actuary-will-be asked in 科學數學 · 1 decade ago

有關Normal Distribution的問題..10點

這邊有個問題想要請教各位大大…10點給最用心的人…

如果

Ri ~ Normal (μi, σi的平方)

Rj ~ Normal (μj, σj的平方)

Rp = wiRi + wjRj

wi + wj = 1

試證明

Rp也是一個Normal Distribution

1 Answer

Rating
  • Fwos
    Lv 4
    1 decade ago
    Favorite Answer

    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 )

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