Probability筆記28 - Negative Binomial RV 的誤區

1. 但是小心誤區，以下這種情況不是negative binomials

Roll a 6 sided die until you have seen all of the sides as a result. Let X$X$denote the number of rolls required.

X  = X1 + X2 + X3 + X4 + X5 + X6
trial要成功的條件是“擲出過未見過的點數”，所以事實上這是一個without replacement的實驗，所以每個Xi即便是geometric RV，但是每次成功trial的機率並不相同，例如先擲出了一個點數之後，“擲出過未見過的點數”的成功機率變下降了，所以不能組成一個negative binomial（要求所有成員Xi有相同的mass)。

X如果真的要拆成六個Xi的和的話，Xi要定義成“擲出第i個未見過的點數”所需要的trial數目，所以：

X1 = 擲出第一個未見過點數的trial數目，這是一個geometric RV，p = 6/6，因為尚未擲出過任何點數

X2 = 剩下五個未見過點數中，擲出其中一個，p = 5/6
.
.
.

X6: p =1/6

所以E(X) = E(X1) + E(X2) + ... + E(X6) = 1/p1 + 1/p2 + ... = 1 + 6/5 + 6/4 + 6/3 + 6/2 + 6/1

2. 辨識真正語意

“

”

P(X+Y > 10) = ?
由於X,Y是一樣mass的independent geometric RVs，所以可以令U = X+Y為一negative binomial(2, 0.6)，則P(U>10)的語意為“前十次必須fail”，fail的定義則是不能讓U滿足2次的success，因為其r = 2。

所以前十次：
(1) all fail with probability 0.4^10
(2) 只有一次success: 0.4^9 * 0.6 * choose(10,1)


如果加上Z近來攪和，問P(X+Y+Z > 10) = ?
一樣的邏輯，令U = X+Y+Z 為一個NB(r = 3, 0.6)，所以P(U>10) 的話，前十次的trial不能出現3次success，因為其r = 3。

(1) P(all fail): 0.4^10
(2) P(1 success): 0.4^9 * 0.6 * choose(10,1)

(3) P(2 successes): 0.4^8 * 0.6^2 * choose(10,2)

3.  只有Geometric RV有memoryless特性！！！

“Suppose that 60% of people in Chicago are fans of da Bears. Assume that the fans' preferences are independent. We interview fans until we find the 3rd person who is a fan of da Bears. (This is different than the setup from question #1 of Problem Set 12. Here we need to interview at least 3 people, but in that former question, we interviewed exactly 3 people.) Let $X$ X denote the number of people we interview altogether.”

問P(X>6) = ?

X是一個negative binomial(3, 0.6)。
P(X>6)，所以前6次不能出現3個success:

(1) P(all fail) = 0.4^6
(2) P(1 success) = 0.4^5 * 0.6 * choose(6,1)
(3) P(2 success) = 0.4^4 * 0.6^2 * choose(6,2)


再問P(X>6 | X>4)= ?

這邊要注意negative binomial RV並沒有memoryless的特性！！！！
所以乖乖用conditional probability的定義去算吧！