[機統] Exponential Distribution's CDF

看板Math作者 (Carael Macchiato)時間13年前 (2011/03/01 17:51), 編輯推噓2(202)
留言4則, 3人參與, 最新討論串1/1
根據機率課本已知 Pr[X < t] = 1 - exp(-λt) 現在如果改成已知 X1 X2 分別為獨立產生的exponential random number (同λ) 請問 Pr[X1+X2 < t] 的機率該怎麼求呢 一開始想說用 Pr[X1<t]*Pr[X2<t] 後來仔細一想又覺得怪怪的 不好意思麻煩大家了 -- 《Sometimes Love Just Ain't Enough》 Yan-zi "But there's a danger in loving somebody too much. And its sad when you know it's your heart you can't trust. There's a reason why people don't stay where they are. Baby sometimes love just ain't enough." -- ※ 發信站: 批踢踢實業坊(ptt.cc) ◆ From: 140.116.177.6

03/01 18:39, , 1F
X1+X2是Gamma random variable
03/01 18:39, 1F
※ 編輯: CxMacchi 來自: 140.116.177.6 (03/01 18:46)

03/01 18:54, , 2F
謝謝樓上 我再去翻翻書!
03/01 18:54, 2F

03/01 19:03, , 3F
或者叫Erlang也可以
03/01 19:03, 3F

03/02 12:33, , 4F
謝謝,我找到Erlang的作法了 :)
03/02 12:33, 4F
文章代碼(AID): #1DRC6gMO (Math)