lecture12
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MODULE 3
FUNCTION OF A RANDOM VARIABLE AND ITS
DISTRIBUTION
LECTURES 12-16
Topics
3.1 FUNCTION OF A RANDOM VARIABLE
3.2 PROBABILITY DISTRIBUTION OF A FUNCTION OF
A RANDOM VARIABLE
3.3 EXPECTATION AND MOMENTS OF A RANDOM
VARIABLE
3.4 PROPERTIES OF RANDOM VARIABLES HAVING
THE SAME DISTRIBUTION
3.5 PROBABILITY AND MOMENT INEQUALITIES 3.5.1 Markov Inequality 3.5.2 Chebyshev Inequality
3.5.3 Jensen Inequality 3.5.4 ๐จ๐ด-๐ฎ๐ด-๐ฏ๐ด inequality
3.6 DESCRIPTIVE MEASURES OF PROBABILITY
DISTRIBUTIONS 3.6.1 Measures of Central Tendency
3.6.1.1 Mean 3.6.1.2 Median 3.6.1.3 Mode
3.6.2 Measures of Dispersion
3.6.2.1 Standard Deviation 3.6.2.2 Mean Deviation 3.6.2.3 Quartile Deviation 3.6.2.4 Coefficient of Variation
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3.7 MEASURES OF SKEWNESS
3.8 MEASURES OF KURTOSIS
MODULE 3
FUNCTION OF A RANDOM VARIABLE AND ITS
DISTRIBUTION
LECTURE 12
Topics
3.1 FUNCTION OF A RANDOM VARIABLE
3.2 PROBABILITY DISTRIBUTION OF A FUNCTION OF
A RANDOM VARIABLE
3.1 FUNCTION OF A RANDOM VARIABLE
Let ๐บ,โฑ,๐ be a probability space and let ๐ be random variable defined on ๐บ,โฑ,๐ .
Further let โ:โ โ โ be a given function and let ๐:๐บ โ โbe a function of random
variable ๐, defined by ๐ ๐ = โ ๐ ๐ ,๐ โ ๐บ. In many situations it may be of interest
to study the probabilistic properties of ๐, which is a function of random variable ๐. Since
the variable ๐ takes values in โ, to study the probabilistic properties of ๐, it is necessary
that ๐โ1 ๐ต โ โฑ,โ๐ต โ โฌ1, i.e., ๐ is a random variable. Throughout, for a positive integer
๐,โ๐ will denote the ๐-dimensional Euclidean space and โฌ๐ will denote the Borel sigma-
field in โ๐ .
Definition 1.1 Let ๐ and ๐ be positive integers. A function โ:โ๐ โ โ๐ is said to be a Borel function if
โโ1 ๐ต โ โฌ๐ ,โ๐ต โ โฌ๐ . โ
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The following lemma will be useful in deriving conditions on the function โ:โ โ โ so
that ๐:๐บ โ โ, defined by ๐ ๐ = โ ๐ ๐ ,๐ โ ๐บ, is a random variable. Recall that, for
a function ฮจ:๐ท1 โ ๐ท2 and ๐ด โ ๐ท2, ฮจโ1 ๐ด = ๐ โ ๐ท1: ฮจ ๐ โ ๐ด .
Lemma 1.1
Let ๐:๐บ โ โ and โ:โ โ โ be given functions. Define ๐:๐บ โ โ by ๐ ๐ =
โ ๐ ๐ ,๐ โ ๐บ. Then, for any ๐ต โ โ,
๐โ1 ๐ต = ๐โ1 โโ1 ๐ต .
Proof. Fix ๐ต โ โ. Note that โโ1 ๐ต = ๐ฅ โ โ: โ ๐ฅ โ ๐ต . Clearly
โ ๐ ๐ โ ๐ต โ ๐ ๐ โ โโ1 ๐ต .
Therefore,
๐โ1 ๐ต = ๐ โ ๐บ: ๐ ๐ โ ๐ต
= ๐ โ ๐บ: โ ๐ ๐ โ ๐ต
= ๐ โ ๐บ: ๐ ๐ โ โโ1 ๐ต
= ๐โ1 โโ1 ๐ต . โ
Theorem 1.1
Let ๐ be a random variable defined on a probability space ๐บ,โฑ,๐ and let โ:โ โ โ be
a Borel function. Then the function ๐:๐บ โ โ, defined by ๐ ๐ = โ ๐ ๐ ,๐ โ ๐บ, is a
random variable.
