Summary of Specific Discrete R.V.’s
and their Distributions
1. Bernoulli
– Setting: Probability experiment with two possible oucomes: Success
(S) or Failure (F)
– Random Variable:X = 1 if success; X = 0 otherwise. X ∼ Ber(p)
– Sample Space: Im(X) = {0, 1}
– PMF:
(
px (1 − p)1−x if x ∈ {0, 1}
pX (x; p) =
0 otherwise
– CDF:

0 if x < 0

P
bxc x
B1,p (x) =
p (1 − p)1−x if 0 ≤ x < 1
i=0


1 if x ≥ 1
– Mean and Variance: E[X] = p, V ar(X) = p(1 − p)
2. Binomial
– Setting: Count the number of successes in a sequence of n iid Bernoulli
trials.
– Random Variable: X = number of successes in a sequence of n iid
Bernoulli trials. X ∼ Bin(n, p)
– Sample Space: Im(X) = {0, . . . , n}
– PMF:
( n
pX (x; n, p) =
px (1 − p)1−x if x ∈ {0, . . . , n}
0 otherwise
x
– CDF:


0 if x < 0
P
bxc n x
Bn,p (x) =
p (1 − p)1−x if 0 ≤ x < n
i=0
x


1 if x ≥ n
– Mean and Variance: E[X] = np, V ar(X) = np(1 − p)
3. Geometric
– Setting: Perform iid Bernoulli experiments until a success occurs.
– Random Variable: X = trial on which the first success occurs.
X ∼ Geo(p)
– Sample Sapce: Im(X) = {1, 2, 3, . . .}
– PMF:
pX (x; p) =
(
(1 − p)x−1 p if x ∈ {1, . . . , }
0 otherwise.
– CDF:
(
0 if x < 1
Geop (x) =
1 − (1 − p)bxc if x ≥ 1
– Mean and Variance: E[X] = p−1 , V ar(X) =
1−p
p2
4. Poisson
– Setting: Number of occurrences of a relatively rare event, where the
probability of the event is constant across time intervals or spatial
regions
– Random Variable: X = number of events. X ∼ P o(λ).
– Sample Space: Im(X) = {0, 1, 2, . . .}
– PMF:
(
pX (x; λ) =
λx exp(−λ)
x!
if x ∈ {0, 1, . . . , }
0 otherwise.
– CDF:
(
0 if x < 0
P oλ (x) = Pbxc λx exp(−λ)
i=0
x!
if x ≥ 0
– Mean and Variance: E[X] = λ, V ar(X) = λ
Important Connection Between Bernoulli Random
Variables and Binomial Random Variables
• A Binomial r.v. can be represented as the sum of iid Bernoulli
r.v.’s:
– Let X1, . . . Xn be iid Bernoulli r.v.’s. For i = 1 . . . , n,
(
1 if Success
Xi =
0 otherwise.
P (Xi = 1) = p = 1 − P (Xi = 0)
– Let X = X1 + X2 + . . . + Xn.
– Then, X ∼ Bin(n, p).
– Verify, using the representation of X as a sum of iid Bernoulli
r.v.’s, that E[X] = np and V ar(X) = np(1 − p):
∗ By linearity of expectation, E[X] = E[X1] + E[X2] + . . . +
E[Xn] = p + p + . . . + p = np
∗ By independence, V ar(X) = V ar(X1) + V ar(X2) + . . . +
V ar(Xn) = p(1 − p) + p(1 − p) + . . . + p(1 − p) = np(1 − p)
– Note: A Bernoulli r.v. is a Binomial r.v. with n = 1.
Modified Geometric
• The Geometric r.v. tells us the trial on which the first success
occurs in a sequence of iid Bernoulli experiments. Alternatively,
we could keep track of the number of failures before the first
success.
• Let Y = number of failures before the first success in a sequence
of iid Bernoulli experiments, each with success probability p. We
will say that Y has a “Modified Geometric” distribution.
• If X ∼ Geo(p), then Y = X − 1.
• What is Im(Y )?
Im(Y ) = {0, 1, . . .}.
• What is the pmf of Y ?
P (Y = k) = P (X = k + 1) = (1 − p)k+1−1p = (1 − p)k p, for k = 0, 1, . . .
• What is E[Y ]? What is V ar(Y )?
1
1−p
−1=
p
p
1−p
V ar(Y ) = V ar(X − 1) = V ar(X) = 2
p
E[Y ] = E[X − 1] = E[X] − 1 =