Average angle between two randomly chosen vectors in a unit squareConsider two randomly chosen vectors (a,b) and (c,d) within the unit square, where a,b,c, and d are chosen uniformly from [0,1]. What is the expected angle between the vectors?

Question

Siena Bennett · Accepted Answer

Step 1We have       d    x      d    y  =  r      d    r      d    ϕ. Denote the polar coordinates of the two points by       r    1    ,      ϕ    1   and       r    2    ,      ϕ    2  . We need two separate cases according as the points are in the same or different octants.For the same octant, the integral over the region       ϕ    2    &amp;lt;      ϕ    1    &amp;lt;      π    4   is                                          ∫          0                      π            4                                    d                          ϕ          1                          ∫          0                                    ϕ              1                                                d                          ϕ          2                          ∫          0                      1                          cos              ⁡                              ϕ                1                                                                          r                    1                d                  r          1                          ∫          0                      1                          cos              ⁡                              ϕ                2                                                                          r                    2                d                  r          2                          (                      ϕ            1                    −                      ϕ            2                    )                                                  =                              1          4                          ∫          0                      π            4                                    d                          ϕ          1                          1                                    cos              2                        ⁡                          ϕ              1                                                ∫          0                                    ϕ              1                                                d                          ϕ          2                                                    ϕ              1                        −                          ϕ              2                                                          cos              2                        ⁡                          ϕ              2                                                                        =                              1          4                          ∫          0                      π            4                                    d                          ϕ          1                          1                                    cos              2                        ⁡                          ϕ              1                                                            [                          (                              ϕ                1                            −                              ϕ                2                            )                        tan            ⁡                          ϕ              2                        −            log            ⁡            cos            ⁡                          ϕ              2                        ]                    0                                    ϕ              1                                                                        =                    −                  1          4                          ∫          0                      π            4                                    d                          ϕ          1                                      log            ⁡            cos            ⁡                          ϕ              1                                                          cos              2                        ⁡                          ϕ              1                                                                        =                    −                  1          4                                      [            tan            ⁡                          ϕ              1                        (            1            +            log            ⁡            cos            ⁡                          ϕ              1                        )            −                          ϕ              1                        ]                    0                      π            4                                                            =                              π          16                +                  1          8                log        ⁡        2        −                  1          4                        .            Step 2For different octants, the integral over the region       ϕ    2    &amp;lt;      π    4    &amp;lt;      ϕ    1   is                                          ∫                      π            4                                π            2                                    d                          ϕ          1                          ∫          0                      π            4                                    d                          ϕ          2                          ∫          0                      1                          cos              ⁡                              (                                  π                  2                                −                                  ϕ                  1                                )                                                                          r                    1                d                  r          1                          ∫          0                      1                          cos              ⁡                              ϕ                2                                                                          r                    2                d                  r          2                          (                      ϕ            1                    −                      ϕ            2                    )                                                  =                              1          4                          ∫                      π            4                                π            2                                    d                          ϕ          1                          1                                    cos              2                        ⁡                          (                              π                2                            −                              ϕ                1                            )                                                ∫          0                      π            4                                    d                          ϕ          2                                                    ϕ              1                        −                          ϕ              2                                                          cos              2                        ⁡                          ϕ              2                                                                        =                              1          4                          ∫                      π            4                                π            2                                    d                          ϕ          1                          1                                    cos              2                        ⁡                          (                              π                2                            −                              ϕ                1                            )                                                            [                          (                              ϕ                1                            −                              ϕ                2                            )                        tan            ⁡                          ϕ              2                        −            log            ⁡            cos            ⁡                          ϕ              2                        ]                    0                      π            4                                                            =                              1          4                          ∫                      π            4                                π            2                                    d                          ϕ          1                                                    ϕ              1                        −                          π              4                        +                          1              2                        log            ⁡            2                                              cos              2                        ⁡                          (                              π                2                            −                              ϕ                1                            )                                                                        =                              1          4                          ∫          0                      π            4                                    d                ϕ                                            π              4                        +                          1              2                        log            ⁡            2            −            ϕ                                              cos              2                        ⁡            ϕ                                                            =                              1          4                                      [                          (                              π                4                            +                              1                2                            log              ⁡              2              )                        tan            ⁡            ϕ            −                          (              ϕ              tan              ⁡              ϕ              +              log              ⁡              cos              ⁡              ϕ              )                        ]                    0                      π            4                                                            =                              1          4                log        ⁡        2                .            There are 4 symmetric copies of the first region and 2 of the second, for a total of                                           π            4                    +          log          ⁡          2          −          1          ≈          0.4785                        .This is not too different from the value       π    6    ≈  0.5236 if the points are picked from the first quadrant of the unit disk.