Proof. Fix ๐ต โ โฌ1. Since โ is a Borel function, we have โโ1 ๐ต โ โฌ1. Now using the
fact that ๐ is a random variable it follows that
๐โ1 ๐ต = ๐โ1 โโ1 ๐ต โ โฑ.
This proves the result. โ
Remark 1.1
(i) Let โ:โ โ โ be a continuous function. According to a standard result in
calculus inverse image of any open interval ๐, ๐ ,โโ โค ๐ < ๐ โค โ, under
continuous function โ is a countable union of disjoint open intervals. Since โฌ1
contains all open intervals and is closed under countable unions it follows that
โโ1 ๐, ๐ โ โฌ1 , whenever โโ โค ๐ < ๐ โค โ . Now on employing the
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arguments similar to the one used in proving Theorem 1.1, Module 2 (also see
Theorem 1.2, Module 2) we conclude that โโ1 ๐ต โ โฌ1,โ๐ต โ โฌ1. It follows
that any continuous function โ:โ โ โ is a Borel function and thus, in view of
Theorem 1.1, any continuous function of a random variable is a random
variable. In particular if ๐ is a random variable then ๐2 , ๐ , max ๐, 0 , sin๐
and cos๐ are random variables.
(ii) Let โ:โ โ โ be a strictly monotone function. Then, for โโ โค ๐ < ๐ โค โ,
โโ1 ๐, ๐ is a countable union of intervals and therefore โโ1 ๐, ๐ โ โฌ1, i.e.,
โ is a Borel function. It follows that if ๐ is a random variable and if โ:โ โ โ
is strictly monotone then โ(๐) is a random variable. โ
A random variable ๐ takes values in various Borel sets according to some probability law
called the probability distribution of random variable ๐ . Clearly the probability
distribution of a random variable of absolutely continuous/discrete type is described by
its distribution function (d.f.) and/or by its probability density function/probability mass
function (p.d.f/p.m.f.). For a given Borel function โ:โ โ โ, in the following section, we
will derive probability distribution of โ(๐) using the probability distribution of random
variable ๐. โ
3.2 PROBABILITY DISTRIBUTION OF A FUNCTION OF A
RANDOM VARIABLE
In our future discussions when we refer to a random variable, unless otherwise stated, it
will be either of discrete type or of absolutely continuous type. The probability
distribution of a discrete type random variable will be referred to as a discrete
(probability) distribution and the probability distribution of a random variable of
absolutely continuous type will be referred to as an absolutely continuous (probability)
distribution.
The following theorem deals with discrete probability distributions.
Theorem 2.1 Let ๐ be a random variable of discrete type with support ๐๐ and p.m.f. ๐๐(โ ) . Let
โ:โ โ โ be a Borel function and let ๐:๐บ โ โ be defined by ๐ ๐ = โ ๐ ๐ ,๐ โ ฮฉ.
Then ๐ is a random variable of discrete type with support ๐๐ = โ ๐ฅ :๐ฅ โ ๐๐ and p.m.f.
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๐๐ ๐ง = ๐๐ ๐ฅ , if ๐ง โ ๐๐๐ฅโ๐ด๐ง
0, otherwise
= ๐ ๐ โ ๐ด๐ง , if ๐ง โ ๐๐0, otherwise
,
where ๐ด๐ง = ๐ฅ โ ๐๐ : โ ๐ฅ = ๐ง . Proof. Since โ is a Borel function, using Theorem 1.1, it follows that ๐ is a random
variable. Also ๐ is of discrete implies that ๐๐ is countable which further implies that ๐๐
is countable. Fix ๐ง0 โ ๐๐, so that ๐ง0 = โ ๐ฅ0 for some ๐ฅ0 โ ๐๐ .
Then
๐ = ๐ฅ0 = ๐ โ ๐บ:๐ ๐ = ๐ฅ0 โ ๐ โ ๐บ: โ ๐ ๐ = โ ๐ฅ0
= โ ๐ = โ ๐ฅ0
= ๐ = ๐ง0 ,
and
๐ โ ๐๐ = ๐ โ ๐บ: ๐ ๐ โ ๐๐ โ ๐ โ ๐บ: โ ๐ ๐ โ ๐๐
= โ ๐ โ ๐๐
= ๐ โ ๐๐ .
