Re: [機統] mode of lognormal distribution

[sapphireBOB (澄響幸輝絕品泉水)]

※ 引述《cadywu ( )》之銘言：

請問一下為什麼mode of lognormal distribution是exp(μ-σ^2) 不是exp(μ)?
normal distribution機率最大的不是μ嗎?


Let Y be r.v. lognormal (μ,σ^2)

so we have lnY ~ N(μ,σ^2) , with pdf f(y)

μ is the mean of inidcated Normal distribution

it is not the mean of lognormal distribution 

To find E(Y) ,  we must find pdf of Y first , say k(y)

Assume X = lnY , we have J = 1/y

so k(y) = f(lnY)(1/y) = [√2π(σy)]^-1 exp[-(lny-μ)^2/2σ^2]

∵ Y = e^X

∴E(Y) = E(e^X) = E (e^tX)︱    = exp[u+(σ^2)/2]
                            t=1

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◆ From: 140.113.10.40
推 cadywu      :大大 謝謝你對於mean的推導 不過我還是想問一下
→ cadywu      :除了用f(y)的微分求mode的方式以外 還有什麼比較直觀
→ cadywu      :的方式 可以解釋為什麼X的mode是μ, 但是Y的mode不是
→ cadywu      :exp(μ)嗎? 謝謝~~
Observe the Normal distribution first

since the mode m satisfies : f(m) = sup f(x)
                                     x

it is clear that f(x) has sup at x = μ

推 cadywu      :k(y) has sup at x = μ+(σ^2)/2, not μ-σ^2 ?
( my previous assertion is wrong , sorry :p)

it is not clear to say the mode of Y is  exp (μ-σ^2)

       d
take   ─ (ln k(y)) = 0   will be feasible
       dy

then we will have y = exp (μ-σ^2)

so the mode of Y will be exp (μ-σ^2) after verifying second derivative

it is just an arithmetic process

if we fix μ to be constant

pdf of Y will tend to be right-skew under greater variance σ
※ 編輯: sapphireBOB     來自: 140.113.10.40        (09/03 18:36)
→ yhliu       :能用 mean 推 mode? 那為什麼不能用 median 推 mode?