Stochastic

Intuitively this holds since xi 1 means one success

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Unformatted text preview: hen X and Y are independent. In the discrete case this is easy: X P (X + Y = z ) = P (X = x)P (Y = z x) (A.16) x To see the first equality, note that if the sum is z then X must take on some value x and Y must be z x. The first equality is valid for any random variables. The second holds since we have supposed X and Y are independent. 215 A.3. EXPECTED VALUE, MOMENTS Example A.16. If X = binomial(n, p) and Y = binomial(m, p) are independent, then X + Y = binomial(n + m, p). Proof by direct computation. P (X + Y = i) = i X ✓n ◆ j =0 j = p (1 p) i = ✓ p (1 j n+m i ◆ n+m i p (1 i p) nj · ✓ m i j ◆ pi j (1 p)m i+j i X ✓n ◆ ✓ m ◆ · j ij j =0 p)n+m i The last equality follows from the fact that if we pick i individuals from a group of n boys and m girls, which can be done in n+m ways, then we must have j i boys and i j girls for some j with 0 j i. Much easier proof. Consider a sequence of n + m independent trials. Let X be the number of successes in the first n trials an...
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This document was uploaded on 03/06/2014 for the course MATH 4740 at Cornell.

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