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problems binomial random, veriable has the moment generating function, y(t)=E eux 1. A nd+ 1-p)n. Show that EIX|-np and Var(X) np(1-p) using that EIX)-v(0) nd E.X2 =ψ (0). 2. Lex X be uniformly distributed over (a b). Show that ElXI 쌓 and Var(X) = (b and second moments of this random variable where the pdf of X is (x)N of a continuous randonn variable is defined as E[X-广.nf(z)dz. )a using the first Note that the nth moment 3. Show that (z-p)2 exp [ーしご]dz = σ2 Hint. Judu = uu-j vdu and foco e-//2dy- 4. The moment generating function of the normal distribution with parameters μ and σ exp ( μ1+ 2t2 ) for _ oo < t < oo. Show that Els-ψ(0) = μ and Var(X) = ψ(0)-W(0)12 2 is ψ(t) σ2 5. Suppose that Xi. 2, and X3 are independent random variables such that Elx,-0 and EIX ] = 1 for i 12,3. Find the value of EXj(2X1-Xs) 6. Suppose that X and Y are random variables such that Var(X)-Var()-2 and Cov(x, Y)- 1. Find the value of Var(3X -Y +2). 7. Show that E(X)=0 and Var(X)= if Xi, . . . , Xu are independent and identically distributed with E(,)-0 and E(X ) = σ2 for i = 1,···,n
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The Moment Generating Function of the Binomial Distribution Consider the binomial function pan-r with q = 1-p Then the moment generating function is given by n! pq (pe ) (+pet), where the final equality is understood by recognising that it represents the expansion of binomial. If we differentiate the moment generating function with respect to t using the function-of-a-function rule, then we get dt = npe (q + pet)-1 . Evaluating this at t = 0 gives Notice that this result is already familiar and that we have obtained it previ- ously by somewhat simpler means. To find the second moment, we use the product rule duv du du dr to get = npet(e+ pet)-2(1-1)pet + (e+ pet)} Evaluating this at t = 0 gives From this, we see that = npq

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