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Distribution of Sum of Two Weird Random Variables....
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[QUOTE="mjc123, post: 6621268, member: 610180"] For a given value x = X, What is the probability that z has a value Z? What is the probability that y has the value Y = (X -a -bZ)/c? Integrate over z: f(x=X) = ∫f(z=Z)f(Y=(X-a-bZ)/c)dZ (Assuming z and y are distributed independently. If not, you have to use a conditional probability for Y.) Note that the range of z and y may be limited to less than their full possible range, e.g. if z is normal, it can take negative values, but if y is exponential it can only be positive (or zero). Therefore Z is limited to values for which X - a - bZ is nonnegative, ie Z ≤ (X - a)/b. (That's if b and c are both positive, work it out for yourself for other cases.) [/QUOTE]
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Distribution of Sum of Two Weird Random Variables....
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