[商管] [統計] 機率論
the joint density function of X and Y is given by
f(x,y)=(2e^-x)*(e^-2y)
0 < x < 無限
0 < y < 無限
compute
(1) P(X>1,Y<2)
(2) P(X<2Y)
(3) P(X<3)
(where, M = infinity)
At first, we need calcuate p.d.f. of r.v. X and Y
M M
f(x)= S (2e^-x)*(e^-2y)dy = e^-x * S e^(-2y)d(2y) = e^-x*ㄗ(1)= e^-x,
0 0 0<x<M
M M
f(y)=S (2e^-x)*(e^-2y)dx = 2e^(-2y) * S e^(-x)dx = 2e^(-2y)*ㄗ(1)=2e^(-2y),
0 0 0<y<M
1 2
(1) P(X>1,Y<2)= S S (2e^-x)*(e^-2y)dydx
0 0
1 2
= S e^(-x)*[-e^(-2y)] dx
0 0
1
= (1-e^(-4))*[-e^(-x)]
0
= 1-e^(-1)-e^(-4)+e^(-5) #
M
(2) P(X<2Y) = S P(X<2Y | Y=y)*P(Y=y) dy
0
M 2y
= S S e^(-x)dx *2e^(-2y) dy
0 0
M 2y
= S [-e^(-x)] *2e^(-2y) dy
0 0
M
= S (2e^(-2y)-2e^(-4y)) dy
0
M 1 M
= S e^(-2y)d(2y) - --- S e^(-4y)d(4y)
0 2 0
= ㄗ(1)-(1/2)*ㄗ(1)
= 1/2 #
3 3
(3)P(X<3)= S e^(-x)dx = [-e^(-x)] = 1-e^(-3) #
0 0
大概解一下,失眠解得有時會錯...
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◆ From: 140.114.231.103
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03/24 02:55, , 1F
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※ 編輯: loveliver 來自: 140.114.54.33 (03/24 08:48)
※ 編輯: loveliver 來自: 140.114.54.33 (03/24 08:50)