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1
Content
 Introduction
 Expectation and variance of continuous random




variables
Normal random variables
Exponential random variables
Other continuous distributions
The distribution of a function of a random variable
2
5.1 Introduction
 Let X be such a random variable. We say that X is a
continuous random variable if there exits a nonenegative
function f, defined for all real 𝑥 ∈ (−∞, ∞), having the
property that, for any set B of real numbers
𝑃 𝑋∈𝐵 =
𝑓 𝑥 𝑑𝑥
(1.1)
𝐵
The function f is called the probability density function of
the random variable X.
 All probability statements about X can be answered in
terms of f. Letting B=[a,b], we obtain
𝑏
𝑃 𝑎≤𝑥≤𝑏 =
𝑓 𝑥 𝑑𝑥
(1.2)
𝑎
3
Example 1a
 Suppose that X is a continuous random variable whose
probability density function is given by
2
𝐶
4𝑥
−
2𝑥
0
<
𝑥
<
2
𝑓 𝑥 =
0
𝑜𝑡ℎ𝑒𝑟𝑤𝑖𝑠𝑒
(a) What is the value of C?
(b) Find P{X>1}.
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Solution. 1a
 We must have
∞
𝑓
−∞
2
𝑥 𝑑𝑥 = 1
4𝑥 − 2𝑥 2 𝑑𝑥 = 1
𝐶
0
𝐶 2𝑥 2 −
3
3
𝐶=
8
𝑃 𝑋>1 =
∞
𝑓
1
𝑥=2
2𝑥 3
𝑥 𝑑𝑥 =
=1
𝑥=0
3 2
8 1
2
4𝑥 − 2𝑥 𝑑𝑥 =
1
2
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The cumulative distribution F
 The relationship between the cumulative distribution F and the
probability density f is expressed by
𝐹 𝑎 = 𝑃 𝑋 ∈ −∞, 𝑎
𝑎
=
𝑓 𝑥 𝑑𝑥
−∞
Differentiating both sides of the preceding equation yields
𝑑
𝐹 𝑎 = 𝑓(𝑎)
𝑑𝑥
That is, the density is the derivative of the cumulative distribution
function. A somewhat more intuitive interpretation of the density
function may be obtained from (1.2) 𝑎+𝜖/2
as follows:
𝜖
𝜖
𝑃 𝑎− ≤𝑋≤𝑎+ =
𝑓 𝑥 𝑑𝑥 ≈ 𝜖𝑓(𝑎)
2
2
𝑎−𝜖/2
when e is small and when f(.) is continuous at x=a. The probability that
X will be contained in an interval of length e around the point a is
approximately ef(a). From this result we see that f(a) is a measure of
how likely it is that the random variable will be near a.
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Example 1b
 The amount of time in hours that a computer
functions before breaking down is a continuous
random variable with probability density function
given by
𝜆𝑒 −𝑥/100 , 𝑥 ≥ 0
𝑓 𝑥 =
0,
𝑥<0
 What is the probability that
 (a) A computer will function between 50 and 150 hours
before breaking down?
 (b) It will function for fewer than 100 hours?
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Solution. 1b
 (a) Since
∞
1=
∞
−∞
 We obtain
1=
𝑒 −𝑥/100 𝑑𝑥
𝑓 𝑥 𝑑𝑥 = 𝜆
𝑥
0
−𝜆(100)𝑒 −100 | ∞0
= 100𝜆 or 𝜆 =
1
100
 Hence, the probability that a computer will function
between 50 and 150 hours before breaking down is given by
150
𝑥
1 −𝑥/100
−100 150
𝑃 50 < 𝑋 < 150 =
𝑒
𝑑𝑥 = −𝑒
| 50
50 100
=
1
−2
𝑒
−
3
−2
𝑒
≈ 0.384
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Solution. 1b
 (b) Similarly,
100
𝑥
1 −𝑥/100
100
−
100
𝑃 𝑋 < 100 =
𝑒
𝑑𝑥 = −𝑒
| 0
100
0
= 1 − 𝑒 −1 ≈ 0.633
 In other words, approximately 63.3 percent of the time,
a computer will fail before registering 100 hours of use.
