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Math Help - Bound probability of sum of two variables

  1. #1
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    Bound probability of sum of two variables

    If x1 and x2 are independent, Pr(a<x1<b)=p, Pr(c<x2<d)=p, and y=x1+x2. can we prove that Pr(a+c<y<b+d)>= p? I have proved it if x1 and x2 are normally distributed. Is this also true for general distributions?
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    Re: Bound probability of sum of two variables

    Hello,

    Consider A=\{a<X_1<b~,~c<X_2<d\} and B=\{a+c<Y<b+d\}

    We have that P(A)=P(a<X_1<b)P(c<X_2<d)=p^2 (by independence).


    We also know that \begin{cases} a<X_1<b \\ c<X_2<d \end{cases}\Rightarrow a+c<Y<b+d

    Then recall that if we have an implication A => B, then P(A)\leq P(B)

    This lets us write that Pr(a+c<y<b+d)>= pē
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    Re: Bound probability of sum of two variables

    Pr(a+c<y<b+d)>= pē is too conservative because pē<p. It can be shown that if x1 and x2 are independently and normally distributed, Pr(a+c<y<b+d)>= p. I am wondering if this is also true for other distributions.
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  4. #4
    Guy
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    Re: Bound probability of sum of two variables

    Quote Originally Posted by g322 View Post
    Pr(a+c<y<b+d)>= pē is too conservative because pē<p. It can be shown that if x1 and x2 are independently and normally distributed, Pr(a+c<y<b+d)>= p. I am wondering if this is also true for other distributions.
    False in general. Take X, Y to be independent exponential rvs and a = c = 0, b = d = 1.

    Moo's work suggests to me that his bound is the best one can expect without more assumptions.
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  5. #5
    Moo
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    Re: Bound probability of sum of two variables

    Quote Originally Posted by Guy View Post
    Moo's work suggests to me that his bound is the best one can expect without more assumptions.
    There is no doubt about it ! It's my work your talking about, constantly skimming perfection and exactness.

    (hey, I'm just kidding)
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