Multivariate probability distribution

👨🏻‍🏫 學術台數學

Apricot

2007-01-22 2:32 am

Y is a Poisson random varible with mean λ, its random behavior is modeled by a gamma distribution

f(λ)=1/[Γ(α)β^α]*λ^(α-1)*e^(-λ/β), λ>0
f(λ)=0, elsewhere

Find the unconditional probability distibution for Y.

更新1:

的確是negative binomial distribution, 謝謝你的提示. 但如果可以用番問題d symbol就更好. p=(1/(beta+1)), 1-p=(beta/(beta+1)) 而k=y, and r=alpha

更新2:

咁條式就變成: (y+alpha-1)/[y!(alpha-1)!] x [beta/(beta+1)]^y x [1/(beta+1)]^alpha...對嗎? why [1/(beta+1)]=p?

更新3:

應該是: (y+alpha-1)!/[y!(alpha-1)!] x [beta/(beta+1)]^y x [1/(beta+1)]^alpha where (y+alpha-1)!/[y!(alpha-1)!] =(y+alpha-1)Cy

更新4:

回myisland8132: 已經有answer, 唔難都唔post出來問...

回答 (1)

myisland8132

2007-01-22 2:54 am

✔ 最佳答案

The answer is negative binomial distribution
you can reference the following proof, although it is not actually the same of your question , but it is useful

圖片參考：http://upload.wikimedia.org/math/2/7/9/279e4307a9fc78b888f8f4ada19c972f.png

圖片參考：http://upload.wikimedia.org/math/5/3/b/53b1ee517dabe9878690e49bdefaec46.png

圖片參考：http://upload.wikimedia.org/math/9/b/5/9b5214c898fc00e6b160ea12cf9f827f.png

圖片參考：http://upload.wikimedia.org/math/0/2/5/02514c64b78adedaa918afcc8e88ee0d.png

圖片參考：http://upload.wikimedia.org/math/d/d/9/dd9fab09bab72cd76d26d92dfb14ee71.png

圖片參考：http://upload.wikimedia.org/math/a/d/e/adec7a0427b0d9539f525b498df149ed.png

2007-01-21 18:59:47 補充：
網址是http://en.wikipedia.org/wiki/Negative_binomial_distribution這裡的Negative_binomial_distribution不是離散個條

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