{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

MoreIndRVs+ConditionalDist-4

# MoreIndRVs+ConditionalDist-4 - Sum of Independent Binomial...

This preview shows pages 1–3. Sign up to view the full content.

1 Sum of Independent Binomial RVs Let X and Y be independent random variables X ~ Bin(n 1 , p) and Y ~ Bin(n 2 , p) X + Y ~ Bin(n 1 + n 2 , p) Intuition: X has n 1 trials and Y has n 2 trials o Each trial has same “success” probability p Define Z to be n 1 + n 2 trials, each with success prob. p Z ~ Bin(n 1 + n 2 , p), and also Z = X + Y More generally: X i ~ Bin(n i , p) for 1 i N p n X N i i n i i , Bin ~ 1 1 Sum of Independent Poisson RVs Let X and Y be independent random variables X ~ Poi( l 1 ) and Y ~ Poi( l 2 ) X + Y ~ Poi( l 1 + l 2 ) Proof: (just for reference) Rewrite (X + Y = n ) as (X = k , Y = n k ) where 0 k n Noting Binomial theorem: so, X + Y = n ~ Poi( l 1 + l 2 ) n k n k k n Y P k X P k n Y k X P n Y X P 0 0 ) ( ) ( ) , ( ) ( n k k n k n k k n k n k k n k k n k n n e k n k e k n e k e 0 2 1 ) ( 0 2 1 ) ( 0 2 1 )! ( ! ! ! )! ( ! )! ( ! 2 1 2 1 2 1 l l l l l l l l l l l l n n e n Y X P 2 1 ) ( ! ) ( 2 1 l l l l n k k n k n k n k n 0 2 1 2 1 )! ( ! ! ) ( l l l l Reference: Sum of Independent RVs Let X and Y be independent Binomial RVs X ~ Bin(n 1 , p) and Y ~ Bin(n 2 , p) X + Y ~ Bin(n 1 + n 2 , p) More generally, let X i ~ Bin(n i , p) for 1 ≤ i ≤ N, then Let X and Y be independent Poisson RVs X ~ Poi( l 1 ) and Y ~ Poi( l 2 ) X + Y ~ Poi( l 1 + l 2 ) More generally, let X i ~ Poi( l i ) for 1 ≤ i ≤ N, then p n X N i i N i i , Bin ~ 1 1 N i i N i i X 1 1 Poi ~ l Expected Values of Sums Let g(X, Y) = X + Y. Compute E[g(X, Y)] = E[X + Y] E[X + Y] = E[X] + E[Y] Generalized: Holds regardless of dependency between X i ’s We’ll prove this next time n i i n i i X E X E 1 1 ] [

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document