有關probability density function

Consider

∞
∫ |x|/(1+|x|)^4 dx
-∞

　∞
=2∫ x/(1+x)^4 dx
　0

　∞
=2∫ (1+x-1)/(1+x)^4 dx
　0

　∞
=2∫ [ 1/(1+x)^3 - 1/(1+x)^4 ] dx
　0

　　　　　　　　　　　　∞
= 2[ -1/2(1+x)^2 + 1/3(1+x)^3 ]
　　　　　　　　　　　　0

= 2 ( 1/2 - 1/3)


= 1/3

Hence
∞
∫ k|x|/(1+|x|)^4 dx = 1
-∞

k/3 = 1

k = 3

圖片參考：http://www.yousendit.com/transfer.php?action=check_download&ufid=TTZuYUlncG9RR2RjR0E9PQ&key=14b3e44eae24799f1370a0602f66a582155c07a2&bid=UmNLTG0rdzhWRDgwTVE9PQ&[email protected]

Do you understand?

2009-03-31 20:33:40 補充：
the picture would be deleted after 7 days.

Download it if you like.

2009-03-31 20:34:51 補充：
[email protected] & joseph.tamm are wrong & poor.

The integral is not quite difficult jar wor

k 冇得計架! 除非你有一個固定嘅 x0 同 f(x0).
佢嘅數值只不過係用嚟模擬某情況下嘅機會率嘅外在因素.

比方說, 我有兩條電線, 我去量度經過呢兩條線嘅雜訊強度嘅機會率. 佢哋雖然用唔同物料做, 但係我又知道佢哋兩條電線都係遵循同樣嘅物理定律.

佢哋嘅pdf可以用同一類數學公式嚟代表(就好似你俾出嚟果條咁), 但係佢哋會因為物理上既唔而形响到 k 參數. 記住同一類, 唔代表同一樣!

就好似 straight line (Y = aX + b) versus parabola (Y = cX^2 + dX + e), 佢哋係兩類型既公式, 但係每一類都有無限咁多條.

你嘅 k 就好似以上Y = aX + b 嘅 a 同 b; Y = cX^2 + dX + e 嘅 c, d, e咁. 佢哋只不過係模擬一種狀況嘅parameters.

2009-03-29 13:34:00 補充：
呀唔好意思, 記錯咗, 係有得計... 因為integrate pdf=1.