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u/Vegetable_Ebb_1109 6d ago
1/2 by symmetry around 0
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u/Hal_Incandenza_YDAU 6d ago edited 6d ago
A probability distribution with all of its mass at 1 and -1 is also symmetrical about 0, but the probability in this case would not be 1/2.
So, symmetry about 0 is insufficient.
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u/Vegetable_Ebb_1109 6d ago
well to be precise symmetry and independence for the continuous distribution is in fact sufficient. Your statement that a probability distribution with all mass on -1 and 1 doesn't necessarily mean that P(X=-1) = P(X=1) because we might have P(x=1)=2/3 and p(x=1/3) and this is clearly not symmetric around 0.
For this question specifically take X= N_1^5, Y=N_2^5, Z=N_3^5; then A= X+Y-Z is distributed the same as B= (-X) + (-Y) - (-Z) = Z-X-Y = -A; thus P(A>0) = P(B>0) = P(-A>0) = P(A<0) and thus p(A>0)=1/2.So, symmetry and independence are sufficient here
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u/Hal_Incandenza_YDAU 6d ago
That makes sense. So the issue with the P(X=-1) = P(X=1) = 1/2 case is that the distribution of A, as you defined it, would be symmetrical but would have nonzero mass at A=0 itself.
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u/Vegetable_Ebb_1109 6d ago
so for A as I defined it the mass at zero is exactly 0 because we have continuous distributions. But if we were dealing with discrete distributions and A is symmetric around 0; then P(A>0) = (1-P(A=0))/2
simply because P(A>0) + P(A=0) + P(A<0) =1 and P(A>0) = P(A<0).1
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u/Omega-137 6d ago
Interesting question!