$E[X]$ derivation in Coupon Collecting Problem in Ross Probability Models

Question

In Ross Probability Models 11th edition on page 307, the following is written (in picture link). Coupon Collecting $E[X]$ derivation. $X=\max(X_j)$ is the time at which we obtain the collection of the coupons, and $X_{j}$ is independent exponential. I am not following why $E[X]=\int_{0}^{\infty} P(X \ge t)dt$?

Answer 1

This isn't specific to this particular random variable $X$. It's a general expression for the expected value of a non-negative random variable that you can derive from the corresponding integral with the density using integration by parts:

$$ E[X]=\int_0^\infty f_X(t)t\mathrm dt=\left[-P(X\ge t)t\right]_0^{\infty}+\int_0^\infty P(X\ge t)\mathrm dt\;. $$

Answer 2

Let $(\Omega,\mathcal F,\mathbb P)$ be a probability space and $X:\Omega\to\mathbb R$ a random variable with $\mathbb P(X\geqslant0)=1$. The map $(\omega,x)\mapsto \mathsf 1_{\{X(\omega)>x\}}$ on $\Omega\times[0,\infty)$ is $\mathcal F\otimes\mathcal B(\mathbb R)$-measurable and is nonnegative. The measure $\mathbb P\otimes m$ ($m$ denoting Lebesgue measure) on $\Omega\times[0,\infty)$ is $\sigma$-finite, and hence by Tonelli's theorem the integral $$ \iint_{\Omega\times[0,\infty)} \mathsf 1_{\{X(\omega)>x\}} \ \mathsf d(\mathbb P\times x) $$ exists and is equal to the iterated integrals in the sequel. One has \begin{align} \int_\Omega\int_{[0,\infty)}\mathsf 1_{\{X(\omega)>x\}}\ \mathsf dx\ \mathsf d\mathbb P(\omega) &= \int_\Omega\int_0^{X(\omega)}\ \mathsf dx\ \mathsf d\mathbb P(\omega)\\ &= \int_\Omega X(\omega)\ \mathsf d\mathbb P(\omega)\\ &= \mathbb E[X] \end{align} by definition of expectation, and further \begin{align} \int_{[0,\infty)}\int_\Omega \mathsf 1_{\{X(\omega)>x\}}\ \mathsf d\mathbb P(\omega)\ \mathsf dx &= \int_{[0,\infty)}\mathbb P(\omega\in\Omega:X(\omega)>x)\ \mathsf dx\\ &=\int_0^\infty \mathbb P(X>x)\ \mathsf dx. \end{align} Hence, $$ \mathbb E[X] = \int_0^\infty \mathbb P(X>x)\ \mathsf dx. $$