Let X and Y be two iid random variables on ( Ω , F , P ). I want to show that X − Y has a symmetric distribution. I know that P X ( A ) = P Y ( A ) for every A ∈ B ( R ). How can I show that P X − Y ( A ) = P Y − X ( A ) for every A ∈ B ( R )? I know that it should be enough to analyze the family of intervals ( ∞ , x ], but I don't see how I can relate P ( Y ≤ x ) = P ( X ≤ x ) to P ( X − Y ≤ x ) = P ( Y − X ≤ x ).

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Let   X and   Y be two iid random variables on   (  Ω  ,      F    ,      P    ). I want to show that   X  −  Y has a symmetric distribution. I know that             P        X    (  A  )  =            P        Y    (  A  ) for every   A  ∈      B    (      R    ). How can I show that             P              X      −      Y        (  A  )  =            P              Y      −      X        (  A  ) for every   A  ∈      B    (      R    )? I know that it should be enough to analyze the family of intervals   (  ∞  ,  x  ], but I don&#039;t see how I can relate       P    (  Y  ≤  x  )  =      P    (  X  ≤  x  ) to       P    (  X  −  Y  ≤  x  )  =  P  (  Y  −  X  ≤  x  ).

robegarj · Accepted Answer

Let       P    X   and       P    Y   denote the distribution of   X and   Y. Then,      P          X      −      Y        (  A  )  =      E        [          1        {    X    −    Y    ∈    A    }    ]    =        ∫          −      ∞              +      ∞            ∫          −      ∞              +      ∞            1    {  x  −  y  ∈  A  }  d      P    X    (  x  )  d      P    Y    (  y  )Since       P    X   and       P    Y   are equal, we can write      P          X      −      Y        (  A  )  =      ∫          −      ∞              +      ∞            ∫          −      ∞              +      ∞            1    {  x  −  y  ∈  A  }  d      P    Y    (  x  )  d      P    X    (  y  ). Then, by the change of variable   x  ←  y and   y  ←  x, we have      P          X      −      Y        (  A  )  =      ∫          −      ∞              +      ∞            ∫          −      ∞              +      ∞            1    {  y  −  x  ∈  A  }  d      P    Y    (  y  )  d      P    X    (  x  )    =      P          Y      −      X        (  A  )

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