CS 485: Mathematical Foundations for the Information Age
Lecture 11 ▪ February 11, 2009
Jihai Chen (jc628) and Jeff Pankewicz (jhp36)
A Review on Probability Distributions
:
1) Binomial Distribution:
The Binomial distribution models the probability of k successes in n independent trials, with
success in each trial having probability p.
Pr(k successes) = C(n,k) p
k
(1p)
nk
(we want exactly k successes, and the rest fail)
Deriving the formula for expected value (E[k]) and variance (
s
2
) of a Binomial distribution:
E[k] = k*Prob(k) for all values of k =
S
k=0~n
k*C(n,k) p
k
(1p)
nk
To simplify, consider (p+q)
n
=
S
k=0~n
C(n,k)p
k
q
nk
. Taking derivative of both sides w.r.t “p”, we
get:
n(p+q)
n1
= (1/p)
S
k=0~n
k*C(n,k) p
k
(1p)
nk
= (1/p)*E[k]
now if we set q=1p, then (p+1p)
n1
=1, so the LHS becomes just n, and we have:
n=(1/p)E[k]=> E[k]= np
Now, to get the variance, we take the second derivative of the equation
(p+q)
n
=
S
k=0~n
C(n,k)p
k
q
nk
:
n(n1)(p+q)
n2
=
S
k=0~n
k(k1) C(n,k) p
k2
q
nk
sub in q=1p, and multiply both sides by p
2
, we get
p
2
n(n1) =
S
k=0~n
k(k1) C(n,k) p
k
q
nk
Factor out the right hand side with k
2
and k, we have
p
2
n(n1) =
S
k=0~n
k
2
C(n,k) p
k
q
nk

S
k=0~n
kC(n,k) p
k
q
nk
We realize that the first term of the RHS is
E[k
2
] , and the second term is E[k] = np, moving the second term on the right to the left, we
receive:
E[k
2
]= p
2
n(n1)+np
Now we also note that