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Suppose that X,Y, and Z are exponentially distributed with mean 3,4, and 5 respectively. Further assume that they are independent. Find the expected value variance of the max(X,Y,Z).

Answer; E[max(X,Y,Z)] = 7.4641 & Var[max(X,Y,Z)]=24.746

I'm having trouble determining how to get the distribution function for this.

I know that $f_W(x)=nf_X(x)(F_X(x))^{n-1}$

Since all the distributions have different parameters I don't know which one to plug in for $f_X(x)$ if I find the joint distribution I get $f(x,y,z)=\frac{1}{3}e^{\frac{-x}{3}}$ * $\frac{1}{4}e^{\frac{-x}{4}}$ * $\frac{1}{5}e^{\frac{-x}{5}}$ = $\frac{1}{60}e^{\frac{-47x}{60}}$

then $f_W(x)$=3$\frac{1}{60}e^{\frac{-47x}{60}}$ $\left(\frac{x}{60}e^{\frac{-47x}{60}}\right)^{2}$

E[max(X,Y,Z)] = $\int_0^\infty $$f_W(x)$ dx = $\frac{2}{934407}$

I really don't even know how to apply this formula to this problem. Can someone help?

Osuynonma
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  • The actual mean and variance seem to be $\dfrac{176833}{23688}$ and $\dfrac{4628598911}{187040448}$ ā€“ Henry Oct 06 '15 at 06:49

2 Answers2

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I know that $f_W (x)=n\,f_X (x)(F_X (x))^{nāˆ’1}$

That's only the case for identical distributions. For what you have use: $$f_W(w) = f_X(w) F_Y(w)F_Z(w) + F_X(w) f_Y(w)F_Z(w) + F_X(w) F_Y(w)f_Z(w)$$

This is the density where: one of the three equals $w$ and the other two are less.

Graham Kemp
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Because the individual variables are not identically distributed, the calculation of the order statistic is not the same as if they were identically distributed.

Define $W = \max(X,Y,Z)$. Then we want $\operatorname{E}[W]$ and $\operatorname{Var}[W]$. To this end, we should try to calculate the CDF $$F_W(w) = \Pr[W \le w] = \Pr[\max(X,Y,Z) \le w].$$ But we know that if the largest of $X$, $Y$, and $Z$ is at most $w$, then each of $X$, $Y$, and $Z$ is at most $w$. That is to say, $$\Pr[\max(X,Y,Z) \le w] = \Pr[(X \le w) \cap (Y \le w) \cap (Z \le w)] \overset{\text{ind}}{=} \Pr[X \le w]\Pr[Y \le w]\Pr[Z \le w],$$ where in the last equality (with "ind" on top), the equality holds because the random variables are independent. Now we recall that for an exponential distribution with mean $\mu$, the CDF is $$F_X(x) = 1 - e^{-x/\mu}, \quad x \ge 0.$$ So for $X$, $\mu = 3$; for $Y$, $\mu = 4$; and for $Z$, $\mu = 5$. Thus $$F_W(w) = (1 - e^{-w/3})(1-e^{-w/4})(1-e^{-w/5}), \quad w \ge 0.$$ This gives the desired CDF of the maximum order statistic. Now to calculate the expectation, we use the shortcut $$\operatorname{E}[W] = \int_{w = 0}^\infty (1 - F_W(w)) \, dw.$$ And for the variance, which is a bit harder, we use the second moment shortcut $$\operatorname{Var}[W] = \operatorname{E}[X^2] - \operatorname{E}[X]^2 = \int_{w=0}^\infty 2w(1-F_W(w)) \, dw - \operatorname{E}[X]^2.$$ This saves us the extra step of having to differentiate $F_W(w)$ to obtain the density, then re-integrate via the customary formulas $$\operatorname{E}[W^k] = \int_{x=0}^\infty w^k f_W(w) \, dw,$$ etc.

heropup
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