Three uniform random variables $X = [2.9,3.1]$, $Y = [2.7,3.1]$, $Z = [2.9,3.3]$. What is the expected value of the maximum of these three variables? $E(\max(X,Y,Z))$.
I have tried to split the problem into three parts P$(X \, \text{is the max})\cdot X$, $P(Y \, \text{is the max})\cdot Y$ and $P(Z \, \text{is the max})\cdot Z$ but I have not been able to figure out the correct implementation of this logic.
(Rather stunningly, this was downvoted. Too much mathematical precision, perhaps? :-) Happy reading!)
First observation: The random variables $X$, $Y$ and $Z$ are independent and uniform on $[2.9,3.1]$, $[2.7,3.1]$ and $[2.9,3.3]$ respectively hence $$X=2.9+0.2U,\qquad Y=2.9+0.2R,\qquad Z=2.9+0.2S,$$ where $U$, $R$ and $S$ are independent and uniform on $(0,1)$, $(-1,1)$ and $(0,2)$ respectively.
Consequence: $E(\max(X,Y,Z))=2.9+0.2\cdot E(\max(U,R,S))$.
Second observation: In turn, $$R=V-A,\qquad S=W+B,$$ where $V$ and $W$ are uniform on $(0,1)$, $A$ and $B$ are Bernoulli random variables uniform on $\{0,1\}$, and all the random variables are again independent.
Consequence: $E(\max(U,R,S))=E(T)$ with $T=\max(U,V-A,W+B)$.
Third observation: If $B=1$, then $T=W+1$. If $B=0$ and $A=1$, then $T=\max(U,W)$. If $B=0$ and $A=0$, then $T=\max(U,V,W)$.
Consequence: $E(T)=\frac12E(W+1)+\frac14E(\max(U,W))+\frac14E(\max(U,V,W))$.
Fourth observation: For every positive integer $n$, if $(U_1,\ldots,U_n)$ are i.i.d. uniform on $(0,1)$, then $M_n=\max(U_1,\ldots,U_n)$ has expectation $$E(M_n)=\int_0^1(1-P(M_n\leqslant x))dx=\int_0^1(1-x^n)dx=\frac{n}{n+1}.$$ Consequence: $E(W)=\frac12$, $E(\max(U,W))=\frac23$, $E(\max(U,V,W))=\frac34$.
Conclusion: $$E(\max(X,Y,Z))=2.9+0.2\left(\tfrac12\left(\tfrac12+1\right)+\tfrac14\left(\tfrac23\right)+\tfrac14\left(\tfrac34\right)\right).$$
Yes, assuming that $X,Y,Z$ are mutually independent, then by the law of iterated expectation:
$$\begin{align}\mathsf E(\max(X,Y,Z)) & = \mathsf E(X\mid X>Y, X>Z)\mathsf P(X>Y, X>Z) \\ & +\mathsf E(Y\mid Y>X, Y>Z)\mathsf P(Y>X, Y>Z) \\ & +\mathsf E(Z\mid Z>X, Z>Y)\mathsf P(Z>X, Z>Y)\end{align}$$
Now $$\begin{align} \mathsf E(X\mid X>Y, X>Z)\mathsf P(X>Y, X>Z) & = \int_{2.9}^{3.1} \frac{x}{3.1-2.9}\int_{2.7}^{x}\frac{1}{3.1-2.7}\operatorname d y\int_{2.9}^{x} \frac{1}{3.3-2.9}\operatorname d z\operatorname d x \\[1ex] & = \frac{125}{4}\int_{2.9}^{3.1} x(x-2.7)(x-2.9)\operatorname d x \\[1ex] & = \frac{19}{30} \end{align}$$
Can you do the rest? (Be wary of the lower interval for $Y$ and upper interval for $Z$.)
Here’s another solution, basically the same as already given, but perhaps easier to follow, and at least more colorful.
The ranges over which $\color{blue}X$, $\color{brown}Y$, and $\color{green}Z$ are uniformly distributed are shown above. Partition what can happen into the following three disjoint cases.
Case 1 ($p=\frac{1}{2}$): $\color{green}{Z\ge3.1}$.
In this case, $E(\max(\color{blue}X,\color{brown}Y,\color{green}Z))=E(\color{green}Z)=3.2=\frac{16}{5}$.
Case 2 ($p=\frac{1}{4}$): $\color{green}{Z\le3.1}$ and $\color{brown}{Y\le2.9}$.
Here, $E(\max(\color{blue}X,\color{brown}Y,\color{green}Z))=E(\max(\color{blue}X,\color{green}Z))$, which equals* $2.9+0.2\cdot\frac{2}{3}=\frac{91}{30}$, because in this case, $\color{blue}X$ and $\color{green}{Z}$ are uniform on $[2.9,3.1]$.
Case 3 ($p=\frac{1}{4}$): $\color{green}{Z\le3.1}$ and $\color{brown}{Y\ge2.9}$.
Here, $\color{blue}X$, $Y$, and $\color{green}Z$ are all uniform on $[2.9,3.1]$, so* $E(\max(\color{blue}X,\color{orange}{Y},\color{green}{Z}))=2.9+0.2\cdot\frac{3}{4}=\frac{61}{20}$.
Therefore $E(\max(\color{blue}X,\color{brown}{Y},\color{green}Z))=\frac{1}{2}\cdot\frac{16}{5}+\frac{1}{4}\cdot\frac{91}{30}+\frac{1}{4}\cdot\frac{61}{20}=\frac{749}{240}$
*The expected value of the maximum of $k$ uniform random variables on $[0,1]$ is $\frac{k}{k+1}$: This is because the $k$ values divide $[0,1]$ into $k+1$ intervals, and the expected length of each (and in particular, the rightmost one) is $\frac{1}{k+1}$.
