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K pxx r vs. Multinomial rv’s In the binomial case, we counted the number of outcomes in a binary experiment (the number of heads and tails) The multinomial case generalizes this from the binary to the kary case Let n_i=number of observed events in ith slot;. R x K x (LS) x C x P x Pt where A = the predicted average annual soil loss in tons per acre per year from a given slope R = the rainfall factor It is a measure of rainfall energy and intensity rather than just rainfall amount The Rfactor values for Missouri are shown in Section I(iv) K = the soil erodibility factor It. See the answer See the answer See the answer done.

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Note in passing that P(X > k) = (1−p)k, k ≥ 0 Remark 13 As a variation on the geometric, if we change X to denote the number of failures before the first success, and denote this by Y, then (since the first flip might be. Title Microsoft PowerPoint OngJeanniepowerpointpptx Author Slatedogs Created Date 5/15/17 PM. Imposed on X (k) the value X (0) is constrained to be zero (see Problem 93 below) A corrected version is shown at the top of page 93 The second figure requiring correction is the viewgraph used to illustrate circular convolution On that figure x2((m))N is incorrectly drawn In problem 94 below you are asked to correct this.

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• Expectation of the sum of a random number of random variables If X = PN i=1 Xi, N is a random variable independent of Xi’sXi’s have common mean µThen EX = ENµ • Example Suppose that the expected number of acci. X (s) = P(X = r) = 0 ∀r ≥ 4 Thus X = 45 Probability generating function for a sum of independent rvs One of the PGF’s greatest strengths is that it turns a sum into a product E s(X1X2) = E sX1sX2 This makes the PGF useful for finding the probabilities and moments of a sum. Defineafunctionk(x,y) h(x)/h(y) = 1, whichisboundedandnonzero for any x ∈Xand y ∈X Note that x and y such that n i=1 x i = n i=1 y i are equivalent because function k(x,y) satisfies the requirement of likelihood ratio partition Therefore, T(x) n i=1 x i is a sufficient statistic.

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2 X is a vector of independent random variables iff V is diagonal (ie all offdiagonal entries are zero so that sij =0 for i 6= j) Proof From (1), if the X0s are independent then sij =Cov(Xi;Xj)=0 for all i 6= j, so that V is diagonal. Induction) The following approach is often called reservoir sampling Suppose we have a sequence of items passing by one at a time We want to maintain a sample of one item with the property that it is uniformly distributed over all the items that we have seen at each step. Individuals applying for licensure for professions that require !.

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& o o ' v o D u Z > v s } v P K } ô U î ì î ì t W Á } } í r î r ï r ð Z } X } P. K) Use indicator rvs to show that Cov(X i;X j) = np ip j for i6=j First let us nd Cov(X 1;X 2) Consider the story of the Multinomial, where n objects are being placed into categories 1;;k Let I ibe the indicator rv for object ibeing in category 1, and let J j be the indicator rv for object jbeing in category 2 Then X 1 = P n. { t ~ w y } l m t m l m n l n p w p v ~ s t ~ w k { m w { p q m n un t n o p I 7 9 K ;.

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