Therefore,
๐ ๐ = ๐ง0 โฅ ๐ ๐ โ ๐ฅ0 > 0, (since ๐ฅ0 โ ๐๐),
and ๐ ๐ โ ๐๐ โฅ ๐ ๐ โ ๐๐ = 1.
It follows that ๐๐ is countable, ๐ ๐ = ๐ง > 0,โ๐ง โ ๐๐ and ๐ ๐ โ ๐๐ = 1, i. e., ๐ is
a discrete type random variable with support ๐๐.
Moreover, for ๐ง โ ๐๐,
๐ ๐ = ๐ง = ๐ ๐ โ ๐บ:โ ๐ ๐ = ๐ง
= ๐ ๐ = ๐ฅ
๐ฅโ๐ด๐ง
= ๐๐ ๐ฅ
๐ฅโ๐ด๐ง
= ๐ ๐ โ ๐ด๐ง .
Hence the result follows. โ
The following corollary is an immediate consequence of the above theorem.
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Corollary 2.1
Under the notation and assumptions of Theorem 2.1, suppose that โ:โ โ โ is one-one
with inverse function โโ1:๐ท โ โ, where ๐ท = โ ๐ฅ : ๐ฅ โ โ . Then ๐ is a discrete type
random variable with support ๐๐ง = โ ๐ฅ :๐ฅ โ ๐๐ and p.m.f.
๐๐ ๐ง = ๐๐ โ
โ1 ๐ง , if ๐ง โ ๐๐0, otherwise
= ๐ ๐ = โโ1 ๐ง , if ๐ง โ ๐๐0, otherwise
. โ
Example 2.1
Let ๐ be a random variable with p.m.f.
๐๐ ๐ฅ =
1
7, if ๐ฅ โ โ2,โ1, 0, 1
3
14, if ๐ฅ โ 2, 3
0, otherwise
.
Show that ๐ = ๐2 is a random variable. Find its p.m.f. and distribution function.
Solution. Since โ ๐ฅ = ๐ฅ2, ๐ฅ โ โ, is a continuous function and ๐ is a random variable,
using Remark 1.1 (i) it follows that ๐ = โ ๐ = ๐2 is a random variable. Clearly
๐๐ = โ2,โ1, 0, 1, 2, 3 and ๐๐ = 0, 1, 4, 9 . Moreover,
๐ ๐ = 0 = ๐ ๐2 = 0 = ๐ ๐ = 0 =1
7,
๐ ๐ = 1 = ๐ ๐2 = 1 = ๐ ๐ โ โ1, 1 =1
7+
1
7=
2
7,
๐ ๐ = 4 = ๐ ๐2 = 4 = ๐ ๐ โ โ2, 2 =1
7+
3
14=
5
14,
and ๐ ๐ = 9 = ๐ ๐2 = 9 = ๐ ๐ โ โ3, 3 = 0 +3
14=
3
14โ
Therefore the p.m.f. of ๐ is
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๐๐ ๐ง =
1
7, if ๐ง = 0
2
7, if ๐ง = 1
5
14, if ๐ง = 4
3
14, if ๐ง = 9
0, otherwise
,
and the distribution function of ๐ is
๐น๐ ๐ง =
0, if ๐ง < 01
7, if 0 โค ๐ง < 1
3
7, if 1 โค ๐ง < 4
11
14, if 4 โค ๐ง < 9
1, if z โฅ 9
โ โ
Example 2.2
Let ๐ be a random variable with p.m.f.
๐๐ ๐ฅ = ๐ฅ
2550, if ๐ฅ โ ยฑ 1, ยฑ2,โฏ , ยฑ50
0, otherwise
.
Show that ๐ = ๐ is a random variable. Find its p.m.f., and distribution function.
