设随机变量X和Y相互独立,且X~N(0,1),Y的概率分布为P(Y=0)=P(Y=1)=12,则P(X+Y≤12)=( ) 设随机变量X与Y相互独立,且X服从标准正态分布N(0,1),...
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因为X~N(0,1),所以绛旓細X-Y~N(-1锛5)銆傜敱宸茬煡寰桬X=0锛孌X=1锛孍Y=1锛孌Y=4锛屼簬鏄疎(X-Y)=EX-EY=-1锛X锛孻鐩镐簰鐙珛锛鎵浠(X-Y)=DX+D(-Y)=DX+DY=5銆傛晠X-Y~N(-1锛5) 绛旓細X鐨姒傜巼瀵嗗害鍑芥暟涓 p(x)= 1 x鈭(0,1)0 鍏朵粬 Y鐨勬鐜囧瘑搴﹀嚱鏁颁负 f(x)= e^(-x) x鈮0 0 鍏朵粬 鍒╃敤鍜岀殑鍒嗗竷鍏紡鍙煡,Z鐨勬鐜囧瘑搴﹀嚱鏁颁负 g(y)=鈭玆 p(x)f(y-x)dx =0 y鈮0 鈭玔0,y]e^(x-y)dx=1-e^(-y) 01 涔熷氨鏄痁鐨勬鐜囧瘑搴︽槸涓垎娈靛嚱鏁般 绛旓細/12=1/3锛孍(X^2)=D(X)+E(X)^2=4/3锛屽洜涓篩~N(2,4)锛屾墍浠(Y)=2,D(Y)=4锛孍(Y^2)=D(Y)+E(Y)^2=8锛屾牴鎹湡鏈涗笌鏂瑰樊鐨勬ц川鍙緱锛欵(XY)=E(X)E(Y)=2锛孍((XY)^2)=E((X^2)(Y^2))=E(X^2)E(Y^2)=32/3锛孌(XY)=E((XY)^2)-[E(XY)]^2=20/3銆 绛旓細闅忔満鍙橀噺X涓嶻鐩镐簰鐙珛,閭d箞 D(X-2Y+3)=DX+ 2² *DY 鑰 X锝濨(16,0.5),Y鏈嶄粠鍙傛暟涓9鐨勬硦鏉惧垎甯 鎵浠 DX=16*0.5*(1-0.5)=4,鑰孻鐨勬柟宸氨绛変簬娉婃澗鍒嗘暟鐨勫弬鏁,鍗矰Y=9,浜庢槸 D(X-2Y+3)=DX+ 2² *DY =4 + 4 *9 =40 ... 绛旓細2019-05-30 璁鹃殢鏈哄彉閲廥涓嶻鐩镐簰鐙珛,涓擷~N(0,1),N(1,1)鍒... 13 2012-12-14 璁鹃殢鏈哄彉閲廥鍜孻鐩镐簰鐙珛涓擷~E(1),Y~E(2),姹俍=... 10 2016-12-02 璁鹃殢鏈哄彉閲廥,Y鐩镐簰鐙珛,涓擷~N(1,2),Y~N(0,1... 2016-07-05 璁鹃殢鏈哄彉閲廥涓嶻鐩镐簰鐙珛,涓旈兘鏈嶄粠N(0,1/2),鍒橢(|... 147 ... 绛旓細浣犲ソ锛佸彲浠ュ鍥惧埄鐢ㄦ硦鏉惧垎甯冨叕甯冭绠楋紝绛旀鏄18e^(-6)銆傜粡娴庢暟瀛﹀洟闃熷府浣犺В绛旓紝璇峰強鏃堕噰绾炽傝阿璋紒 绛旓細绛旀鏄疊锛屽彲浠ュ弽鐢ㄦ柟宸绠楀叕寮忔潵璁$畻骞虫柟鐨勬湡鏈涖傝閲囩撼锛岃阿璋紒 绛旓細妤间笂姝hВ. 绛旓細E(X*Y)=1/2 鐢变簬X锛孻鐩镐簰鐙珛锛鍥犳E(X锛孻)=鈭埆xyf(x,y)dxdy =鈭埆xyf1(x)f2(y)dxdy =鈭玿f1(x)dx鈭珁f2(y)dy =E(X)E(Y)=1/2*1 =1/2 绛旓細銆愮瓟妗堛戯細f(x,y)=(1/锛4蟺锛)*e^[-x^2/2-(y-1)^2/8]F(x,y)=FX(x)*FY(y),F(0,1)=FX(0)*FY(1)=0.5*0.5=0.25 本站交流只代表网友个人观点,与本站立场无关 |