# CS代考程序代写 University of California, Los Angeles Department of Statistics

University of California, Los Angeles Department of Statistics
Statistics 100B Instructor: Nicolas Christou
Noncentral χ2 and noncentral F distributions
Let Y1,Y2,…,Yn be i.i.d. random variables with Yi ∼ N(μi,σ2),i = 1,2,…,n. If each μi = 0 then 􏰃n Y2
Q= i=1 i ∼χ2.Whatifeachμ ̸=0? σ2 n i
The m.g.f. of Q is given by 􏰏􏰍n 􏰎􏰐
􏰈 Yi2 􏰋
􏰄 Y22 􏰅􏰌 􏰋
MQ(t)=E exp t
Let’s examine one of these expectations:
×E exp tσ2
i=1
σ2
=E exp tσ2
􏰋 􏰄Yi2􏰅􏰌 􏰀∞ 1 􏰋tyi2 (yi−μi)2􏰌 E exp tσ2 = σ√2πexp σ2 − 2σ2
−∞
􏰄 Y12 􏰅􏰌 􏰋
􏰄 Yn2 􏰅􏰌 ×…×E exp tσ2
dyi.
􏰅2
.
Back to the expectation
􏰋􏰄Yi2􏰅􏰌􏰋tμ2i􏰌􏰀∞1 􏰏1−2t􏰄μi􏰅2􏰐
E exp tσ2 =exp σ2(1−2t) −∞ σ√2π×exp − 2σ2 yi−1−2t dyi. If we multiply and divide by √1 − 2t we have the integral of a normal pd.f. with mean μi
Evaluate the integral using:
tyi2 (yi −μi)2 yi2(1−2t) 2μiyi μ2i
σ2−2σ2 =−2σ2+2σ2−2σ2 tμ2i 1 − 2t 􏰄
μi = σ2(1 − 2t) − 2σ2 yi − 1 − 2t
σ2 (and therefore it is equal to 1, to finally get 1−2t
􏰋 􏰄Yi2􏰅􏰌 1 􏰋 tμ2i 􏰌 E exp tσ2 = √1−2texp σ2(1−2t) .
􏰃n Now we can find the moment generating function of Q = i=1
−n 􏰋t􏰃ni=1μ2i 􏰌 MQ(t) = (1−2t) 2 exp σ2(1−2t) .
Yi2
.
1−2t
and variance
M (t)=(1−2t)−neθ t
follows the χ2 distribution with noncentrality parameter θ. We write Q ∼ χ2(n, θ). Therefore
σ2
In general, a random variable Q that has m.g.f. of the form Q 2 1−2t
n􏰍n􏰎 􏰃 Y2 􏰈μ2
Q= i=1 i ∼χ2 n, i .
σ2
Note: If the noncentrality parameter is zero then Q ∼ χ2n.
i=1
σ2