Type
Description
Probability Distribution
Func.
Cumulative Distribution
Func.
Expected Value, µ
Variance, 𝜎2
Std Dev = 𝜎
Finite
Random
Variable
Finite discrete-it can
take on only finitely
many possible
values (ex: X =
0,1,2, or 3). In this
case you can list all
possible values
Probability mass function
(p.m.f.), fX (x) where
fX (x) = P(X = x)
= height of p.m.f.
1)(0 xfX
sum of p.m.f. values = 1
Cumulative density
function (c.d.f.),
()
X
F x P X x
non-decreasing
max value of 1
step function
x
X
x
xfx
xXPx
XE
all
all
)(
)(
)(
𝑉(𝑋)=
∑(𝑥−𝜇)2𝑓𝑋(𝑥)
Binomia
l
Random
Variable
Binomial Setting:
1.You have n
repeated trials of an
experiment.
2. On a single trial,
there are two
possible outcomes,
success or failure.
3.The probability of
success, p , is the
same from trial to
trial.
4.The outcome of
each trial is
independent.
Probability mass function
(p.m.f.), fX (x) where
fX (x) = P(X = x)
In this case, a histogram
associated with
()P X x
Cumulative density
function (c.d.f.),
()
X
F x P X x
non-decreasing
max value of 1
step function
pnXE )(
𝑉(𝑋)=
𝑛𝑝(1− 𝑝)
Continu
ous
Random
Variable
Continuous-if the
possible values form
an entire interval of
numbers (ex: any
positive number)
Probability Density
function (p.d.f.), The
value of the p.d.f., fX (x)
()P X x
.
The value of fX (x) is
simply the height of the
density curve at the value of
x.
fX (x) > 0
( ) 0P X x
for all x
Cumulative density
function (c.d.f.),
)()( xXPxFX
)()()( aFbFbXaP XX
)(1)( xFxXP X
non-decreasing
max value of 1
continuous function,
usually
𝐸(𝑋)=
∫ 𝑥∙ 𝑓𝑋(𝑥)𝑑𝑥
∞
−∞
𝑉(𝑋)=
∫(𝑥−𝜇)2𝑓𝑋(𝑥)𝑑𝑥
Expone
ntial
Random
Variable
Exponential random
variables are
continuous random
variables and
usually describe the
waiting time
between consecutive
events.
Probability Density
function (p.d.f.),
Cumulative density
function (c.d.f.),
EX
𝑉(𝑋)= 𝛼2
Uniform
Random
Variable
If X is uniform on
the interval [a,b]
then we have a
continuous uniform
random variable
Probability Density
function (p.d.f.),
bx
bxa
ab
ax
xf X
if 0
if
1
if 0
)(
Cumulative density
function (c.d.f.),
bx if
bxa if
ab
ax
ax if
xFX
1
0
)(
2
)( ab
XE
𝑉(𝑥)
=(𝑏− 𝑎)2
12
xe
x
xf x
X0if
10if0
)( /
xe
x
xF x
X0if1
0if0
)( /