Expectation of area of rectangle built from two random variables

The question:

A 1 meter long stick is broken at a random point chosen uniformly along its length, the short part $$X$$ is used as the length of a side of the rectangle. The longer part of the stick is broken again at a random point uniformly chosen along it’s length - the shorter part of the second breaking $$Y$$ is used as the length of a second side of the same rectangle. find the expectation of the rectangle’s area

my attempt so far: $$X = \text{The shorter part of the broken stick, chosen uniformly so I can say it’s length is } \sim U_{[0,\tfrac12]}$$ $$Y = \text{The shorter part of the broken longer stick. I know the longer stick’s length is }1+x \text{ so I can say the length of Y is } Y|X\sim U_{[0,\frac{1+x}{2}]}$$ (because Y’s the smallest part of a stick with length 1-x so it can at most be the midpoint). \begin{aligned} E(S_\square) &= E(XY) = E(E(XY|X)) = E(XE(X|Y)) = E\left(X\left(\frac{\frac{(1+X)-0}{2}}{2}\right)\right) \\ &=0.25E(X)+0.25E(X^2) = 0.25(0.5)+0.25(\textsf{var}(X)+E(X)^2) \\ &=0.125+0.25(1/12 +0.25) \\ &=\frac{5}{24}\end{aligned} The real answer is $$\frac{1}{24}$$ - I don’t see what I got wrong. Is my understanding and definition of the random variables wrong?

• i found my error - if I’m going by lengths I should’ve defined Y as $U \sim[0,\frac{1-x}{2}]$ since Y can only reach 1/2 of the longer stick. I wish to see if there’s a different- more elegant way to define the random variables though Oct 6, 2022 at 13:28

Here is a more concise approach. The distribution of the rectangle's area is $$\min(X,1-X)(\min(Y,1-Y)\max(X,1-X))$$ where $$X,Y$$ are iid $$U(0,1)$$ random variables. Since this function in $$X$$ and $$Y$$ is symmetric about the lines $$X=\frac12$$ and $$Y=\frac12$$ and the joint distribution of $$X,Y$$ is $$1$$ over the support, the expected area is $$4\int_0^{1/2}\int_0^{1/2}xy(1-x)\,dy\,dx=\frac1{24}$$

• why is there a max(X,1-X) in the distribution of the rectangle’s area? can you please explain how you got the expression for the distribution? Oct 6, 2022 at 13:44
• Oh I understand now! because you took Y from 0 to 1, you had to multiply it by the remaining longer stick’s length: max(x,1-x) to get the correct length! Oct 6, 2022 at 14:06
• can you explain to me though how do we get from $E(XY)$ to your integral? I understand we’re integrating over $min(X,X-1) \cdot (min(Y,1-Y)max(X,1-X))$ which we then convert to your integral with the times four due to symmetry twice, but i’m struggling with getting from E(XY) to the integral with the pdf. Oct 6, 2022 at 14:12
• I think i understand how - imgur.com/a/nHsVmMW - is this formalism accurate to your solution? Oct 6, 2022 at 14:18
• @Lonimous Yes.. Oct 6, 2022 at 14:25