Re: [機統] 右偏分配平均與中位數的關係
※ 引述《scitamehtam (scitamehtam)》之銘言：
: 在右偏的機率分配下
: 平均數 > 中位數
: 直觀理解是因為極端值
: 對平均數的影響也比較大嗎？
: 還是有其他直觀的解釋呢？
: 有證明嗎？
: 謝謝
: ----
: Sent from BePTT on my iPhone 12
https://reurl.cc/pLyO2Q
https://www.tandfonline.com/doi/full/10.1080/10691898.2005.11910556
這結果很多情況都不是對的...
尤其是discrete distribution:
All Poisson distributions have an infinite right tail and positive skew
(equal to ) – yet for more than 30% of parameter values, for example μ =
.75 (Figure 5), the mean is less than the median.
即使是continous distribution:
To construct a multimodal violation, simply take a discrete violation (e.g.,
Figure 2 or Figure 5) and add random normal “noise” to each value of X. The
noise makes the distribution continuous, but if the noise variance is small
there will be little change to the mean, median, mode, or skew
不過...
Unimodal continuous densities are more cooperative. CitationGroeneveld and
Meeden (1977) prove that the skew gives the relative positions of mean,
median and mode for the F, beta, and gamma densities (the gamma includes the
exponential and the chi-square). More generally, CitationMacGillivray (1981)
proves the relationship for a large class of continuous unimodal densities
including the entire Pearson family.
eg: X~ exp(a)
∞ -x/a ∞ -u -u ∞ ∞ -u u u ∞
μ = ∫xe /a dx = a∫ ue du = a(-ue | +∫ e du) = a(-lim ---- -e |)
0 0 0 0 u->∞ e^u 0
1
= a(-lim ----- - (-1)) = a
u->∞ e^u
2 ∞ 2 -x/a 2 ∞ 2 -u 2 2 -u ∞ ∞ -u
E(X ) = ∫ x e /a dx = a ∫ u e du = a (-u e | + 2∫ue du)
0 0
0 0
2 u^2 2E(X)
= a (-lim ---- + ------- )
u->∞ e^u a
2 2
= a (-lim ----- + 2)
u->∞ e^u
2
= 2a
2 2 2 2 2 2
=> σ = E(X )-μ = 2a - a = a
X-μ 3 1 3 2 2 3
skew = E[(------) ] = ------E(X -3X μ+3Xμ -μ )
σ σ^3
1 3 2 2 3
= ------(E(X )-3μE(X ) +3μ E(X)-μ )
σ^3
1 3 3 3 3
= -----(E(X )-6a +3a -a )
a^3
1 3 3
= -----(E(X )-4a )
a^3
3 ∞ 3 -x/a 3 ∞ 3 -u 3 3 -u ∞ ∞ 2 -u
E(X ) = ∫ x e /a dx = a ∫ u e du = a (-u e | + 3∫ u e )
0 0 0 0
3 u^3 3E(X^2)
= a (-lim ----- + --------)
u->∞ e^u a^2
3 6
= a (-lim ------ + 6)
u->∞ e^u
3
= 6a
=> skew = 2 > 0
m -x/a
∫ e /a dx = 1/2
0
-x/a m
=> -e | = 1/2
0
-m/a
=> 1-e = 1/2
=> -m/a = ln(1/2)
=> m = aln2 < alne = a = μ
這個情況下的確skew > 0, μ > m
--
※ 發信站: 批踢踢實業坊(ptt.cc), 來自: 114.47.95.208 (臺灣)
※ 文章網址: https://www.ptt.cc/bbs/Math/M.1676758261.A.959.html
討論串 (同標題文章)
本文引述了以下文章的的內容：
完整討論串 (本文為第 2 之 2 篇)：