Expected Value of the Maximum of 3 Independent Exponential Random Variables

Question

I have 3 independent exponential random variables, X, Y, Z, with lambda parameters λ = 1, 2, 3 respectively.

I need to find E[max{X,Y,Z}] and I'm not exactly sure how to approach this.

Let max{X,Y,Z} = T.

I have found that the CDF for max{X,Y,Z} is:

$ F(t) = P(T \leq t) = (1-e^{-t})(1-e^{-2t})(1-e^{-3t}) = 1 - e^{-t} - e^{-2t} + e^{-4t} +e^{-5t} - e^{-6t}$

I am thinking that the expectation can be found as follows:

$\int_{0}^{\infty} tf(t),dt$ where f(t) is the PDF found by deriving the CDF. However, when I do this I find that I get a very complicated integral which does not compute nicely.

When looking online I found that people calculate calculate the expectation as $E[T] = \int_{0}^{\infty} P(T>t,dt)$. I don't understand why this is correct - could someone please explain?

I also know that I could write the max{X,Y,Z} in terms of a minimum as the minimum of exponential random variables is also exponential. However, I'm not sure how to approach this with 3 random variables.

Thanks

Answer 1

Maybe it helps your intuition if you consider the discrete case to understand $$ E[T] =\int_0^\infty P(T > t) dt $$ Assume T is discrete and nonnegative. Then one can rewrite the above as: \begin{align} E[T] &= \sum_{t=0}^\infty P(T > t) \\ &= \sum_{t=0}^\infty \sum_{k=t+1}^\infty P(T = k) \\ &= [P(T = 1) + P(T= 2) + P(T= 3) + \dots] + [P(T= 2) + P(T= 3) + \dots] + [P(T= 3) + \dots] + \dots \\ &= 1 \cdot P(T=1) + 2 \cdot P(T=2) + \dots \\ &= \sum_{t=0}^\infty t \cdot P(T=t) = E[T] \end{align}

Answer 2

Using the follow defintion of mean you get that

$$\mathbb{E}[T]=\int_0^{\infty}\left[ 1-F_T(t)\right]dt$$

thus your integral is immediate being

$$\mathbb{E}[T]=\int_0^{\infty}\left[ e^{-t}+e^{-2t}-e^{-4t}-e^{-5t}+e^{-6t}\right]dt=1+\frac{1}{2}-\frac{1}{4}-\frac{1}{5}+\frac{1}{6}$$

no calculations are needed to solve the integral once you realize that

$$\int_0^{\infty} \theta e^{-\theta x}dx=1$$

thus, as an example

$$\int_0^{\infty}e^{-kx}dx=\frac{1}{k} \int_0^{\infty}k e^{-k x}dx=\frac{1}{k}$$