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Example 1d
 If X is continuous with distribution function FX and
density function fX, find the density function of Y=2X.
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Solution. 1d
 First way
 Derive
 Differrentiate
 Derive
𝑎
𝑎
𝐹𝑌 𝑎 = 𝑃 𝑌 ≤ 𝑎 = 𝑃 2𝑋 ≤ 𝑎 = 𝑃 𝑋 ≤
= 𝐹𝑋 ( )
2
2
 Differentiation
1
𝑎
𝑓𝑌 𝑎 = 𝑓𝑋 ( )
2
2
 Another way to determine fY is to note that
𝜖
𝜖
𝜖𝑓𝑌 𝑎 ≈ 𝑃 𝑎 − ≤ 𝑋 ≤ 𝑎 + =
2
𝜖
𝜖
𝑎2 𝜖
𝑎
𝜖
𝜖
𝑎
𝑃 𝑎 − ≤ 2𝑌 ≤ 𝑎 + = 𝑃{ − ≤ 𝑋 ≤ + } ≈ 𝑓𝑋 ( )
2
2
2
4
2
4
2
2
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5.2 Expectation and variance of
continuous random variables
 Expected value
∞
𝐸𝑋 =
𝑥𝑓 𝑥 𝑑𝑥
−∞
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Example 2a
 Find E[X] when the density function of X is
2𝑥,
𝑓 𝑥 =
0,
𝑖𝑓 0 ≤ 𝑥 ≤ 1
𝑜𝑡ℎ𝑒𝑟𝑤𝑖𝑠𝑒
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Solution. 2a
𝐸 𝑋 =
𝑥𝑓 𝑥 𝑑𝑥 =
1
2 𝑑𝑥
2𝑥
0
= 2/3
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Example 2b
 The density function of X is given by
𝑓 𝑥 =
Find E[eX].
1 𝑖𝑓 0 ≤ 𝑥 ≤ 1
0
𝑜𝑡ℎ𝑒𝑟𝑤𝑖𝑠𝑒
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Solution. 2b
 Let Y=eX
The probability distribution function of Y
𝐹𝑌 𝑥 = 𝑃 𝑌 ≤ 𝑥 = 𝑃 𝑒 𝑋 ≤ 𝑥 = 𝑃 𝑋 ≤ log 𝑥
log(𝑥)
=
𝑓 𝑦 𝑑𝑦 = log(𝑥)
0
By differentiating FY(x), we can conclude that the
probability density function of Y is given by
1
𝑓𝑌 𝑥 =
1≤𝑥≤𝑒
𝑥
Hence,
𝐸 𝑒𝑋 = 𝐸 𝑌 =
∞
𝑒
𝑥𝑓𝑌 𝑥 𝑑𝑥 =
−∞
𝑑𝑥 = 𝑒 − 1
1
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Proposition 2.1
 If X is a continuous random variable with probability
density function f(x), then, for any real-valued
function g,
∞
𝐸𝑔 𝑥
=
𝑔 𝑥 𝑓 𝑥 𝑑𝑥
−∞
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Lemma 2.1
 For a nonnegative random variable Y,
∞
𝐸𝑌 =
𝑃 𝑌 > 𝑦 𝑑𝑦
0
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Example 2c
 A stick of length 1 is split at a point U that is uniformly
distributed over (0,1). Determine the expected length
of the piece that contains the point p, 0≦p≦1.
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Solution. 2c
 Let Lp(U) denote the length of the substick that contains
the point p, and note that
1−𝑈 𝑈 <𝑝
𝐿𝑝 𝑈 =
𝑈
𝑈>𝑝
Hence, from Proposition
2.1
1
𝑝
𝐸 𝐿𝑝 𝑈
=
0
𝐿𝑝 𝑢 𝑑𝑢 =
2
1
1 − 𝑢 𝑑𝑢 +
0 2
𝑝
𝑢𝑑𝑢
𝑝
1
1−𝑝
1
1
= −
+ −
= + 𝑝(1 − 𝑝)
2
2
2 2
2
 Since p(1-p) is maximized when p=1/2, it is interesting to
note that the expected length of the substick containing
the point p is maximized when p is the midpoint of the
original stick.
