Download Week 8: Expected value of a binomial distribution.

Expected Value of a Binomial Distribution
Arthur White∗
14th November 2016
Recall that we say a random variable X ∼ Binom(n, π) follows a binomial distribution
if n independent trials occur, with a constant probability of success P(Success) = π, and X
corresponds to the total number of observed successes. In particular, then
n x
P(X = x) = P(x) =
π (1 − π)n−x ,
x = 0, . . . , n.
x
Then the expected value of X, E[X], is found to be:
E[X] =
n
X
xP(x)
x=0
n
X
n x
=
x
π (1 − π)n−x
x
x=1
=
n
X
x=1
= nπ
n!
π x (1 − π)n−x
(x − 1)!(n − x)!
n
X
x=1
n−1
X
(n − 1)!
π x−1 (1 − π)n−x
(x − 1)!(n − x)!
(n − 1)!
π y (1 − π)n−1−y
y!(n
−
1
−
y)!
y=0 |
{z
}
Binom(n − 1, π)
= nπ.
= nπ
Here we have substituted y = x − 1, and,Pnoticing that, for the problem at hand Y ∼
n−1
Binom(n − 1, π), made use of the fact that y=0
P(Y = y) = 1.
∗
Based extensively on material previously taught by Eamonn Mullins.
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Variance of a Binomial Distribution
Next we calculate the variance, Var[X]. We will do so in a somewhat indirect way:
E[X(X − 1)] =
n
X
x(x − 1)P(x)
x=0
n
X
n x
=
x(x − 1)
π (1 − π)n−x
x
x=2
=
n
X
x=2
n!
π x (1 − π)n−x
(x − 2)!(n − x)!
= n(n − 1)π
2
= n(n − 1)π 2
n
X
x=2
n−2
X
(n − 2)!
π x−2 (1 − π)n−x
(x − 2)!(n − x)!
(n − 2)!
π y (1 − π)n−2−y
y!(n
−
2
−
y)!
y=0 |
{z
}
Binom(n − 2, π)
= n(n − 1)π 2 .
This time we have substituted y = x − 2, and, again noticed that one of the terms on the
right involves the summation of a probability distribution over its sample space.
Recall that an alternative form for the variance is Var[X] = E[X 2 ] − E[X]2 . Then we can
write:
Var[X] = E[X 2 ] − E[X]2
= E[X 2 ] − E[X] +E[X] − E[X]2
|
{z
}
=E[X(X−1)]
= n(n − 1)π 2 + nπ − n2 π 2
= nπ − nπ 2
= nπ(1 − π).
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