] > MATH 321: Solutions for Homework 3

MATH 321 - Probability & Statistical Inference



Homework 3 Solution

  1. The formula C k n ⁢ p k ⁢ q n − k suffices to compute the probability associated with any integer output k from 0 … n of the random variable Y. However, in the case of event sets similar to Y < n − 2 , it becomes necessary to write the probability as P⁡ Y < n − 2 = 1 − P⁡ Y ≥ n − 2 .

    Specifically, the event Y < n − 2 ∩ Y ≤ n − 1 requires us to compute P⁡ Y ≤ n − 3 = 1− P⁡ n − 2 ≤ Y ≤ n = 1 − p n − n ⁢ p n − 1 ⁢ q − ½ ⁢ n ⁢ n − 1 ⁢ p n − 2 ⁢ q 2 . This is the numerator of the formula proposed for this exercise. The denominator is just P ⁡ Y ≤ n − 1 = 1 − P ⁡ Y = n = 1 − p n . The fraction whose terms we have just mentioned is the conditional probability that is described in the formula.


  2. Because S is a binomial ⁡ n p random variable, we can use the previous calculations E ⁡ S 2 = n ⁢ n − 1 ⁢ p 2 + n ⁢ p and E ⁡ S = n ⁢ p .

    The calculation for expected value of the three-factor product is:

    E ⁡ S ⁢ S − 1 ⁢ S − 2 = ∑ k = 0 n ⁡ k ⁢ k − 1 ⁢ k − 2 ⁢ C n k ⁢ p k ⁢ q n − k = ∑ k = 3 n ⁡ k ⁢ k − 1 ⁢ k − 2 ⁢ n ⁢ n − 1 ⁢ n − 2 ⁢ n − 3 ! k ⁢ k − 1 ⁢ k − 2 ⁢ k − 3 ! ⁢ n − k ! ⁢ p 3 ⁢ p k − 3 ⁢ q n − 3 − k − 3 = n ⁢ n − 1 ⁢ n − 2 ⁢ p 3 ⁢ ∑ k = 3 n ⁡ n − 3 ! k − 3 ! ⁢ n − k ! ⁢ p k − 3 ⁢ q n − 3 − k − 3 .

    The sum is then re-indexed using l=k−3. Previously, k varied from 3 up to n. Now the indexing variable l runs from 0 up to n−3, and the calculation becomes:

    E ⁡ S ⁢ S − 1 ⁢ S − 2 = n ⁢ n − 1 ⁢ n − 2 ⁢ p 3 ⁢ ∑ l = 0 n − 3 ⁡ n − 3 ! l ! ⁢ n − 3 − l ! ⁢ p l ⁢ q n − 3 − l   = n ⁢ n − 1 ⁢ n − 2 ⁢ p 3 ⁢ ∑ l = 0 n − 3 ⁡ C l n − 3 ⁢ p l ⁢ q n − 3 − l = n ⁢ n − 1 ⁢ n − 2 ⁢ p 3 .

    The last equality comes about because the previous sum is the total of the n−2 probabilities associated with a binomial ⁡ n − 3 p random variable. That sum is one. Therefore

    E ⁡ S 3 = E ⁡ S ⁢ S − 1 ⁢ S − 2 + 3 ⁢ E ⁡ S 2 − 2 ⁢ E ⁡ S = n ⁢ n − 1 ⁢ n − 2 ⁢ p 3 + 3 ⁢ n ⁢ n − 1 ⁢ p 2 + n ⁢ p .