Finding the PDF of a Second Order Statistic Suppose that $Y_1,Y_2 ~ i.i.d$ Expo(b) 
Find the pdf of the second order statistic, $U_2 =max(Y_1,Y_2)$
$F_{U_2}(u)= P(U_2<u) = P\{Max(Y_1,Y_2)<u\}= P(Y_1<u,Y_2<u)=P(Y_1<u)P(Y_2<u)= P(Y_1<u)^2 =(1-e^{-u/b})^2$
Then you do some differentiation and get the answer.
The problem is that I am confused about how we go from $P\{Max(Y_1,Y_2)<u\}$ to $P(Y_1<u,Y_2<u)$ and from $P(Y_1<u)P(Y_2<u)$ to $P(Y_1<u)^2$ 
could someone please explain?
 A: Distribution of the maximum of two exponential random variables.
With the help of Comments by @Callculus and @JusttoAnswer, I hope the
derivation of the CDF of $W = \max(Y_1,Y_2)$ is clear, and that
you have PDF $f_W(u) = (2/b)(1 - \exp(-u/b))\exp(-u/b),$ for $u > 0.$
This is an important relationship in applied probability theory,
particularly in reliability and queueing applications.
From the PDF, you can find that $E(W) = b/2 + b.$ Here is an intuitive rationale
for the mean in two steps:
(1) The first order statistic is $V = \min(Y_1, Y_2) \sim Expo(b/2),$
which has mean $E(V) = b/2.$ That makes sense intuitively. If a system
consists of two components of exponential lifetime connected in series, then the system fails
when the first component fails. With two similar components at risk, the
time to failure should be half as long as with only one at risk.
(2) After the first component fails, we wait for the second to fail in order
to get the average lifetime of the second order statistic. But, by the
no-memory property of exponential distributions,  the second component
doesn't 'remember' it has been going for time $V$ already, so its additional
lifetime averages $b$.  Hence, $E(W) = b/2 + b = 1.5b.$
When two components are connected in parallel, the average lifetime of the parallel
system is longer $W$ than the average lifetime $Y$ of a single component. 
Sometimes communications satellites fail because the CPU of the onboard
computer gets zapped by a cosmic ray. So the exponential distribution is
a good model for lifetimes. If two CPUs are connected in parallel, so
that the second can take over when the first gets zapped, then the average
lifetime of the computer may be extended from $b$ years to $1.5b$ years.
It is easy to illustrate the distribution of $W$ using R statistical software.
In R, the parameter for an exponential distribution is the failure rate $1/b$ instead of the average lifetime $b.$ In the simulation, I let $b = 5.$
 m = 10^6;  b = 5     # do a million iterations with means 5
 y1 = rexp(m, 1/b);  y2 = rexp(m, 1/b)
 w = pmax(y1, y2)     # vector of a million maximums
 mean(w)
 ## 7.505647          # aprx E(W) = 7.5

 hist(w, prob=T, br=50, col="skyblue2", ylim=c(0,.1))
 curve((2/b)*(1 - exp(-x/b))*exp(-x/b), lwd=2, add=T)

The histogram below, based on a million realizations of $W$,
approximates the PDF of $W$. The black curve is just a graph of the PDF of $W$,
which you derived.
 
