Pr{X-Y<a}的分佈為何?

Question

若X,Y為兩個independent random variables
且為exponential distribution
一般來說課本會教X+Y的pdf和cdf如何計算
那 X-Y的pdf and cdf要如何算呢?
也就是說Pr{X-Y

請問最後得到的式子一定要是兩個嗎？可不可以合併成一個，然後正負值的a都適用呢?

Answer 1

設 X ~ Exponential(λ1), Y ~ Exponential(λ2) 獨立。

Pr{X-Y ≤ a} = ∫-∞∞ Pr{X-Y ≤ a | Y = y}fY(y) dy= ∫-∞∞ Pr{X ≤ a+y}fY(y) dy = ∫-∞∞ FX(a+y)fY(y) dy



若 a ≥ 0，則FX-Y(a) = Pr{X-Y ≤ a} = ∫0∞ (1 - e-λ1(a+y))λ2e-λ2y dy

= λ2∫0∞ (e-λ2y - e-λ1a-y(λ1+λ2)) dy = λ2 limb→∞(-e-λ2y/λ2 + e-λ1a-y(λ1+λ2)/(λ1+λ2))|y=0b = λ2 limb→∞(-e-λ2b/λ2 + e-λ1a-b(λ1+λ2)/(λ1+λ2) + 1/λ2 - e-λ1a/(λ1+λ2)) = λ2(1/λ2 - e-λ1a/(λ1+λ2)) = 1 -  e-λ1a λ2/(λ1+λ2) fX-Y(a) = d/da FX-Y(a) = e-λ1a λ1λ2/(λ1+λ2)

若 a < 0，則FX-Y(a) = Pr{X-Y ≤ a} = ∫-a∞ (1 - e-λ1(a+y))λ2e-λ2y dy


= λ2∫-a∞ (e-λ2y - e-λ1a-y(λ1+λ2)) dy = λ2 limb→∞(-e-λ2y/λ2 + e-λ1a-y(λ1+λ2)/(λ1+λ2))|y=-ab = λ2 limb→∞(-e-λ2b/λ2 + e-λ1a-b(λ1+λ2)/(λ1+λ2) + eλ2a/λ2 - e-λ1a+a(λ1+λ2)/(λ1+λ2)) = λ2(eλ2a/λ2 - eλ2a/(λ1+λ2)) = eλ2a (1 - λ2/(λ1+λ2))  = eλ2a λ1/(λ1+λ2)fX-Y(a) = d/da FX-Y(a) = eλ2a λ1λ2/(λ1+λ2)


2006-09-27 13:29:20 補充：
無法合併...除非定義 u(x) = 0 if x < 0, u(x) = 1 if x ≥ 0，則P(X-Y≤a) = exp(λ2a) λ1/(λ1+λ2) u(-a) + (1 - e-λ1a λ2/(λ1+λ2)) u(a)

Answer 2

Answer 3

1、令Z=-Y，作變數變換求Z的pdf
2、用convolution求X+Z的pdf
3、用積分求X+Z的cdf