Ashlynn Hale · Accepted Answer

Step 1For this sort of problem, the first thing to do is minimize the numbers of variable you need to work with as much as possible. In this case, you can work with the angles of the two vectors directly.Let-       θ    1    =      tan          −      1        ⁡      b    a   and       θ    2    =      tan          −      1        ⁡      d    c  -       θ    m    =  min  (      θ    1    ,      θ    1    ) and       θ    M    =  max  (      θ    1    ,      θ    2    )-   f  (  θ  ) be the common CDF for       θ    1    ,      θ    2  , i.e.  f  (  θ  )  =      CDF                  θ        k              (  θ  )                    =                    d        e        f              P  [      θ    k    ≤  θ  ]     for     k  =  1  ,  2The average angle you want is the expected value of       |        θ    1    −      θ    2        |    =      θ    M    −      θ    m  ,To compute this, we need the CDF for       θ    m   and       θ    M                                CDF                                    θ              m                                      (        θ        )                                                                =                                      d              e              f                                      P        [                  θ          m                ≤        θ        ]        =        P        [                  θ          1                ≤        θ        ∨                  θ          2                ≤        θ        ]                                          =        1        −        P        [                  θ          1                &amp;gt;        θ        ∧                  θ          2                &amp;gt;        θ        ]        =        1        −        P        [                  θ          1                &amp;gt;        θ        ]        P        [                  θ          2                &amp;gt;        θ        ]                                          =        1        −        (        1        −        f        (        θ        )                  )          2                                              CDF                                    θ              M                                      (        θ        )                                                                =                                      d              e              f                                      P        [                  θ          M                ≤        θ        ]        =        P        [                  θ          1                ≤        θ        ∧                  θ          2                ≤        θ        ]        =        P        [                  θ          1                ≤        θ        ]        P        [                  θ          2                ≤        θ        ]                                          =        f        (        θ                  )          2                    Step 2This allows us to express the average angle as an integral                              A                                                    =                                      d              e              f                                      E                              [                                    |                          θ          1                −                  θ          2                          |                                      ]                                              =        E        [                  θ          M                −                  θ          m                ]        =                  ∫          0                      π            2                          θ        (                  CDF                                    θ              M                                ′                −                  CDF                                    θ              m                                ′                )        d        θ                                          =                  ∫          0                                    π              2                                      θ        [                  f          2                (        θ        )        −        1        +        (        f        (        θ        )        −        1                  )          2                          ]          ′                d        θ                                          =        2                  ∫          0                                    π              2                                      θ        [        f        (        θ        )        (        f        (        θ        )        −        1        )                  ]          ′                d        θ            Integrate by part and notice   f  (  θ  )  +  f  (      π    2    −  θ  )  =  1, we find      A    =  2      ∫    0          π              /            2        f  (  θ  )  (  1  −  f  (  θ  )  )  d  θ  =  4      ∫    0          π              /            4        f  (  θ  )  (  1  −  f  (  θ  )  )  d  θIt is easy to see     f  (  θ  )  =      1    2    tan  ⁡  θ   for     θ  ∈  [  0  ,      π    4    ]. Change variable to   t  =  tan  ⁡  θ and integrate, the end result is:      A    =      ∫    0    1              t      (      2      −      t      )              1      +              t        2              d  t  =      π    4    +  log  ⁡  (  2  )  −  1

Consider two randomly chosen vectors (a,b) and (c,d) within the unit square, where a,b,c, and d are chosen uniformly from [0,1]. What is the expected angle between the vectors?

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