STAT 418HW14 SolutionsSections 7.1-7.41, 3, 6, 16, 18, 21, 26, 33, 36; TE 1, 21. Let X = 1 if the coin toss lands heads, and let it equal 0 otherwise. Also, let Y denotethe value that shows up on the die. Then, with p(i, j) = P (X = i, Y = j),E(return) =6Xj=12jp(1, j) +6Xj=1j2p(0, j) =112(42 + 10.5) = 4.375.3.E [|X − Y |a] =Z10Z10|x − y|ady dxZ10|x − y|ady =Zx0(x − y)ady +Z1x(y −x)ady =Zx0uadu +Z1−x0uadu =xa+1+ (1 − x)a+1a + 1E [|X − Y |a] =1a + 1Z10xa+1+ (1 − x)a+1dx =2(a + 1)(a + 2).6. Let Xi={value of die on a single roll}.E10Xi=1Xi=10Xi=1EXi= 10 ·72= 35.16. By u-subsitution with u = y2/2,E[X] =Zy>xy1√2πe−y2/2dy =e−x2/2√2π.18.Let X = {number of m atches} and let Ii= 1 for a match on card i and 0 otherwise.Note that E(Ii) = 1/13.E(X) = E52Xi=1Ii= 52113= 4.21.a.36510031365336436597= .9301b. Let X = {number of distinct birthdays for a group of 100 people}. Let Ij= 1, if day jis someone’s birthday and = 0, otherwise. ThenE[X] = E"365Xi=1Ij#=365X1=1E[Ij] = 365"1 −364365100#= 87.58.126.a.E[max] =Z10P (max > t)dt=Z10(1 − P (max ≤ t))dt=Z101 − tndt =nn + 1.b.E[min] =Z10P (min > t)dt=Z10(1 − t)ndt =1n + 1.33.a. E[X2+ 4X + 4] = E[X2] + 4E[X] + 4 = V ar(X) + (E[X])2+ 4E[X] + 4 = 14b. V ar(4 + 3X) = V ar(3X) = 9V ar(X) = 4536. Let Xi= 1 if roll i lands on 1 and 0 otherwise. Let Yi= 1 if roll i lands on 2 and 0otherwise.Cov(Xi, Yj) = E[XiYj] − E[Xi]E[Yj] =−136i = j136−136= 0 i 6= jCov(X, Y ) = CovnXi=1Xi,nXj=1Yj=XiXjCov(Xi, Yj) = −n36TE1. For any a,E(X − a)2 = E[X2− 2aX + a2] = E[X2] − 2aE[X] + a2ddaE(X − a)2 =dda{E[X2] − 2aE[X] + a2} = −2E[X] + 2a−2E[X] + 2a = 0 ⇒ a = E[X].TE2.E[|X − a|] =Zx<a(a − x)f(x)dx +Zx>a(x − a)f(x)dx= aF (a) −Zx<axf(x)dx +Zx>axf(x)dx − a[1 − F (a)]= 2aF (a) − 2Za−∞xf(x)dx − addaE[|X − a|] = 2af(a) + 2F (a) − 2af(a) − 12F (a) − 1 = 0 ⇒ F (a) =
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