Expected value of the Max of three exponential random variables

Question

So the question asks:

Let $X_1,X_2,X_3\sim \operatorname{Exp}(\lambda)$ be independent (exponential) random variables (with $\lambda> 0$).
(a) Find the probability density function of the random variable $Z = \max \{X_1,X_2,X_3\}$.
(b) Let $T = X_1+X_2/2+X_3/3$, use moment generating functions to prove $Z\sim T$ (same distribution). Find $E[Z]$ and $\operatorname{Var}[Z]$.

So far I got:

(a)$F(x) = 1-e^{-\lambda x}$

$F_Z(z) = P (Z \leq z) = P(\max(X_1,X_2,X_3) ≤ z) = P(X_1\leq z, X_2 \leq z, X_3 \leq z)= P(X_1\leq z)P(X_2\leq z) P(X_3\leq z) = (1-e^{-\lambda z})^3$

$f_Z(z) = F_Z'(z) = (1-e^{-\lambda z})^3 =3\lambda e^{-3\lambda z}(e^{\lambda z}-1)^2$

(b) for this part, I did not quite understand what it wanted me to prove actually...

I got: $M_X(t) = λ/(λ-t )$

$M_Z(t) = M_{X_1}(t)M_{X_2}(t) M_{X_3}(t) = [\lambda/(\lambda-t )] [\lambda /(2(\lambda-t) ] [\lambda /(3(\lambda-t) ]= [\lambda/(\lambda-t)]^3/6$

So what does it mean by proving $Z\sim T$ (same distribution) ?

And for the $E[Z]$ and $\text{Var} [Z]$, I actually tried to do it using the standard method which is $$ E[Z]=\int z\cdot3\lambda e^{-3\lambda z}(e^{\lambda z}-1)^2dz $$

which becomes super complicated...

So is there a simple way to calculate the $E[Z]$ and $\text{Var} [Z]$ without literally solving the integration?

Answer 1

(b) Just means use MGFs to show that $Z$ and $T$ have the same distribution.

So is there a simple way to calculate the $E[Z]$ and $\text{Var} [Z]$ without literally solving the integration?

Yes. Intuitively, if you have $3$ light bulbs with lifetimes that are iid exponential $\lambda$, then $Z$ is the time until the last light bulb burns out. In other words, we wait for the first of three to burn out ($Y_3$), then we wait for the first of two to burn out $(Y_2)$, then we finally wait for the last to burn out $(Y_1)$.

Notice (prove for yourself), that the minimum of $n$ iid exponential random variables follows an exponential distribution with mean $1/(n\lambda)$. Hence, we have

$$E[Z] =E[Y_3+Y_2+Y_1] = E[Y_3]+E[Y_2]+E[Y_1] = \frac{1}{3\lambda}+\frac{1}{2\lambda}+\frac{1}{\lambda} = \frac{1}{\lambda}\sum_{k=1}^3\frac{1}{k}.$$

Use independence and argue similarly for the variance.

In fact, this is the exact same thing as the $T$ they describe.; $$X_1+X_2/2+X_3/3 \overset{d}{=} Y_3+Y_2+Y_1.$$

Answer 2

First you would look at the distribution of $Z$. Notice that the moment generating function for $Z$ is equal to $$ \int_0^{\infty } 3 \lambda e^{t x} e^{-3 \lambda x} \left(e^{\lambda x}-1\right)^2 \, dx=\int_0^{\infty } 3 \lambda e^{t x} e^{-3 \lambda x} e^{2\lambda x} \, dx - \int_0^{\infty } 3 \lambda e^{t x} e^{-3 \lambda x} e^{\lambda x} \, dx + \int_0^{\infty } 3 \lambda e^{t x} e^{-3 \lambda x} \, dx $$ and the right hand side terms would correspondingly be equal to $$ \int_0^{\infty } 3 \lambda e^{t x} e^{-3 \lambda x} \left(e^{\lambda x}-1\right)^2 \, dx=\frac{3 \lambda }{\lambda -t} + -\frac{6 \lambda }{2 \lambda -t} + \frac{3 \lambda }{3 \lambda -t} $$ which would simplify as $$ \int_0^{\infty } 3 \lambda e^{t x} e^{-3 \lambda x} \left(e^{\lambda x}-1\right)^2 \, dx=\frac{6 \lambda ^3}{(\lambda -t) (2 \lambda -t) (3 \lambda -t)}. $$ Therefore, the moment generating function of $T$ and the moment generating function of $Z$ are equal to each other, which would imply that they have the same distribution.

To find the expectations, notice that $$ E[X^n]=\left. \frac{d^n M(t)}{dt^n}\right|_{t=0} $$