How was this step in integration done?

Question

So there’s a sum, about light bulbs. I have $n$ lightbulbs, and the failure times of each lightbulb follow an exponential distribution with parameter $\theta$. I have to find the expected failure time of the first and last light bulbs. I can see that $X_{(1)}$ is the failure time of the lightbulb that is the first to fail, and $X_{(n)}$ is that of the last one to fail. It turns out, on evaluating the pdf, that $X_{(1)}$ has an exponential distribution with parameter $n\theta$, which easily gives me the expectation as $\frac{1}{n\theta}$. This was easy enough to understand. But, there’s a step given later on, that evaluates the CDF of $X_{(n)}$ $=Y,$say, to be $F_Y(y)=( 1 - e^{-\theta y})^n$ (I know this formula) And then evaluates the expectation as $E[Y] = \int_{0}^\infty{ (1 - F_Y(y) )} dy $ (I don’t understand this step).

Answer 1

For a nonnegative random variable $Y$ there is an equivalent formula for the expectation that can sometimes be easier to compute.

$$E[Y] = \int_0^\infty P(Y \ge y) \, dy$$