概率学,研究随机变量平方和的,麻烦有专研精神的强人们能发发慈悲进来...
发布网友
发布时间：2024-09-03 00:11
共3个回答
热心网友
时间：2024-09-18 14:27
所以 X^2的MGF Mx^2(t) = ∫ exp(tx^2) f(x) dx = ∫ exp(tx^2) [1/√(2π) ] exp( - x^2/2) dx
= ∫ [1/√(2π) ] exp[ - (1-2t)x^2/2 ] dx
令 x√(1-2t) = y ，则 dy = √(1-2t) dx ，所以
Mx^2(t) = ∫ [1/√(2π) ] exp[ - (1-2t)x^2/2 ] dx = [1/√(1-2t) ] ∫ [1/√(2π) ] exp( - y^2/2) dy
= 1/√(1-2t) = (1-2t)^(-1/2)
注意 积分 ∫ [1/√(2π) ] exp( - y^2/2) dy 是一个N(0,1) 的全概率，等于1.
(b) X1，X2，。。。，Xd iid，Sd=Σ_( j=1)^(d) Xj^2，所以
MSd(t) = Mx1^2(t) Mx2^2(t) ..... Mxd^2(t) = [Mx1^2(t)]^d
第一个“=”因为独立（independence），第二个“=”因为同分布（identical）。
由（a），Mx1^2(t) = (1-2t)^(-1/2)
所以，MSd(t) = [Mx1^2(t)]^d = (1-2t)^( - d/2)
(c) Chi-square (χ^2) 分布的pdf是 f(x) = (1/2)^(d/2) [1/Γ(d/2)] x^(d/2 - 1) exp(- x/2)
（自由度为d）
所以 MGF 是 Mχ^2(t) = ∫ exp(tx) f(x) dx
= ∫ exp(tx) (1/2)^(d/2) [1/Γ(d/2)] x^(d/2 - 1) exp(- x/2) dx
= (1/2)^(d/2) ∫ [1/Γ(d/2)] x^(d/2 - 1) exp[- x(1/2-t)] dx
=[(1/2)^(d/2) / (1/2 - t)^(d/2) ] ∫ (1/2 - t)^(d/2) [1/Γ(d/2)] x^(d/2 - 1) exp[- x(1/2-t)] dx
= 1/ (1-2t)^(d/2) = (1-2t)^(-d/2)
注意 (1/2 - t)^(d/2) [1/Γ(d/2)] x^(d/2 - 1) exp[- x(1/2-t)] 是 Gamma(d/2，(1/2-t) ) 的pdf，所以，∫ (1/2 - t)^(d/2) [1/Γ(d/2)] x^(d/2 - 1) exp[- x(1/2-t)] dx 是 Gamma(d/2，(1/2-t) )分布随机变量的全概率，等于1 。
最后可以看到Sd的MGF 和 自由度是d的χ^2 分布的MGF一样，所以，Sd就是一个自由度为d的χ^2 分布的随机变量。
Comments: pls just give me the English ones next time, it took me a minute to think about what 矩量生成函数 is. If u tell me MGF, i'd get it in a second. I'm not bragging myself, and no offense to u, but i would prefer to read the original English. I don't even know how to translate some statistical terms to Chinese, tbh. anyways, this one is solved. for the other 2, i dont have time right now. I got some ideas about the probability generating function one, and am confident it's not hard to deal w/. For the branching process one, it looks like a conditional probability problem, been a while since i last time touched this process, may take me a while.
BTW, about your name, just curious, is there a Miss_Spectral, who has to be your gf, right? lol
热心网友
时间：2024-09-18 14:27
a)先求出X^2的moment generating function:Mx^2(t)
Mx^2(t)=∫exp(jtx^2)f0(x)dx=(1-2jt)^(1/2) f0(x)是标准正态分布的密度函数
b)Sd的moment generating function:MSd（t)=[ Mx^2(t)]^d=(1-2jt)^(d/2)
C)当a=1/2 v=d/2时求出伽马分布的moment generating function:
MF(t)=∫exp(jtx)f(x)dx=(1-2jt)^(d/2) f(x)是伽马分布的密度函数
从上面的moment generating function看MF(t)=MSd(t)
所以Sd是服从当a=1/2 v=d/2时的伽马分布,也就是自由度为d的Chi-square分布.
S'd与Sd是相同的,都是d个独立的都是标准正态分布平方的和所以也服从Chi-square分布.
注:以上积分的区间都是(-∞到+∞)
热心网友
时间：2024-09-18 14:29