Solution. As โ ๐ฅ = ๐ฅ ,๐ฅ โ โ, is a continuous function and ๐ is a random variable,
using Remark 1.1 (i), ๐ = ๐ is a random variable. We have ๐๐ = ยฑ 1, ยฑ2,โฏ , ยฑ50
and ๐๐ = 1, 2,โฏ , 50 . Moreover, for ๐ง โ ๐๐,
๐ ๐ = ๐ง = ๐ ๐ = ๐ง = ๐ ๐ โ โ๐ง, ๐ง = โ๐ง
2550+
๐ง
2550=
๐ง
1275โ
Therefore the p.m.f. of ๐ is
๐๐ ๐ง =
๐ง
1275, if ๐ง โ 1, 2,โฏ , 50
0, otherwise,
and the distribution function of ๐ is
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๐น๐ ๐ง =
0, if ๐ง < 11
1275, if 1 โค ๐ง < 2
๐ ๐ + 1
2550, if ๐ โค ๐ง < ๐ + 1, ๐ = 2, 3,โฏ ,49
1, if ๐ง โฅ 50
. โ
Example 2.3
Let ๐ be a random variable with p.m.f.
๐๐ ๐ฅ = ๐๐ฅ ๐๐ฅ 1 โ ๐ ๐โ๐ฅ , if ๐ฅ โ 0, 1,โฏ ,๐
0, otherwise
,
where ๐ is a positive integer and ๐ โ 0,1 . Show that ๐ = ๐ โ ๐ is a random variable.
Find its p.m.f. and distribution function.
Solution. Note that ๐๐ = ๐๐ = 0, 1,โฏ , ๐ and โ ๐ฅ = ๐ โ ๐ฅ, ๐ฅ โ โ , is a continuous
function. Therefore ๐ = ๐ โ ๐ is a random variable. For ๐ฆ โ ๐๐
๐ ๐ = ๐ฆ = ๐ ๐ = ๐ โ ๐ฆ = ๐
๐ โ ๐ฆ ๐๐โ๐ฆ 1 โ ๐ ๐ฆ =
๐๐ฆ 1 โ ๐ ๐ฆ๐๐โ๐ฆ .
Thus the p.m.f. of ๐ is
๐๐ ๐ฆ = ๐๐ฆ 1 โ ๐ ๐ฆ๐๐โ๐ฆ , if ๐ฆ โ 0, 1,โฏ ,๐
0, otherwise,
and the distribution function of ๐ is
๐น๐ ๐ฆ =
0, if ๐ฆ < 0๐๐ , if 0 โค ๐ฆ < 1
๐๐ 1 โ ๐ ๐๐๐โ๐
๐
๐=0
, if ๐ โค ๐ฆ < ๐ + 1, ๐ = 1,2,โฏ ,๐ โ 1
1, if ๐ฆ โฅ ๐
. โ
The following theorem deals with probability distribution of absolutely continuous type
random variables.
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Theorem 2.2
Let ๐ be a random variable of absolutely continuous type with p.d.f. ๐๐(โ ) and support
๐๐ . Let ๐1, ๐2,โฏ , ๐๐ , be open intervals in โ such that ๐๐ โฉ ๐๐ = ๐, if ๐ โ ๐ and ๐๐ =๐๐=1
๐๐ . Let โ:โ โ โ be a Borel function such that, on each ๐๐ ๐ = 1,โฆ ,๐ ,โ: ๐๐ โ โ is
strictly monotone and continuously differentiable with inverse function โ๐โ1(โ ) . Let
โ ๐๐ = โ ๐ฅ :๐ฅ โ ๐๐ so that โ ๐๐ ๐ = 1,โฆ ,๐ is an open interval in โ . Then the
random variable ๐ = โ ๐ is of absolutely continuous type with p.d.f.
๐๐ ๐ก = ๐๐
๐
๐=1
โ๐โ1 ๐ก
๐
๐๐กโ๐โ1 ๐ก ๐ผโ ๐๐ (๐ก).
Proof. We will provide an outline of the proof which may not be rigorous. Let ๐น๐(โ ) be
the distribution function of ๐. For ๐ก โ โ and ฮ > 0,
๐น๐ ๐ก + ฮ โ ๐น๐ ๐ก
ฮ=๐ ๐ก < โ ๐ โค ๐ก + ฮ
ฮ
= ๐ ๐ก < โ ๐ โค ๐ก + ฮ,๐ โ ๐๐
ฮ
๐
๐=1
โ
Fix ๐ โ 1,โฆ ,๐ . First suppose that โ๐ (โ ) is strictly decreasing on ๐๐ . Note that ๐ โ
๐๐ = โ ๐ โ โ ๐๐ and โ ๐๐ is an open interval. Thus, for ๐ก belonging to the exterior
of โ ๐๐ and sufficiently small ฮ > 0 , we have ๐ ๐ก < โ ๐ โค ๐ก + ฮ,๐ โ ๐๐ = 0.
