Section 3.3
If the space of a random variable X consists of discrete points, then X is
said to be a random variable of the discrete type. If the space of a
random variable X is a set S consisting of an interval (possibly
unbounded) of real numbers or a union of such intervals, then X is said
to be a random variable of the continuous type. The probability density
function (p.d.f.) of a continuous type random variable X is an integrable
function f(x) satisfying the following conditions:
(1) f(x) > 0 for x  S
(Note: for convenience, we may allow S to
contain certain discrete points where f(x) = 0.)
f(x) dx = 1
(2)
b
S
(3) P(A) = P(X  A) =
f(x) dx
A
i.e, P(a < X < b) =
f(x) dx
a
Suppose X is a continuous-type random variable with outcome space
S and p.d.f f(x).
x f(x) dx = E(X) =  .
The mean of X is
S
The variance of X is
(x – )2f(x) dx = E[(X – )2] =
S
E[X2 – 2X + 2] = E(X2) – 2E(X) + E(2) =
E(X2) – 22 + 2 = E(X2) – 2 = 2 = Var(X) .
The standard deviation of X is  = Var(X) .
Recall that for any random variable X, the cumulative distribution
x
function (c.d.f.) of X is defined to be F(x) = P(X  x). If X is a
continuous-type random variable, we may write
F(x) = f(t) dt
The Fundamental Theorem of Calculus implies F /(x) = f(x).
–
For a continuous-type random variable X with outcome space S and
p.d.f f(x), we can define the following:
etx f(x) dx
The m.g.f. of X (if it exists) is M(t) = E(etX) =
S
and (just as with discrete-type random variables) M(n)(0) = E(Xn).
The (100p)th percentile of the distribution of X, where 0 < p < 1,
is defined to be a number p such that F(p) = p.
The median of the distribution of X is defined to be 0.5 .
The first, second, and third quartiles of the distribution of X are
defined to be 0.25 , 0.5 , and 0.75 respectively.
3–y
1. A random variable Y has p.d.f. f(y) = —— if 1 < y < 3 .
2
(1,2)
Find each of the following:
1
(a) P(2 < Y < 3)
3
3
3
f(y) dy
(3 – y)
——— dy
2
=
=
3y
y2
— – —
2
4
2
2
3
=
1
—
4
y=2
(b) P(1.5 < Y < 2.5)
2.5
f(y) dy
1.5
2.5
2.5
(3 – y)
——— dy
2
=
1.5
=
3y
y2
— – —
2
4
=
y = 1.5
1
—
2
(c) P(Y < 2)
2
2
2
f(y) dy
–
(3 – y)
——— dy
2
=
=
3y
y2
— – —
2
4
1
y=1

(d) E(Y)
–
y f(y) dy
=
3
—
4
=
1
3
3
(3 – y)
y ——— dy
2
(3y – y2)
———— dy =
2
=
1
3
3y2
y3
— – —
4
6
=
y=1
5
—
3

(e) Var(Y)
E(Y2)
=
3
y2
f(y) dy
=
–
y2
(3 – y)
——— dy =
2
1
3
3
(3y2 – y3)
———— dy =
2
3y3
y4
— – —
6
8
1
Var(Y) =
=
y=1
3 –
5
—
3
2
2
= —
9
3
(f) the m.g.f. (moment generating function) of Y
3

