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1. (8%)求 Z ∞
1
dx x3 。 Sol:Z ∞
1
dx x3 =−1
2x−2¯¯¯∞
1
=1 2
2. (12%)令c , λ為正的常數 。 設f (t) = ce−λt, t≥ 0 ,是隨機變數X 之機率密度函數 。 求出c之值及E(X) , Var(X) (用λ表出)。(各4%)
Sol:
1 = Z ∞
0
ce−λtdt =−c λe−λt¯¯¯∞
0 = c λ, c = λ E(X) =
Z ∞
0
xλe−λxdx = Z ∞
0
−xde−λx=−xe−λx¯¯¯∞
0 + Z ∞
0
e−λxdx = 0 + 1 λ
Z ∞
0
λe−λxdx = 1 λ E(X2) =
Z ∞
0
x2λe−λxdx = Z ∞
0
−x2de−λx =−x2e−λx¯¯¯∞
0
+ Z ∞
0
2xe−λxdx = 0 +2
λEX = 2 λ2 V ar(X) = E(X2)− (EX)2= 1
λ2
1
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