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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?

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  • $\begingroup$ 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 $\endgroup$
    – kal_elk122
    Oct 6, 2022 at 13:28

1 Answer 1

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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 $$4\int_0^{1/2}\int_0^{1/2}xy(1-x)\,dy\,dx=\frac1{24}$$

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  • $\begingroup$ 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? $\endgroup$
    – kal_elk122
    Oct 6, 2022 at 13:44
  • $\begingroup$ 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! $\endgroup$
    – kal_elk122
    Oct 6, 2022 at 14:06
  • $\begingroup$ 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. $\endgroup$
    – kal_elk122
    Oct 6, 2022 at 14:12
  • $\begingroup$ I think i understand how - imgur.com/a/nHsVmMW - is this formalism accurate to your solution? $\endgroup$
    – kal_elk122
    Oct 6, 2022 at 14:18
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    $\begingroup$ @Lonimous Yes.. $\endgroup$ Oct 6, 2022 at 14:25

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