Joint CDF and PDF of the maximum and minimum of three i.i.d. Unif(0,1) random variables

Question

I have a question about the following problem from Introduction to Probability.

Problem

Let $U_1, U_2, U_3$ be i.i.d. $\text{Unif}(0, 1)$, and let $L = \min(U_1, U_2, U_3)$, $M =\max(U_1, U_2, U_3)$. (a) Find the marginal CDF and marginal PDF of $M$, and the joint CDF and joint PDF of $L, M$. Hint: for the latter, start by considering $P(L<l,M>m)$.

Attempt at a Solution

Following the hint,

$$P(L<l,M>m) = (1-m)(l)(1)$$

because at least one of the random variables must exceed $m$, which occurs with probability $1-m$, at least one must be less than $l$, which occurs with probability $l$, and the remaining random variable can have any value.

$$P(L>l,M>m) = (1-m)(1-l)(1)$$

because at least one of the random variables must exceed $m$ and of the remaining two at least one must exceed $l$.

Similarly, $$P(L>l,M<m) = (m-l)^3$$ since all the random variables must be between $l$ and $m$.

If these approaches are analogous, why is the expression for $P(L>l,M<m)$ correct but those for $P(L<l,M>m)$ and $P(L>l,M>m)$ incorrect?

I also tried other unsuccessful approaches, for example,

$$P(L>l,M>m) = (1-m)(1-l^2)(1)$$

because at least one of the random variables must exceed $m$ and of the remaining two at least one must exceed $l$ and

$$P(\text{ both of the remaining two RVs }<l) = l^2. $$

so

$$P(\text{ at least one of the remaining two RVs }>l) =1-l^2. $$

Answer 1

NB: assuming that $0<\ell<m<1$.

Following the hint, $$P(L<l,M>m) = (1-m)(l)(1)$$

because at least one of the random variables must exceed m, which occurs with probability 1-m, at least one must be less than l, which occurs with probability l, and the remaining random variable can have any value.

You have to count distinct arrangements of the samples.

So among the three samples you need to measure probabilities that:

• Two are below $\ell$ and one above $m$,
• One is below $\ell$, one above $m$, and one between those values.
• Two are above $m$ and one is below $\ell$.

$\begin{align}{\mathsf P(L<\ell, M>m)}&={\tbinom 32 \ell^2(1-m)+3! \ell(m-\ell)(1-m)+\tbinom 32 \ell(1-m)^2}\\[1ex]&=3\,\ell\,(1-m)\,(\ell+2(m-\ell)+(1-m))\\[1ex]&=3\,\ell\,(1-m)\,(1+m-\ell)\end{align}$

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For the joint pdf, I would suggest adjusting your argument to say "$\{L=\ell,M=m\}$ is any outcome where: the samples are arranged ($3!$ ways) with the least at $\ell$, the greatest at $m$, and the middle anywhere between those two.".

$$\begin{align}f_{\small L,M}(\ell,m)&= 3! f(\ell)\,(F(m)-F(\ell))\,f(m)\\[1ex]&=6 (m-\ell) \end{align}$$

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So by integration we also get...

$$\begin{align}\mathsf P(L<\ell,M>m)&=\int_0^\ell\int_m^1 6\,(y-x)\,\mathrm d y\,\mathrm d x\\[1ex]&= \int_0^\ell 3\,(1^2-m^2)-6\,x\,(1-m)\,\mathrm d x\\&=3\,\ell\,(1-m^2)-3\,\ell^2\,(1-m)\\&=3\,\ell\,(1-m)\,(1+m-\ell) \end{align}$$

Answer 2

(a) General formula for max/min of n i.i.d random variables:

$F_{min(X_{1}, \dots, X_{n})}(x) = 1 - (1 - F_{X_{i}}(x))^n$

$F_{max(X_{1}, \dots, X_{n})}(x) = F_{X_{i}}(x)^n$

$p_{min(X_{1}, \dots, X_{n})}(x) = F_{min(X_{1}, \dots, X_{n})}(x)^{\prime} = n(1 - F_{X_{i}}(x))^{n-1} \cdot p_{X_{i}}(x)$

$p_{max(X_{1}, \dots, X_{n})}(x) = F_{max(X_{1}, \dots, X_{n})}(x)^{\prime} = nF_{X_{i}}(x)^{n-1} \cdot p_{X_{i}}(x)$