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On each side of a square with unit length sides choose a point uniformly. Connect these points to form a quadrilateral. What is the expected perimeter of this quadrilateral?

I am able to simulate and also solve the integral to get the expected area of $\frac12$.

$$1-\frac12\int_{d=0}^{1}\int_{c=0}^{1}\int_{b=0}^{1}\int_{a=0}^{1}a(1-b)+b(1-c)+c(1-d)+d(1-a) da db dc dd=\frac12$$

The corresponding integral for the perimeter does not appear to have a straightforward solution but simulation shows the expected value to be somewhere around $3.06$.

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  • $\begingroup$ The expected area is also solved here in this question. $\endgroup$ – Goldbug Dec 11 '19 at 15:15
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The expected length of a side is $$\int_0^1 \int_0^1 \sqrt{x^2+y^2}\ dx \ dy$$ I was able to get Alpha do do the two integrals, the $x$ one giving $$\frac 14 \left(2 (\sqrt{y^2 + 1} + y^2 \log(\sqrt{y^2 + 1} + 1)) - y^2 \log(y^2)\right)$$ and the second giving $$\frac 13 (\sqrt2 + \sinh^{-1}(1))≈0.76520$$ The expected perimeter is $4$ times this, or about $3.0608$

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  • $\begingroup$ I find it surprising that we can treat each side independently as it would seem the lengths of the sides are correlated. However, your solution does appear to yield the correct answer. Why can we treat the sides independently? $\endgroup$ – Goldbug Dec 11 '19 at 15:48
  • $\begingroup$ It is the linearity of expectation. That works even if the variables are correlated. $\endgroup$ – Ross Millikan Dec 11 '19 at 15:49
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Your vertices lie on $(a,0),(1,b),(c,1),(0,d)$ hence by the Pythagorean Theorem, the perimeter is $$ P(a,b,c,d) = \sqrt{(a-1)^2+b^2} + \sqrt{(b-1)^2+(c-1)^2} + \sqrt{c^2+(d-1)^2} + \sqrt{a^2+d^2} $$ and you need to compute $\mathbb{E}[P]$ over a joint distribution of 4 independent uniforms, i.e. $$ \int_{[0,1]^4} P(a,b,c,d) da\ db\ dc\ dd \approx 3.0676, $$ with the approximation provided by Wolfram Alpha.

To be more precise, note that each of the roots basically yield the same result (since $1-U$ and $U$ have the same distribution if $U \sim \mathcal{U}(0,1)$). So we get $$ \mathbb{E}[P] = 4\int_0^1 \int_0^1 \sqrt{a^2 + d^2}\ da\ dd = \frac{4}{3} \left[ \sqrt2 + \sinh^{-1}(1) \right] = \frac{4}{3} \left[ \sqrt2 + \ln \left(1 + \sqrt2 \right) \right] $$


With that approach, area becomes simple as well. Notice that the area of your quadrilateral is exactly the area of the unit square less areas of 4 right triangles, each of which has legs sides, which are uniformly distributed and hence has an expected area of $$ \int_0^1 \int_0^1 \frac{ad}{2} \ da\ dd = \frac12 \left(\int_0^1 a\ da\right) \left(\int_0^1 d\ dd\right) = \frac18, $$ so the total area is $A = 1 - \frac48= \frac12$ as you computed.

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