第4章随机变量的数字特征习题答案

14.1—4.2数学期望习题答案1解：(1)()EX=(2)0.1(1)0.400.310.20.4(2)E(3X+1)=3E(X)＋1＝3(0.4)10.2(3)E(2X)=40.120.400.310.212解：(1)dxxf)(=112Adx1x=Aarctanx|11=A[arctan1arctan(1)]=A2=12A(2)E(X)=11)(xxfdx=11221)xdxx（=1ln(1+2x)|11=010000013.(),2()1XyyyyYEXxfxdxxdxEYyfydxyedyyeedye解：13(23)21322EXY(2)随机变量X与Y相互独立，E(XY)＝E(X)E(Y)=214解：P(X=0)=0.3+aP(X=1)=0.3+b,2()0.3EXbP(Y=0)=0.2P(Y=1)=0.2+bP(Y=2)=0.2+a222222()0.22(0.2)14()1.3422.4EYbbabEXYEXEYab即4a+2b=1.1又由分布律的性质，得0.3+a+0.3+b=1,即a+b=0.4a=0.15,b=0.254.3方差习题答案1解：设X表示在取得合格品以前已经取出的废品数，则X＝0，1，2，3。P（X＝0）＝912＝43，P（X＝1）＝399121144P（X＝2）＝3299,121110220P（X＝3）＝333121220CCX0123P43449220922012()EX＝0×43+1×449＋2×2209＋3×2201＝1032()EX=0×43+1×449＋4×2209＋9×2201=229,故22()()[()]DXEXEX=1100351.2解：012323011010(1)()()(1)(1)()()02323xxxxEXxfxdxxxdxxxdx01343401222210101(2)()())(1)(1)()()34346xxxxDXEXEXxxdxxxdx（3证明：*)(XE＝E()()(XDXEX)=)(1XD[()]EXEX=0D(X*)＝D(DXXEX)()=)(1XD()DX=14解：E(X)=n111niiEXnn,D(X)=21n2211niiDXnn＝2n.5解：()()2()3227()(2)()4()1415(7,5),ZEZEXEYDZDXYDXDYZN故的概率分布密度为：fZ(z)=2(7)10110xe，()xR22211()0,()2,(),()2313(3)()3()03221131[()]()[()](2)(0)3212EXDXEYDYEXYEXEYEXYDXYEXY6.解:由题意：4.4-4.5协方差与相关系数习题答案1解：Cov(X,Y)=XYDXDY=0.456=12;3D(X+Y)=D(X)+D(Y)+2cov(X,Y)=25+36+24=85D(XY)=D(X)+D(Y)2cov(X,Y)=25+3624=372解：(1)P(X+Y=3)=P(0,3)+P(1,2)＋P(2,1)=0.15+0.20+0.05＝0.40(2)X的边缘分布：P{X=0}=P{X=0,Y=1}+P{X=0,Y=2}+P{X=0,Y=3}=0.4P{X=1}=P{X=1,Y=1}+P{X=1,Y=2}+P{X=1,Y=3}=0.4P{X=2}=P{X=2,Y=1}+P{X=2,Y=2}+P{X=2,Y=3}=0.2即X012P0.40.40.2Y的边缘分布：P{Y=1}=P{X=0,Y=1}+P{X=1,Y=1}+P{X=2,Y=1}=0.2P{Y=2}=P{X=0,Y=2}+P{X=1,Y=2}+P{X=2,Y=2}=0.5P{Y=3}=P{X=0,Y=3}+P{X=1,Y=3}+P{X=2,Y=3}=0.3即Y123P0.20.50.3(3)XY012346P0.400.100.250.100.100.05E(X)=0.4+0.4=0.8,E(Y)=0.2+1+0.9=2.1,E(XY)=0.1+0.5+0.3+0.4＋0.3=1.6.Cov(X,Y)=E(XY)EXEY=1.61.68=0.08E(X2)=0.4+0.8=1.2,E(Y2)=0.2+2+2.7=4.9D(X)=E(X2)[E(X)]2=1.20.64=0.56,D(Y)=E(Y2)E[2(Y)]=4.94.41=0.49故X与Y的相关系数为：XY=(,)()()CovXYDXDY＝0.080.70.560.1527.(4)X与Y不是相互独立的。因为0XY，所以X与Y线性相关，故X与Y必不相互独立。或者P{X＝0,Y=1}=0.05，P{X=0}=0.4，P{Y=1}=0.2P{X=0}P{Y=1}≠P{X＝0,Y=1}可以得知X与Y不相互独立。4第四章随机变量的数字特征复习题答案一选择题BDBDC二填空题1．18.42．13．0.94．6三计算题1.解:dxxf)(=20axdx+42()2621bxcdxabc24243320202856()()()()6233233abcEXxfxdxxaxdxxbxcdxxxxabcP(1x3)=21axdx+32)(dxcbx=23a+25b+c=4311,,144abc2解:E(Z)=21E(X)+31E(Y)=67,Cov(X,Y)=XYDXDY=1,D(Z)=41D(X)+91D(Y)+31cov(X,Y)=3637Cov(X,Z)=cov(X,2X+3Y)=21D(X)+31cov(X,Y)=65