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First, I need to find the function that finds the average chord length between $(t,0)$ and $(x,1)$

Then I can integrate this function to get the average distance of 2 points on opposite sides of a unit square

The distance between $(t,0)$ and $(x,1)$ is $\sqrt{(t-x)^2 +1}$

Thus, the average chord length between $(t,0)$ and $(x,1)$ is defined as

$$C(t) = \int_0^{1} (\sqrt{(t-x)^2 +1} )dx $$

My question(s) are, is this work correct? I integrated $C(t)$ from $0$ to $1$ and I got close to ~1.076

Knocker379
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  • I don't know yet, but its square has to be greater than the average squared distances between two such points, which is given by $\iint_{[0,1]^2}\left(1+(t-x)^2\right)dxdy=\frac72$. So you should check that the result is greater than $\sqrt{\frac72}$ – Adren May 29 '22 at 04:56
  • @Adren your $\frac72$ should be $\frac76$. And shouldn't the inequality go the other way? – Greg Martin May 29 '22 at 06:05
  • I get a similar answer to the OP—seems right. A random sampling also supports that number. – Greg Martin May 29 '22 at 06:06
  • @GregMartin yes 7/6 of course. And yes, the inequality is not the right one. Indeed, if we denote by X the random variable equal to the distance between two random points (each one being randomly chosen on a side), then $E(X^2)\ge E(X)^2$. Thank you for pointing the bugs and sorry for the noise … – Adren May 29 '22 at 06:40
  • Here the exact value, computed by maple : $\iint_{[0,1]^2}\sqrt{1+(t-x)^2},dt,dx=\frac23-\frac{\sqrt2}3+\ln(2)\simeq1.07663$ – Adren May 29 '22 at 06:46
  • @Adren There's a typo in your expression for the integral. Calculating the integral by hand, I get $\ \ln\big(\color{red}{1+\sqrt{2}}\big)+\frac{2}{3}-\frac{\sqrt{2}}{3}\ $, and this evaluates to $\ 1.076635\dots\ $. Your expression evaluates to only $\ 0.8884\dots\ $. – lonza leggiera May 31 '22 at 01:08
  • @Knocker379. The working you've shown is correct, and your approximation to the value of the integral of $\ C(t)\ $ is correct to three decimal places (although the fourth digit should be rounded up to $\ 7\ $). However, you don't give any details of how you obtained your approximation to the integral. Therefore, while your method of evaluation is likely to be correct, I wouldn't be prepared to offer a definite opinion on its correctness. – lonza leggiera May 31 '22 at 01:37

1 Answers1

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Your average distance is given by $$ \iint_{(0,1)^2}\sqrt{1+(x-y)^2}\,dx\,dy < \iint_{(0,1)^2}\left(1+\frac{(x-y)^2}{2}\right)\,dx\,dy=\frac{13}{12}.$$ In exact terms, if $X,Y$ are independent and uniformly distributed over $[0,1]$, the probability density function of $X-Y$ is supported on $[-1,1]$ and given by $f(z)=1-|z|$, so our initial integral equals

$$ \int_{-1}^{1}(1-|z|)\sqrt{1+z^2}\,dz = 2\int_{0}^{1}(1-z)\sqrt{1+z^2}\,dz =2\int_{0}^{\text{arcsinh}(1)}(1-\sinh(u))\cosh^2(u)\,du$$ or $$ \left[u+\frac{1}{2}\sinh(2u)-\frac{2}{3}\cosh^3(u)\right]_{0}^{\text{arcsinh}(1)}=\frac{2-\sqrt{2}}{3}+\text{arcsinh}(1)=\frac{2-\sqrt{2}}{3}+\ln(1+\sqrt{2})\approx 1.076635733.$$

Jack D'Aurizio
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