设随机变量X,Y相互独立,且X~N(1,5),Y~N(2,3),试求Z=2X-3Y+1的概率密度,求。。。。 若随机变量X~N(-2,4),Y~N(3,9),且X与Y相互...
\u8bbe\u968f\u673a\u53d8\u91cfX\u4e0eY\u76f8\u4e92\u72ec\u7acb\uff0cX~N\uff082,1),Y~N(1,2),\u5219Z=2X-Y+3\u7684\u5bc6\u5ea6\u51fd\u6570\u8868\u8fbe\u5f0f\u4e3aE(Z)=? ,D(Z)=? ,f\uff08Z)=?\uff081\uff09\u56e0\u4e3aX~N\uff082,1),Y~N(1,2),\u6545;EX=2,DX=1,EY=1,DY=2
\u518d\u7531\u671f\u671b\u4e0e\u65b9\u5dee\u7684\u6027\u8d28\uff1a
E(Z)=E(2X-Y+3)=2E(X)-E(Y)+3=2*2-1+3=6
D(Z)=D(2X-Y+3)=2^2D(X)+\uff08-1)^2D(Y)=4*1+2=6
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\uff082\uff09\u7531\u79bb\u6563\u578b\u968f\u673a\u53d8\u91cf\u5206\u5e03\u5217\u7684\u6027\u8d28\uff0c\u6240\u6709\u70b9\u5bf9\u5e94\u7684\u6982\u7387\u4e4b\u548c\u4e3a 1\uff0c
\u6240\u4ee5\uff1a0.1+0.3+0.1+a+0.2=1
\u7531\u6b64\u6c42\u5f97\uff1aa=0.3
\u800c 0\uff1cX\u22642\u65f6\uff0cx\u53ea\u53d61\u4e0e2\u4e24\u4e2a\u70b9\uff0c\u8fd9\u4e24\u4e2a\u70b9\u5bf9\u5e94\u7684\u6982\u7387\u5206\u522b\u662f0.1\u4e0e0.3\uff0c
\u6545\uff1aP\uff080\uff1cX\u22642\uff09=0.1+0.3=0.4
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E(X)=1,E(Y)=2,D(X)=5,D(Y)=3
所以,Z=2X-3Y+1,服从正态分布
E(Z)=2E(X)-3E(Y)+1=-3,
D(Z)=4D(X)+9D(Y)=47
所以,Z~N(-3,47)
概率密度就好写了
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锛孻鐩镐簰鐙珛锛屾墍浠(X-Y)=DX+D(-Y)=DX+DY=5銆傛晠X-Y~N(-1锛5)
绛旓細銆愮瓟妗堛戯細E(X)=1,D(X)=1;E(Y)=-2,D(Y)=1.鍥犳E(2X+Y)=2E(X)+E(Y)=2脳1+(-2)=0,D(2X+Y)=22D(X)+D(Y)=4脳1+1=5.
绛旓細銆愮瓟妗堛戯細
绛旓細e(z)=e(2x-y)=2e(x)-e(y)=2-0=2 d(z)=d(2x-y)=4d(x)+d(y)=4(2)+1=9 鎵浠=2X-Y+3=(2,9)涓涓簩缁存鎬2113鍒嗗竷鐨勮竟缂樺垎甯冪殑鍜屾绘槸姝f佸垎甯冦傜壒鍒殑涓や釜鐙珛姝f佸垎甯冪殑鍜屾绘槸姝f佸垎甯冦
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绛旓細E(X+Y) = E(X)+E(Y) = -1+1 =0 D(X+Y) = D(X)+D(Y) = 1+3 = 4 X+Y ~ N (0, 4)P(X+Y>2)=P(Z > (2-0)/2 )=P(Z >1)
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