Derive mean and standard deviation of binomial distribution
Answers
Answer:
Step-by-step explanation:
Mean and Variance of Binomial Random Variables
The probability function for a binomial random variable is b(x; n, p) = nxpx(1 − p)n−x
This is the probability of having x successes in a series of n independent trials when the probability of success in any one of the trials is p. If X is a random variable with this probability distribution,
n
E ( X ) = x nx p x ( 1 − p ) n − x
x=0 n
= x
x=0 n
= x=1
n! px(1 − p)n−x x!(n−x)!
n! (x−1)!(n−x)!
px(1 − p)n−x
sincethex=0termvanishes. Lety=x−1andm=n−1. Subbingx=y+1andn=m+1 into the last sum (and using the fact that the limits x = 1 and x = n correspond to y = 0 and y = n − 1 = m, respectively)
m
E(X)= (m+1)! py+1(1−p)m−y
y!(m−y)! y=0
m
= (m + 1)p y=0
m
= np y=0
m! py (1 − p)m−y y!(m−y)!
The binomial theorem says that
(a+b)m =
m! y!(m−y)!
py (1 − p)m−y
m
y!(m−y)!
m! aybm−y
Setting a = p and b = 1 − p
mm
m! y!(m−y)!
y=0
m! y!(m−y)!
py(1−p)m−y = y=0
aybm−y =(a+b)m =(p+1−p)m =1
y=0
⃝c Joel Feldman. 2000. All rights reserved.
1
so that
Similarly, but this time using y = x − 2 and m = n − 2
n
EX(X − 1) = x(x − 1)nxpx(1 − p)n−x
So the variance of X is
x=0 n
= x(x − 1)
x=0 n
= n! (x−2)!(n−x)!
x=2
= n(n − 1)p2
x=2 m
= n(n − 1)p2 y=0
(n−2)! px−2(1 − p)n−x m! py (1 − p)m−y
E(X) = np
n! px(1 − p)n−x x!(n−x)!
px(1 − p)n−x
n
(x−2)!(n−x)!
y!(m−y)!
= n(n − 1)p2p + (1 − p)m
= n(n − 1)p2
E(X2)−E(X)2 =EX(X−1)+E(X)−E(X)2 =n(n−1)p2 +np−(np)2
=
np(1 − p)