Topics in Mathematics 201-BNJ-05
Vincent Carrier
Continuous Random Variables
Let X be a random variable taking values in some set DX . If DX is an uncountable set,
typically an interval, then we say that X is a continuous random variable.
Example: The following random variables X are continuous random variables.
a) X : waiting time at a bank
DX = [0, ∞)
b) X : height of a random individual
DX = R
To each random variable X is associated a density function f : DX → [0, ∞) that
satisfies the conditions
Z
f (x) dx = 1.
1) f (x) ≥ 0 for x ∈ DX
and
2)
DX
The second condition shows that with a continuous random variable, a probability corresponds to an area under the curve. Thus,
Z
f (x) dx
for A ⊂ DX .
P (X ∈ A) =
A
Example: Let X be a random variable with density function
f (x) = c(3x2 − x3 )
for x ∈ [0, 3].
a) Find c.
3
Z 3
x4
81
27
2
3
3
= c 27 −
− (0 − 0) = c
= 1.
(3x − x ) dx = c x −
c
4 0
4
4
0
Therefore,
c=
4
.
27
b) Find P (1 ≤ X ≤ 2).
4
P (1 ≤ X ≤ 2) =
27
Z
1
2
2
4
x4
3
(3x − x ) dx =
x −
27
4 1
2
3
4
1
4 13
13
=
(8 − 4) − 1 −
=
·
=
.
27
4
27 4
27
The graph of the density function is given below.
y
6
0.6
0.4
0.2
-
0
1
2
3
x
Remark: If X is a continuous random variable, then P (X = x) = 0.
Thus,
P (1 ≤ X ≤ 2) = P (1 ≤ X < 2) = P (1 < X ≤ 2) = P (1 < X < 2).
The following definitions are similar to the discrete case, with summation signs being
replaced by integrals.
Z
2
xf (x) dx,
E(X) =
Z
E(X ) =
x2 f (x) dx,
DX
DX
Example: Find µ = E(X), σ 2 = Var(X), and σ =
4
E(X) =
27
Z
4
E(X ) =
27
Z
2
r
3
0
3
p
Var(X).
3
4 3x4 x5
4 35 35
9
(3x − x ) dx =
−
= 3
−
=
27 4
5 0
3
4
5
5
3
0
18
Var(X) =
−
5
σ =
Var(X) = E(X 2 ) − [E(X)]2 .
4
3
4 3x5 x6
4 36 36
18
(3x − x ) dx =
−
= 3
−
=
27 5
6 0
3
5
6
5
4
5
2
9
9
=
5
25
9
3
= .
25
5