# Find joint distribution of minimum and maximum of iid random variables

$$(X_n)$$ 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

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 $$f_{m,M}(s,t)=f_m(s)f_M(t)$$?

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

• How can I do these when $X_i$ have continuous distributions? $P(x \le t)=P(x < t)$, right? Feb 2 '18 at 1:12
• I can use only the first part you wrote, right? And the density will be $nt^{n-1}-n(n-1)(t-s)^{n-2}$? Feb 2 '18 at 1:42
• Ah, right. Then indeed, you just need the first and differentiate. Feb 2 '18 at 5:58
• @GrahamKemp the last case $1\leq s$ and $1 \leq t$ should be $1$, right? Apr 5 '19 at 17:12
• @DanielOrdoñez Good catch. Apr 6 '19 at 0:53

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 $P[A^c]=1-P[A]$.

• "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?) Feb 2 '18 at 0:00
• "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. Feb 2 '18 at 0:01
• What you should have written is Feb 2 '18 at 2:36
• 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...) Feb 2 '18 at 2:46