Expected value for geometric probabilityPoints P = ( X P , Y P ) and Q = ( X Q , Y Q ) were independently chosen from square (−1,0),(0,-1),(1,0),(0,1) with geometric probability. How does one find E | X P − X Q | 2 ?How do you even define expected value here?

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Expected value for geometric probabilityPoints   P  =  (      X    P    ,      Y    P    ) and   Q  =  (      X    Q    ,      Y    Q    ) were independently chosen from square (−1,0),(0,-1),(1,0),(0,1) with geometric probability. How does one find       E              |            X    P    −      X    Q                      |              2       ?How do you even define expected value here?

Jason Petersen · Accepted Answer

Step 1By independence,   A  =  E  (  (      X    P    −      X    Q        )    2    ) is   A  =  E  (      X    P    2    )  +  E  (      X    Q    2    )  −  2  E  (      X    P    )  E  (      X    Q    ). Since       X    P   and       X    Q   are identically distributed,   A  =  2  E  (      X    P    2    )  −  2  E  (      X    P        )    2  .Step 2The density of       X    P   is   f  :  x  ↦  (  1  −      |    x      |        )    +   and f is even hence   E  (      X    P    )  =  0 and   E  (      X    P    2    )  =  2            ∫      0      1              x      2        (    1    −    x    )          d        x    =                  1        6             and   A  =  2  ⋅      1    6    =      1    3  .

Points P=(X_P, Y_P) and Q=(X_Q, Y_Q) were independently chosen from square (-1,0),(0,-1),(1,0),(0,1) with geometric probability. How does one find E∣X_P-X_Q∣^2?

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