Transcription of Stat 515 Final Exam Practice Problems + Solution
1 Prof. H. K. HsiehStat 515 Final Exam Practice Problems + SolutionStudent s Name:Section:.NOTE: Write main steps of your work clearly and circle your Suppose that the joint ofXandYis given byf(x,y) = 8xy,0<x<y<1= 0elsewhere(a) Verify that thef(x,y) given above is indeed a : 10 y08xydxdy= 104yx2]y0dy= 104[y3]dy=y4]10= ,f(x,y) is a pdf.(b) Find the marginal probability density ofX,f1(x).Answer:f1(x) = 1x8xydy= 4xy2]1x= 4x(1 x2),0<x<1.(c) Find the marginal probability density ofY,f2(y).Answer:f2(y) = y08xydy= 4yx2]y0= 4y3,0<y<1.
2 (d) AreXandYindependent?Answer:XandYare not independent asf(x,y)6=f1(x)f2(y). We can alsoargue thatXandYcan not be independent asxandyin the support off(x,y)depend each other.(e) Find the expected value ofXand variance :E(X) = 10xf1(x)dx= 10x4x(1 x2)dx= 4[x3/3 x5/5]10= 4(1/3 1/5) = 8 (X) =E(X2) [E(X)]2= (work out by yourself).(f) Find the conditional densityfY|x(y) ofYgivenX= :f(y|x) = 8xy/[4x(1 x2)] = 2x/(1 x2),0<x<y<12. Suppose thatXandYare independent and each has (t) = 2t0<t<11(a) Write down the joint ofXandY,f(x,y) =?
3 , including the support of (x,y) = (2x)(2y) = 4xy,0<x<1,0<y<1.(b) Draw the linex+y=12, indicating the origin and coordinatesxandy. Thenindicate the area ofx+y<12in the support off(x,y).(Draw by yourself.)(c) Find Pr{X+Y<12}.Pr{X+Y<12}= 120 12 y04xydxdy= 1202y[x2]12 y0dy= 1202y[(12 y)2]dy= 1202y[1/4 y+y2]dy= 2[y28 y33+y44]120=1963. Consider gainsX1,X2(in dollars) of two investiments. Assume that the two gains areindependent and normally distributed withX1 N(100,25),X2 N( 50,16)noting that the second mean is a negative number.
4 Denote the total gain byT=X1+X2.(a) What is the distribution of the total gain? That is, give the name, mean andvariance of the distribution the sum of two normal variables,Tis also a normal variable withE(T) = 100 + ( 50) = 50;V(T) = 25 + 16 = 41(b) What is the chance that the total gain will be greater than $60?P[T>60] =P[Z>(60 50)/ 41] =P[Z> ] = (c) What are the name, mean, variance, and standard deviation of the distributionofY= 2X1 3X2, respectively?Yhas a normal distribution as it is a linear function of normal variavles.
5 Themean and variance ofYareE(Y) =E[2X1 3X2] = 2(100) 3( 50) = 200 + 150 = 350V(Y) =V[2X1 3X2] = 22(25) + ( 3)2(16) = 2444. SupposeXandYare independent and each has a uniform distribution in the interval(0,1).(Omit this problem)(a) Write down the joint pdf ofg(x,y) (b) Find the (t) ofX.(c) Compute the ofU, whereU=X SupposeXandYare independent and each has a distribution with :f(t) = 2e 2t,fort>0.(a) Find the pdf ofW=X2. (Noting thatXhas positive support.)w=x2which is one-to-one transformation asXis non-negative.
6 The inverse isx=w12, andJ=dxdw=12w 12So the pdf ofWisg(w) = 2e 2 w12w 12=w 12e 2 w,forw>0.(b) Write down the joint pdf ofg(x,y) (x,y) = 2e 2x2e 2y= 4e 2(x+y),forx>0,y>0.(c) DefineU=X+ 3Y, andV=Y, then find the joint + 3y,v=y;x=u 3v,y=v,J= 1u 3v>0,v>0 u>3v>0h(u,v) = 4e 2[(u 3v)+v]]= 4e 2(u 2v)u>3v>0(d) Find the (u) = 4 13u0e 2(u 2v)dv= 4e 2u 13u0e4vdv=e 2u[e4v]13u0=e 2u[e43u 1],u>06. SupposeX1,X2,X3,X4are independent and each has a distribution with :f(x) = 2x,for 0<x< ,Y2,Y3,Y4be their order statistics (Y1 Y2 Y3 Y4).
7 (a) i. Pr[X1<t] = 0t2xdx=t2,0<x<1ii. Pr[t1<X1<t2] =t22 t21,0<t1 t2<1iii. Pr[X1>t] = 1 t23(b) Find pdf (y1) =4!1!3!f(y1)[P(X>y1]3= 4(2y1)[1 y21]3= 8y1[1 y21]3,0<y1<1(c) Find pdf (y4) =4!3!1![F(y4)]3f(y4) = 4[y24]3(2y4) = 8y74,0<y4<1(d) Find joint pdf (y1,y3) =4!1!1!1!1!(2y1)[y23 y21](2y3)[1 y23],0<y1<y3<17. Problems in HW#7. (See Solution for HW#7)4)
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