Also, for ๐ก โ โ ๐๐ and sufficiently small ฮ > 0,
๐ ๐ก < โ ๐ โค ๐ก + ฮ,๐ โ ๐๐ = ๐ โ๐โ1 ๐ก + ฮ โค ๐ < โ๐
โ1 ๐ก .
Thus, for all ๐ก โ โ, we have
๐ ๐ก < โ ๐ โค ๐ก + ฮ,๐ โ ๐๐
ฮ=๐ โ๐
โ1 ๐ก + ฮ โค ๐ < โ๐โ1 ๐ก ๐ผโ ๐๐ (๐ก)
ฮ
=1
ฮ ๐๐ ๐ง ๐๐ง
โ๐โ1(๐ก)
โ๐โ1(๐ก+ฮ)
๐ผโ ๐๐ (๐ก)
ฮโ0 โ๐๐ โ๐
โ1 ๐ก ๐
๐๐กโ๐โ1 ๐ก ๐ผโ ๐๐ ๐ก . (2.1)
Similarly if โ๐ is strictly increasing on ๐๐ then, for all ๐ก โ โ, we have
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๐ ๐ก < โ ๐ โค ๐ก + ฮ,๐ โ ๐๐
ฮ=๐ โ๐
โ1 ๐ก < ๐ โค โ๐โ1 ๐ก + ฮ ๐ผโ ๐๐ (๐ก)
ฮ
=1
ฮ ๐๐ ๐ง ๐๐ง
โ๐โ1(๐ก+ฮ)
โ๐โ1(๐ก)
๐ผโ ๐๐ (๐ก)
ฮโ0 ๐๐ โ๐
โ1 ๐ก ๐
๐๐กโ๐โ1 ๐ก ๐ผโ ๐๐ ๐ก . (2.2)
Note that if โ is strictly decreasing (increasing) on ๐๐ then ๐
๐๐กโ๐โ1 ๐ก < > 0 on ๐๐ . Now
on combining (2.1) and (2.2) we get, for all ๐ก โ โ,
๐ ๐ก < โ ๐ โค ๐ก + ฮ,๐ โ ๐๐
ฮ
ฮโ0 ๐๐ โ๐
โ1 ๐ก ๐
๐๐กโ๐โ1 ๐ก ๐ผโ ๐๐ ๐ก ,
โ๐น๐ ๐ก + ฮ โ ๐น๐ ๐ก
ฮ
ฮโ0 ๐๐ โ๐
โ1 ๐ก
๐
๐=1
๐
๐๐กโ๐โ1 ๐ก ๐ผโ ๐๐ ๐ก .
Similarly one can show that, for all ๐ก โ โ,
limฮโ0
๐น๐ ๐ก + ฮ โ ๐น๐ ๐ก
ฮ= ๐๐ โ๐
โ1 ๐ก
๐
๐=1
๐
๐๐กโ๐โ1 ๐ก ๐ผโ ๐๐ ๐ก . (2.3)
It follows that the distribution function of ๐ is differentiable everywhere on โ except
possibly at a finite number of points (on boundaries of intervals โ ๐1 ,โฏ , โ ๐๐ of ๐๐).
Now the result follows from Remark 4.2 (vii) of Module 2 and using (2.3). โ
The following corollary to the above theorem is immediate.
Corollary 2.2
Let ๐ be a random variable of absolutely continuous type with p.d.f. ๐๐(โ ) and support
๐๐ . Suppose that ๐๐ is a finite union of disjoint open intervals in โ and let โ:โ โ โ be a
Borel function such that โ is differentiable and strictly monotone on ๐๐ (i.e., either
โโฒ ๐ฅ < 0,โ๐ฅ โ ๐๐ or โโฒ ๐ฅ > 0,โ๐ฅ โ ๐๐). Let ๐๐ = โ ๐ฅ :๐ฅ โ ๐๐ . Then ๐ = โ ๐ is a
random variable of absolutely continuous type with p.d.f.
