Show that if X and Y are independent random variables, then for all .pdf

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Show that if X and Y are independent random variables, then for all random variables Z, Cov(X, Y+Z) = Cov(X,Z) Solution Cov(X,Y+Z) =E[(x-ux)(y+z-uy-uz)] = E(x-ux)(y-uy)+(x-ux)(z-uz)] as x and y are independent so, E(x-ux)(y-uy)]=0 so, Cov(X,Y+Z) = E[(x-ux)(z-uz)] = cov(X,Z).

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Show that if A is a fixed event of positive probability, then the function Q[B]=P[B|A] taking events B into R satisfies the three defining axioms of probability. Here is the three defining axioms of probability. A probability measure P is a function taking the family of events H to the real numbers such that (i) P[Pi]=1 (ii) For all A includes in H, P[A] 0. (iii) If A1, A2,.....is a sequence of pairwise disjoint events then P[A1 U A2 U.....]=P[Ai] Solution say A1 , A2 , . . ., are called mutually disjoint or pairwise disjoint if Ai n A j = 0 for any two of the events Ai and A j ; that is, no two of the events overlap. According to Kolmogorov’s axioms, each event A has a probability P(A), which is a number. These numbers satisfy three axioms: Axiom 1: For any event A, we have P(A) = 0. Axiom 2: P(S ) = 1. 4 Axiom 3: If the events A1 , A2 , . . . are pairwise disjoint, then CHAPTER 1. BASIC IDEAS P(A1 ? A2 ? · · ·) = P(A1 ) + P(A2 ) + · · · Note that in Axiom 3, we have the union of events and the sum of numbers. Don’t mix these up; never write P(A1 ) ? P(A2 ), for example. Sometimes we sep- arate Axiom 3 into two parts: Axiom 3a if there are only ?nitely many events A1 , A2 , . . . , An , so that we have P(A1 ? · · · ? An ) = ? P(Ai ), n i=1 and Axiom 3b for in?nitely many. We will only use Axiom 3a, but 3b is important later on. Notice that we write n ? P(Ai) i=1 for P(A1 ) + P(A2 ) + · · · + P(An ). 1.4 Proving things from the axioms You can prove simple properties of probability from the axioms. That means, every step must be justi?ed by appealing to an axiom. These properties seem obvious, just as obvious as the axioms; but the point of this game is that we assume only the axioms, and build everything else from that. Here are some examples of things proved from the axioms. There is really no difference between a theorem, a proposition, and a corollary; they all have to be proved. Usually, a theorem is a big, important statement; a proposition a rather smaller statement; and a corollary is something that follows quite easily from a theorem or proposition that came before. Proposition 1.1 If the event A contains only a ?nite number of outcomes, say A = {a1 , a2 , . . . , an }, then P(A) = P(a1 ) + P(a2 ) + · · · + P(an ). To prove the proposition, we de?ne a new event Ai containing only the out- come ai , that is, Ai = {ai }, for i = 1, . . . , n. Then A1 , . . . , An are mutually disjoint 1.4. PROVING THINGS FROM THE AXIOMS (each contains only one element which is in none of the others), and A1 ? A2 ? · · · ? An = A; so by Axiom 3a, we have P(A) = P(a1 ) + P(a2 ) + · · · + P(an ). Corollary 1.2 If the sample space S is ?nite, say S = {a1 , . . . , an }, then P(a1 ) + P(a2 ) + · · · + P(an ) = 1. For P(a1 ) + P(a2 ) + · · · + P(an ) = P(S ) by Proposition 1.1, and P(S ) = 1 by Axiom 2. Notice that once we have proved something, we can use it on the same basis as an axiom to prove further facts. Now we see that, if all the n outcomes are equa.

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