208 Some Special Distributionsellipmake(p=.95,b=matrix(c(1,.75,.75,1),nrow=2),mu=c(5,2)).This R function can be found at the site listed in the Preface.
(a)Run the above code.(b)Change the code so the probability is 0.50.(c)Change the code to obtain an overlay plot of the 0.50 and 0.95 regions.(d)Using a loop, obtain the overlay plot for a vector of probabilities.3.5.6. LetU andV be independent random variables, each having a standard
normal distribution. Show that the mgfE(et(UV)) of the random variableUVis
(1−t^2 )−^1 /^2 ,− 1 <t<1.
Hint: CompareE(etU V) with the integral of a bivariate normal pdf that has means
equal to zero.
3.5.7. LetX andY have a bivariate normal distribution with parametersμ 1 =
5 ,μ 2 =10,σ^21 =1,σ^22 = 25, andρ>0. IfP(4<Y < 16 |X =5)=0.954,
determineρ.
3.5.8.LetX andY have a bivariate normal distribution with parametersμ 1 =
20 ,μ 2 =40,σ 12 =9,σ 22 =4,andρ=0.6. Find the shortest interval for which 0.90
is the conditional probability thatY is in the interval, given thatX= 22.3.5.9.Say the correlation coefficient between the heights of husbands and wives is
0.70 and the mean male height is 5 feet 10 inches with standard deviation 2 inches,
and the mean female height is 5 feet 4 inches with standard deviation 1^12 inches.
Assuming a bivariate normal distribution, what is the best guess of the height of
a woman whose husband’s height is 6 feet? Find a 95% prediction interval for her
height.3.5.10.Letf(x, y)=(1/ 2 π)exp[
−1
2(x^2 +y^2 )]{
1+xyexp[
−1
2(x^2 +y^2 −2)]}
,where−∞<x<∞,−∞<y<∞.Iff(x, y)isajointpdf,itisnotanormal
bivariate pdf. Show thatf(x, y) actually is a joint pdf and that each marginal pdf
is normal. Thus the fact that each marginal pdf is normal does not imply that the
joint pdf is bivariate normal.
3.5.11.LetX, Y,andZhave the joint pdf
(
1
2 π) 3 / 2
exp(
−x^2 +y^2 +z^2
2)[
1+xyzexp(
−x^2 +y^2 +z^2
2)]
,where−∞<x<∞,−∞<y<∞,and−∞<z<∞. WhileX, Y,andZare
obviously dependent, show thatX, Y,andZare pairwise independent and that
each pair has a bivariate normal distribution.