๐๐ ๐ก = ๐๐ โโ1 ๐ก
๐
๐๐กโโ1 ๐ก , if ๐ก โ ๐๐
0, otherwiseโ โ
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It may be worth mentioning here that, in view of Remark 4.2 (vii) of Module 2, Theorem
2.2 and Corollary 2.2 can be applied even in situations where the function โ is
differentiable everywhere on ๐๐ except possibly at a finite number of points.
Example 2.4
Let ๐ be random variable with p.d.f.
๐๐ ๐ฅ = ๐โ๐ฅ , if ๐ฅ > 00, otherwise
,
and let ๐ = ๐2
(i) Show that ๐ is a random variable of absolutely continuous type;
(ii) Find the distribution function of ๐ and hence find its p.d.f.;
(iii) Find the p.d.f. of ๐ directly (i.e., without finding the distribution function of
๐).
Solution. (i) and (iii). Clearly ๐ = ๐2 is a random variable (being a continuous function
of random variable ๐). We have ๐๐ = ๐๐ = 0, โ . Also โ ๐ฅ = ๐ฅ2, ๐ฅ โ ๐๐ , is strictly
increasing on ๐๐ with inverse function โโ1 ๐ฅ = ๐ฅ, ๐ฅ โ ๐๐ . Using Corollary 2.1 it
follows that ๐ = ๐2 is a random variable of absolutely continuous type with p.d.f.
๐๐ ๐ก = ๐๐ ๐ก ๐
๐๐ก ๐ก , if ๐ก > 0
0, otherwise
= ๐โ ๐ก
2 ๐ก, if ๐ก > 0
0, otherwise
.
(ii) We have ๐น๐ ๐ก = ๐ ๐2 โค ๐ก , ๐ก โ โ. Clearly, for ๐ก < 0,๐น๐ ๐ก = 0. For ๐ก โฅ 0,
๐น๐ ๐ก = ๐ โ ๐ก โค ๐ โค ๐ก
= ๐๐ ๐ฅ ๐๐ฅ
๐ก
โ ๐ก
= ๐โ๐ฅ๐๐ฅ
๐ก
0
= 1 โ ๐โ ๐ก .
NPTEL- Probability and Distributions
Dept. of Mathematics and Statistics Indian Institute of Technology, Kanpur 12
Therefore the distribution function of ๐ is
๐น๐ ๐ก = 0, if ๐ก < 0
1 โ ๐โ ๐ก , if ๐ก โฅ 0 .
Clearly ๐น๐ is differentiable everywhere except at ๐ก = 0. Therefore, using Remark 4.2
(vii) of Module 2, we conclude that the random variable ๐ is of absolutely continuous
type with p.d.f. ๐๐ ๐ก = ๐น๐โฒ ๐ก , if ๐ก โ 0 . At ๐ก = 0 we may assign any arbitrary non-
negative value to ๐๐ 0 . Thus a p.d.f. of ๐ is
๐๐ ๐ก = ๐โ ๐ก
2 ๐ก, if ๐ก > 0
0, otherwise
. โ
Example 2.5
Let ๐ be a random variable with p.d.f.
๐๐ ๐ฅ =
๐ฅ
2, if โ 1 < ๐ฅ < 1
๐ฅ
3, if 1 โค ๐ฅ < 2
0, otherwise
,
and let ๐ = ๐2
(i) Show that ๐ is a random variable of absolutely continuous type;
(ii) Find the distribution function of ๐ and hence find its p.d.f;
(iii) Find the p.d.f. of ๐ directly (i.e., without finding the distribution function of
๐).
Solution. (i) and (iii). Clearly ๐ = ๐2 is a random variable (being a continuous function
of random variable ๐ ). We have ๐๐ = โ1, 0 โช 0, 2 = ๐1 โช ๐2 , say. Also โ ๐ฅ =
๐ฅ2 , ๐ฅ โ ๐๐ , is strictly decreasing in ๐1 = โ1, 0 with inverse function โ1โ1 ๐ก =
โ ๐ก;โ ๐ฅ = ๐ฅ2 , ๐ฅ โ ๐๐ , is strictly increasing in ๐2 = 0, 2 , with inverse function
โ2โ1 ๐ก = ๐ก; โ ๐1 = 0, 1 and โ ๐2 = 0, 4 . Using Theorem 2.2 it follows that
๐ = ๐2 is a random variable of absolutely continuous type with p.d.f.
