Simpler way to compute $E[Z]$ where $Z = \min(A,B)$ and $A \sim U(0, 5)$, $B \sim U(0,10)$

Question

I can compute it the following way:

\begin{align} E[Z] = E[Z | A < B]P(A < B) + E[Z | B < A]P(B < A) \\ P(A < B) = 0.75 \\ P(B < A) = 0.25 \\ E[Z | A < B] = E[A | A < B] \\ = \int_{0}^5 \int_{a}^{5} a\frac{1}{37.5} dbda \\ + \int_{0}^5 \int_{5}^{10} a\frac{1}{37.5} dbda \\ = 2\frac{2}{9} \\ E[Z | B < A] = \int_0^5 \int_{0}^a b \frac{1}{12.5} dbda \\ = \int_0^5 \frac{a^2}{2} da \\ = \frac{5}{3} \\ \therefore E[Z] = 2\frac{2}{9} * 0.75 + \frac{5}{3}*0.25 = 2 \frac{1}{12} \end{align}

I am wondering if there is a simpler approach based on symmetry? This problem seems to be nicely structured that there should be a very simple symmetry argument, but I'm not sure what it is.

Answer 1

This is not a symmetry argument, but it seems slightly simpler:

$$E[Z]=\int_0^\infty P(Z>t)\,dt = \int_0^\infty P(A>t\text{ and }B>t)\,dt=\int_0^\infty P(A>t)P(B>t)\,dt.$$ But $$P(A>t)=\begin{cases}1-\frac t 5& \text{if } 0\le t\le 5\\ 0&\text{if }t>5;\end{cases}$$ and similarly $$P(B>t)=\begin{cases}1-\frac t{10}& \text{if } 0\le t\le 10\\ 0&\text{if }t>10.\end{cases}$$ So $$E[Z] = \int_0^5 (1-\frac t 5)(1-\frac t{10})\,dt = \frac{25}{12}.$$

Answer 2

Yes, there is a simpler argument based on symmetry, if you are comfortable believing that if $X$ is uniform in [0,10], then conditioned on $X < 5$, $X$ is uniform on $[0,5]$. (You can easily show this with Bayes rule if you want)

So now, just compute

$$E[Z] = E[Z | B > 5] P(B > 5) + E[Z | B < 5] P(B < 5)$$

Note that $E[Z | B > 5] = E[A | B > 5] = 2.5$

and $E[Z | B < 5] = 5/3$, which is the minimum of 2 uniform [0,5] random variables. (There is a symmetry argument for that as well, which essentially reduces down to picking three points on a circle, and labeling one as a "cutting point", one as "A" and the other as "B". You can also just use CDFs to show it)

Now, you can combine all of these to get $$ 2.5 \times \frac{1}{2} + \frac{5}{3} \frac{1}{2} = 2 \frac{1}{12} $$ as you wanted.

Answer 3

$$\begin{align} \operatorname{E}[Z] &= \frac{1}{50} \int_{a=0}^5 \int_{b=0}^{10} \min(a,b) \, db \, da \\ &= \frac{1}{50} \left( \int_{a=0}^5 \int_{b=0}^a b \, db \, da + \int_{a=0}^5 \int_{b=a}^{10} a \, db \, da \right) \\ &= \frac{1}{50} \left(\int_{a=0}^5 \frac{a^2}{2} \, da + \int_{a=0}^5 a (10-a) \, da \right) \\ &= \frac{1}{50} \left( \frac{125}{6} + 125 - \frac{125}{3} \right) \\ &= \frac{25}{12}. \end{align}$$

By symmetry:

Let $C = B \mid B \le 5 \sim \operatorname{Uniform}(0,5)$. Then

$$\begin{align} \operatorname{E}[Z] &= \operatorname{E}[Z \mid B \le 5]\Pr[B \le 5] + \operatorname{E}[Z \mid B > 5]\Pr[B > 5] \\ &= \frac{1}{2} \left(\operatorname{E}[\min(A,C)] + \operatorname{E}[A] \right) \\ &= \frac{1}{2} \left(\operatorname{E}[A \mid A \le C]\Pr[A \le C] + \operatorname{E}[C \mid A > C]\Pr[A > C] + \frac{5}{2} \right) \\ &= \frac{\operatorname{E}[A \mid A \le C]}{2} + \frac{5}{4} \\ &= \frac{1}{2} \int_{c=0}^5 \int_{a=0}^c \frac{2a}{25} \, da \, dc + \frac{5}{4} \\ &= \frac{1}{50} \int_{c=0}^5 c^2 \, dc + \frac{5}{4} \\ &= \frac{5}{6} + \frac{5}{4} \\ &= \frac{25}{12}. \end{align}$$ I wouldn't consider the second approach any easier than the first.