actuary-will-be 發問時間： 科學數學 · 1 0 年前

# 有關Normal Distribution的問題..10點

Ri ~ Normal (μi, σi的平方)

Rj ~ Normal (μj, σj的平方)

Rp = wiRi + wjRj

wi + wj = 1

Rp也是一個Normal Distribution

### 1 個解答

• Fwos
Lv 4
1 0 年前
最佳解答

Ri ~ Normal (μi, σi^ 2)

Rj ~ Normal (μj, σj^ 2)

the joint distribution of Ri and Rj is a bivariate normal distribution with parameters

μi、μj、σi^ 2、σj^ 2、ρ，

where ρis the coefficient of correlation (相關係數) of Ri and Rj ,

if ρ= 0 , then Ri and Rj are independent (獨立)

the m.g.f. (動差母函數) of Ri and Rj is

Mij ( t1 , t2 )

= E[exp ( t1 * Ri + t2 * Rj) ]

= exp (μi * t 1+μj * t 2+σi^ 2 * t 1^ 2 / 2+σj^ 2 * t 2^ 2 / 2+ρ*σi *σj *t 1 * t 2)

the m.g.f. of Rp is

Mp (t)

= E[exp ( t * Rp)]

= E[exp ( t * (wi * Ri + wj * Rj) ) ]

= E[exp ( t wi * Ri + t wj * Rj)]

= Mij ( t wi , t wj)

= exp [μi *(t wi) +μj * (t wj)

+σi^ 2 * (t wi)^ 2 / 2+σj^ 2 * (t wj)^ 2 / 2 +ρ*σi *σj * (t wi) * (t wj) ]

= exp [ (μi * wi +μj * wj ) * t

+ (σi^ 2 * wi^ 2 +σj^ 2 * wj^ 2 + 2 *ρ*σi *σj *wi * wj ) * t^ 2 / 2 ]

since a m.g.f. is uniquely determined the distribution , we have that

Rp ~ N( μi * wi +μj * wj , σi^ 2 * wi^ 2 +σj^ 2 * wj^ 2 + 2 *ρ*σi *σj *wi * wj )

note that if Ri and Rj are independent (ρ= 0 ), we have that

Rp ~ N( μi * wi +μj * wj , σi^ 2 * wi^ 2 +σj^ 2 * wj^ 2 )