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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Show that Greenwoods formula reduces to the binomial variance formula. Solution Let S(t) be the probability that an item from a given population will have a lifetime exceeding t. For a sample from this population of size N let the observed times until death of N sample members be Corresponding to each ti is ni, the number \"at risk\" just prior to time ti, and di, the number of deaths at time ti. Note that the intervals between each time typically are not uniform. For example, a small data set might begin with 10 cases, have a death at Day 3, a loss (censored case) at Day 9, and another death at Day 11. Then we have (t1 = 3, t2 = 11), (n1 = 10, n2 = 8), and (d1 = 1, d2 = 1). The Kaplan–Meier estimator is the nonparametric maximum likelihood estimate of S(t). It is a product of the form When there is no censoring, ni is just the number of survivors just prior to time ti. With censoring, ni is the number of survivors less the number of losses (censored cases). It is only those surviving cases that are still being observed (have not yet been censored) that are \"at risk\" of an (observed) death.[3] There is an alternative definition that is sometimes used, namely The two definitions differ only at the observed event times. The latter definition is right-continuous whereas the former definition is left-continuous. Let T be the random variable that measures the time of failure and let F(t) be its cumulative distribution function. Note that Consequently, the right-continuous definition of may be preferred in order to make the estimate compatible with a right-continuous estimate of F(t)..

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Show that every subset of a totally bounded set is totally bounded. Solution Let S be a totally bounded set and F a subset of it. AS S is totally bounded, by definition, for every epsilon >0, there exists a final collection of open balls in S of radius epsilon such that their union contain S. Or there exists a finite epsilon net. As S is totally bounded we have S is the union of small subsets Consider F, a subset of S S is contained in A1UA2UA3...An where each Ai is a subset with epsilon radius. Select Ais such that they contain all elements of F Let Al, Am, As,...contain elements of F Then F is obviously contained in AlUAmU... From this it follows that F has an epilon net for itself Hence F is totally bounded..

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