Re: [機統] mode of lognormal distribution
※ 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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※ 發信站: 批踢踢實業坊(ptt.cc)
◆ From: 140.113.10.40
推
09/03 15:15, , 1F
09/03 15:15, 1F
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09/03 15:16, , 2F
09/03 15:16, 2F
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09/03 15:17, , 3F
09/03 15:17, 3F
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09/03 15:18, , 4F
09/03 15:18, 4F
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
09/03 16:35, , 5F
09/03 16:35, 5F
( 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)
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09/03 18:26, , 6F
09/03 18:26, 6F
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