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Based on a similar problem in Euclidean metric I believe the expected distance should be given by: $$ \int_0^1 \int_0^1 \int_0^1 \int_0^1 |z-w|+|x-y| dwdzdydx $$ If this is correct, how does one go about solving it though? Should I do it by considering cases?

bajun65537
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  • A key aspect of such problems is the probability distribution of pairs of points. Here you seem to assume the two points are independently and uniformly sampled. It would improve your Question to say so in the problem setup, or to clarify if this is not your assumption. – hardmath Jul 25 '21 at 14:28
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    You can write this as the sum of two double integrals. You can do them by considering cases. – saulspatz Jul 25 '21 at 14:28
  • Could you elaborate on that? What I thought of was something along the line: let $A(x,z)$ and $B(y,w)$. Then we can write the distance as:

    $$\int_0^1\int_0^1\int_0^x\int_z^1 (x-y)+(w-z) dwdydxdz+ \int_0^1\int_0^1\int_0^x\int_0^z (x-y)+(z-w) dwdydxdz+ \int_0^1\int_0^1\int_x^1\int_z^1 (y-x)+(w-z) dwdydxdz+ \int_0^1\int_0^1\int_x^1\int_0^z (y-x)+(z-w) dwdydxdz$$

     – bajun65537 Jul 27 '21 at 16:12

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