Solution Found!
a We observe a sequence of independent identical trials
Chapter 3, Problem 101E(choose chapter or problem)
Problem 101E
a We observe a sequence of independent identical trials with two possible outcomes on each trial, S and F, and with P(S) = p. The number of the trial on which we observe the fifth success, Y, has a negative binomial distribution with parameters r = 5 and p. Suppose that we observe the fifth success on the eleventh trial. Find the value of p that maximizes P(Y = 11).
b Generalize the result from part (a) to find the value of p that maximizes P(Y = y0) when Y has a negative binomial distribution with parameters r (known) and p.
Questions & Answers
QUESTION:
Problem 101E
a We observe a sequence of independent identical trials with two possible outcomes on each trial, S and F, and with P(S) = p. The number of the trial on which we observe the fifth success, Y, has a negative binomial distribution with parameters r = 5 and p. Suppose that we observe the fifth success on the eleventh trial. Find the value of p that maximizes P(Y = 11).
b Generalize the result from part (a) to find the value of p that maximizes P(Y = y0) when Y has a negative binomial distribution with parameters r (known) and p.
ANSWER:
Solution:
Step 1 of 2:
a)
We have a sequence of independent identical trials with two possible outcomes on each trial, S and F, with P(S) = p.
Let, Y follows the Negative Binomial distribution
P(Y = y) =
Where, r = 5
The claim is to find the value of p that maximizes P(Y = 11)
Let, P(Y = y) =
= 210
Take log on both sides
log( ) = 5 log(p) + 6 log(
)
Differentiate the above equation
[ 5 log(p) + 6 log(q)] =
-
The function is maximized when the first derivative is zero
-
= 0
y =
11 =
p =
Hence, function is maximized when the first derivative is zero and the value of p is