Expected value for continuous random variable

As follows~~~
As follows~~~


圖片參考：http://i182.photobucket.com/albums/x4/A_Hepburn_1990/A_Hepburn01Mar192036.jpg?t=1205930228

f(x)=1/9xe^(-x/3) for x>0

the expected water consumption of water for any given day
=∫x[1/9xe^(-x/3)] dx [from 0 to infinity]
==∫1/9x^2e^(-x/3) dx
=(1/9) Gamma[3]/(1/3)^3
=(1/9)(2)(27)
=54/9
=6

我明天再正式報答啦﹐因為無手機認證﹐今日已經用盡quota

2008-03-20 09:54:40 補充：
f(x)=1/9xe^(-x/3) for x>0

the expected water consumption of water for any given day
=∫x[1/9xe^(-x/3)] dx [from 0 to infinity]
==∫1/9x^2e^(-x/3) dx
=(1/9) Gamma[3]/(1/3)^3
=(1/9)(2)(27)
=54/9
=6

2008-03-20 10:00:09 補充：
當然Audrey Hepburn沒做錯﹐不過可惜中六七沒教Gamma distribution﹐所以用了2次integration by parts。不過我覺得朋友你一定識(因為你之前問過些好深的統計題)﹐所以用了些快方法答啦(而且條題目是精算題?)

我說過補答啦﹐所以我一定要「言出必行」啦