๐๐ ๐ก = ๐๐ โ ๐ก ๐
๐๐ก โ ๐ก ๐ผ 0,1
๐ก + ๐๐ ๐ก ๐
๐๐ก ๐ก ๐ผ 0,4
๐ก
=
1
2, if 0 < ๐ก < 1
1
6, if 1 < ๐ก < 4
0, otherwise
โ
NPTEL- Probability and Distributions
Dept. of Mathematics and Statistics Indian Institute of Technology, Kanpur 13
(ii) We have ๐น๐ ๐ก = ๐ ๐2 โค ๐ก , ๐ก โ โ . Since ๐ ๐ โ โ1, 2 = 1 , we have
๐ ๐ โ 0, 4 = 1.
Therefore, for ๐ก < 0, ๐น๐ ๐ก = ๐ ๐ โค ๐ก = 0 and, for ๐ก โฅ 4, ๐น๐ ๐ก = ๐ ๐ โค ๐ก = 1.
For ๐ก โ [0,4), we have
๐น๐ ๐ก = ๐ โ ๐ก โค ๐ โค ๐ก
= ๐๐ ๐ฅ
๐ก
โ ๐ก
๐๐ฅ
=
|๐ฅ|
2๐๐ฅ, if 0 โค ๐ก < 1
๐ก
โ ๐ก
|๐ฅ|
2๐๐ฅ
1
โ1
+ ๐ฅ
3๐๐ฅ,
๐ก
1
if 1 โค ๐ก < 4
.
Therefore, the distribution function of ๐ is
๐น๐ ๐ก =
0, if ๐ก < 0๐ก
2, if 0 โค ๐ก < 1
๐ก + 2
6, if 1 โค ๐ก < 4
1, if ๐ก โฅ 4
โ
Clearly ๐น๐ is differentiable everywhere except at points 0, 1 and 4. Using Remark 4.2
(vii) of Module 2 it follows that the random variable ๐ is of absolutely continuous type
with a p.d.f.
๐๐ ๐ก =
1
2, if 0 < ๐ก < 1
1
6, if 1 < ๐ก < 4
0, otherwise
โ โ
Note that a Borel function of a discrete type random variable is a random variable of
discrete type (see Theorem 1.1). Theorem 2.2 provides sufficient conditions under which
a Borel function of an absolutely continuous type random variable is of absolutely
continuous type. The following example illustrates that, in general, a Borel function of an
absolutely continuous type random variable may not be of absolutely continuous type.
NPTEL- Probability and Distributions
Dept. of Mathematics and Statistics Indian Institute of Technology, Kanpur 14
Example 2.6
Let ๐ be a random variable of absolutely continuous type with p.d.f.
๐๐ ๐ฅ = ๐โ๐ฅ , if ๐ฅ > 00, ๐ otherwise
,
and let ๐ = ๐ฅ , where, for ๐ฅ โ โ, ๐ฅ denotes the largest integer not exceeding ๐ฅ. Show
that ๐ is a random variable of discrete type and find its p.m.f.
Solution. For ๐ โ โ, we have
๐โ1 โโ,๐ = โโ, ๐ + 1 โ โฌ1.
It follows that ๐ is a random variable. Also ๐๐ = 0, โ . Since ๐ ๐ โ ๐๐ = 1 , we
have ๐ ๐ โ 0, 1, 2,โฏ = 1. Also, for ๐ โ 0, 1, 2,โฆ .
๐ ๐ = ๐ = ๐( ๐ โค ๐ < ๐ + 1 )
= ๐๐ ๐ฅ ๐๐ฅ
๐+1
๐
= ๐โ๐ฅ๐๐ฅ
๐+1
๐
= 1 โ ๐โ1 ๐โ๐
> 0.
Consequently the random variable ๐ is of discrete type with support ๐๐ = 0, 1, 2,โฆ and
p.m.f.
๐๐ ๐ก = ๐ ๐ = ๐ก = 1 โ ๐โ1 ๐โ๐ก , if ๐ก โ 0 ,1, 2,โฆ 0 , otherwise
. โ