(3 – y)
tY
ty
ty
M(t) = E(e ) =
e f(y) dy =
e ——— dy =
2
–
1
3
3y
y2
— – —
2
4
=
1
if t = 0
3
–
————— dy
2
(3ety
yety)
y=1
=
3
1
3tety – (ytety – ety)
———————
2t2
e3t – (2t + 1)et
= ——————
2t2
y=1
if t  0
(g) the c.d.f. (cumulative distribution function) of Y
y
F(y) = P(Y  y) =
f(t) dt
=
–
0
if y < 1
3y
y2
5
= — – — – —
2
4
4
if 1  y < 3
y
y
(3 – t)
——— dt =
2
3t
t2
— – —
2
4
1
t=1
if 3  y
1
1
1/2
1
0
1
2
3
(h) the quartiles of the distribution of Y
F(0.25) = P(Y  0.25) = 0.25
3
2
5
— – — – — = 0.25
2
4
4
6 – 2 – 5
= 1
6 – 2 – 6
= 0
2 – 6 + 6
= 0
From the quadratic formula, we find
 = 3 – 3  1.268
or
 = 3 + 3  4.732
Considering the space of Y, we see that 0.25 = 3 – 3  1.268
F(0.50) = P(Y  0.50) = 0.5
3
2
5
— – — – —
2
4
4
= 0.5
6 – 2 – 5
= 2
6 – 2 – 7
= 0
2 – 6 + 7 = 0
From the quadratic formula, we find
 = 3 – 2  1.586
or
 = 3 + 2  4.414
Considering the space of Y, we see that 0.50  1.586
F(0.75) = P(Y  0.75) = 0.75
3
2
5
— – — – —
2
4
4
= 0.75
6 – 2 – 5
= 3
6 – 2 – 8
= 0
2 – 6 + 8 = 0
From the quadratic formula, we find  = 2 or  = 4.
Considering the space of Y, we see that 0.75 = 2
2. A random variable X has p.d.f. f(x) =
Find each of the following:
(a) P(–0.5 < X < 0.5)
0.5
0.5
f(x) dx =
– 0.5
– 0.5
1
— dx =
6
1/6
if –2 < x < 1
1/2
if 1  x < 2
(1, 1/2)
(–2, 1/6)
–2
0.5
x
1
— = —
6
6
x = –0.5
1
2
(b) P(X > 0.75)

2
f(x) dx =
0.75
2
1
— dx +
6
1
— dx =
2
f(x) dx =
0.75
1
x
—
6
1
0.75
1
2
+
x = 0.75
x
—
2
x=1
=
1
—
24
+
1
—
2
=
13
—
24
(c) P(0.75 < X < 1.25)
1.25
1
f(x) dx =
0.75
0.75
1.25
1
— dx +
6
1
1
— dx = —
2
6
1
(d) E(X)

x f(x) dx
–
=
1
2
x
— dx +
6
x
— dx =
2
–2
1
x2
—
12
1
2
+
x=–2
x2
—
4
x=1
=
–3
—
12
+
3
—
4
=
1
—
2

(e) Var(X)
E(X2) =
x2 f(x) dx
=
–
–2
1
x3
—
18
1
x2
— dx +
6
2
x2
— dx =
2
1
2
+
x=–2
x3
—
6
=
1
—
2
x=1
2
Var(X) =
+
5
1
— – —
3
2
17
= —
12
7
—
6
=
5
—
3
(f) the m.g.f. (moment generating function) of X

1
2
tx
tx
e
e
M(t) = E(etX) =
etx f(x) dx =
— dx +
— dx =
6
2
–
–2
1
1
2
x
—
6
x
—
2
+
x=–2
1
if t = 0
x=1
1
etx
—
6t
=
2
+
x=–2
etx
—
2t
x=1
3e2t – 2et – e–2t
= —————— if t  0
6t
(g) the c.d.f. (cumulative distribution function) of X
x
F(x) = P(X  x) =
f(t) dt
–
0
=
x
x
1
— dt =
6
–2
1
1
— dt +
6
–2
1
x
if x < – 2
t
—
6
x+2
= —— if – 2  x < 1
6
t=–2
x
1
1
t
— dt = — + —
2
2
2
x
= —
2
if 1  x < 2
t=1
1
if 2  x
(g) the c.d.f. (cumulative distribution function) of X
if x < – 2
0
x
F(x) = P(X  x) =
f(t) dt
–
=
x+2
—— if – 2  x < 1
6
x
—
2
if
1
if 2  x
1x<2
1
1/2
2
1
0
1
2
(h) the quartiles of the distribution of X
P(X  0.25) = F(0.25) = 0.25
+2
——–
6
 + 2
= 0.25
= 3/2
0.25 = –1/2
Obviously, 0.50 =
1
P(X  0.75) = F(0.75) = 0.75

— = 0.75
2
0.75 = 3/2