作者Fwos (悼念~娃娃)
看板Statistics
标题Re: [请益] 北科工管91考题
时间Thu May 4 15:49:17 2006
※ 引述《wolf035 (阿荣)》之铭言：
: 1. prove that if X~N(u,sigma^2) and let Z= x-u/sigma then Z have
: mean = 0 and standard deviation = 1
: 2. If the random variable X has an exponential distribution with
: parameter lamda. Find the probability that X greater than its mean.
1. X ~ N(μ,σ^2) ,
the m.g.f. of X is Mx(t)= E[exp (tX)]
= exp (μt+σ^2 *t^2 / 2)
Z = (X-μ)/σ,
the m.g.f. of Z is Mz(t)= E[ exp (tZ) ]
= E{ exp [t(X-μ)/σ] }
= E{ exp(tX/σ) *exp(-tμ/σ) }
= E{ exp [(t/σ)X] } *exp(-μt/σ)
= Mx(t/σ) *exp(-μt/σ)
= exp[μ*(t/σ)+σ^2 *(t/σ)^2 / 2 ] *exp(-μt/σ)
= exp(t^2 / 2)
since the m.g.f. uniquely determines the distribution,
we know that Z ~ N(0,1)
2. X ~ Exp(λ),
Mx(t)= 1 / (1- t/λ), t < λ
Mx'(t) = (1/λ) / (1- t/λ)^2
E[X]= Mx'(0)= 1/λ
Pr(X > 1/λ) = ∫(下界为 1/λ，上界为∞) λ *exp(-λx) dx
= [-exp(-λx)] (下界为 1/λ，上界为∞)
= 1 / e
第一题也可用d.f.法来求，只是要用到积分式的微分觉得表示起来不方便，
所以用m.g.f.法，比较好表示也比较快
这是我第一次在 BBS上 po数学解法，可能有点乱，希望您看得懂
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