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Solution. 2c

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Corollary 2.1
 If a and b are constants, then
E[aX+b]=aE[X]+b
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Example 2e
 Find Var(X) for X as given in Example 2a.
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Solution. 2e
 We first compute E[X2].
𝐸
𝑋2
∞
=
−∞
𝑥 2𝑓
1
2𝑥 3 𝑑𝑥
𝑥 𝑑𝑥 =
0
1
=
2
Hence, since E[X]=2/3, we obtain
2
1
2
1
𝑉𝑎𝑟 𝑋 = −
=
2
3
18
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variance
 If a and b are constants, then
Var(ax+b)=a2Var(x)
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5.3 The uniform random variable
 A random variable is said to be uniformly distributed
over the interval (0,1) if its probability density function
is given by
1 0<𝑥<1
𝑓 𝑥 =
(3.1)
0 𝑜𝑡ℎ𝑒𝑟𝑤𝑖𝑠𝑒
1
𝛼<𝑥<𝛽
𝑓 𝑥 = 𝛽−𝛼
(3.2)
0
𝑜𝑡ℎ𝑒𝑟𝑤𝑖𝑠𝑒
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The uniform random variable
 Since 𝐹 𝑎 =
𝑎
𝑓(𝑥),
−∞
the distribution function of a
uniform random variable on the interval (𝛼, 𝛽) is given
by
0,
𝑎≤𝛼
𝑎−𝛼
,
𝛼<𝑎<𝛽
𝐹 𝑎 =
𝛽−𝛼
1,
𝑎≥𝛼
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The uniform random variable

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Example 3a
 Let X be uniformly distributed over (a,b). Find (a) E[X]
and (b) Var[X]
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Solution. 3a
 (a)𝐸 𝑋 =
∞
𝑥𝑓
−∞
𝑥
𝛽 𝑥
𝑑𝑥 = 𝛼
𝑑𝑥
𝛽−𝛼
𝛽2 − 𝛼 2
𝛽+𝛼
=
=
2(𝛽 − 𝛼)
2
the expected value of a random variable that is uniformly
distributed over some interval is equal to the midpoint of
that interval
 (b) first calculate E[X2]
𝛽
3
3
2
2
1
𝛽
−
𝛼
𝛽
+
𝛼𝛽
+
𝛼
𝐸 𝑋2 =
𝑥 2 𝑑𝑥 =
=
3(𝛽 − 𝛼)
3
𝛼 𝛽−𝛼
Hence
𝛽2 + 𝛼𝛽 + 𝛼 2
𝛼 + 𝛽 2 (𝛽 − 𝛼)2
𝑉𝑎𝑟 𝑋 =
−
=
3
4
12
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Example 3b
 If X is uniformly distributed over (0,10), calculate the
probability that (a)X<3, (b)X>6, and (c)3<X<8.
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Solution. 3b
 (a) 𝑃 𝑋 < 3 =
 (b) 𝑃 𝑋 > 6 =
 (c) 𝑃 3 < 𝑋 < 8
3 1
3
𝑑𝑥
=
0 10
10
10 1
4
𝑑𝑥 =
6 10
10
8 1
1
= 3 𝑑𝑥 =
10
2
35
Example 3c
 Buses arrive at a specified stop at 15-minute intervals
starting at 7 A.M. That is, they arrive at 7, 7:15, 7:30,
7:45, and so on. If a passenger arrives at the stop at a
time that is uniformly distributed between 7 and 7:30,
find the probability that he waits
(a) less than 5 minutes for a bus
(b) more than 10 minutes for a bus
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Solution. 3c
 Let X denote the number of minutes past 7 that the
passenger arrives at the stop.
 (a) 𝑃 10 < 𝑋 < 15 + 𝑃 25 < 𝑋 < 30 =
30 1
𝑑𝑥
25 30
=
1
3
 (b) 𝑃 0 < 𝑋 < 5 + 𝑃 15 < 𝑋 < 20 =
20 1
𝑑𝑥
15 30
=
1
3
15 1
𝑑𝑥
10 30
5 1
𝑑𝑥
0 30
+
+
37
Example 3d – Bertrand’s paradox
 Consider a random chord of a circle. What is the
probability that the length of the cord will be greater
than the side of the equilateral triangle inscribed in
that circle?
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