\begingroup$ @DanielOrdoñez Good catch. $\endgroup$ Apr 6 '19 at 0:53
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First of all, you have an equation where on the left hand side you have a probability of an event - so a number - and on the right hand side you have probabilities multiplied with indicator functions - so a random variable. That's not okay.

Secondly, you are using the independence you want to show, when you say that $P[m<s,M<t,m<M]=P[m<s]P[M<t]$.

Hint on independence: What would you guess is the probability that $P[m>0.8,M<0.2]$? Why does this show that they are not independent?

There is one general trick for these kind of questions, namely $M<c$ if and only if $X_i < c$ for all $i$. Then use independence of the $X_i$ to compute $P[M<c]$. A similar trick is used for the min. Hint: Try the same thing as for the maximum, but use $P[A^c]=1-P[A]$.

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  • $\begingroup$ "Secondly, you are using the independence you want to show, when you say that..." Am I? I used it for $m \ne M$ and I know that $X_i=M$ and $X_j=m$ are independent for $i \ne j$. I take the case where $i=j$ separately. That's why I used indicators (how should I do it without indicators, so that I don't have a random variable on the right side?) $\endgroup$
    – Michalina
    Feb 2 '18 at 0:00
  • $\begingroup$ "There is one general trick for these kind of questions, namely $M<c$ if and only if $Xi<c$ for all $i$. Then use independence of the $Xi$ to compute $P[M<c]$. A similar trick is used for the min. Hint: Try the same thing as for the maximum, but use $P[Ac]=1−P[A]$." - I think I did all of these in my solution. $\endgroup$
    – Michalina
    Feb 2 '18 at 0:01
  • $\begingroup$ What you should have written is $\endgroup$
    – Ecthelion
    Feb 2 '18 at 2:36
  • $\begingroup$ Regarding your first comment: Yes, you are. How else would you get that $P[m<s,M<t,m\neq M]=P[m<s] P[M<t]$? Please go, calculate $P[m>0.8,M<0.2]$ and compare it to $P[m>0.8]$ times $P[M<0.2]$. What I want to tell you is that they cannot be independent because the maximum must be at least as high as the minimum. But I guess Graham gave a 'no additional thinking required' answer, which is probably good for your homework points and the website in general but bad for your understanding. (Sorry for the last comment, I didn't know that pressing enter shoots a comment...) $\endgroup$
    – Ecthelion
    Feb 2 '18 at 2:46

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