1. 3. Random Variables (Fig.3.1)
Let (, F, P) be a probability model for an experiment, and X a function that maps every ζ∈ Ω to a unique point x ∈R the set of real numbers. Since the outcome ζ is not certain, so is the value X(ζ)=x Thus if B is some subset of R, we may want to determine the probability of “ X(ζ) ∈B ”. To determine this probability, we can look at the set A=X (B) ∈Ω that contains all ζ∈ Ω that maps into B under the function X.
(Fig.3.1)

2. Random Variables If belongs to the associated field F, then the
probability of A is well defined .
in that case we can say :
Random Variable (r.v): A finite single valued
function that maps the set of all experimental outcomes
into the set of real numbers R is said to be a r.v, if the set
is an event for every x in R.
(3-1)
However, X(B) may not always belong to F for all B, thus creating difficulties. The notion of random variable (r.v) makes sure that the inverse mapping always results in an event so that we are able to determine the probability for any B ∈R

3. Random Variables
X :r.v,
B represents semi-infinite intervals of the form
The Borel collection B of such subsets of R is the smallest -field of subsets of R that includes all semi-infinite intervals of the above form.
if X is a r.v, then
is an event for every x.
all other sets that can be constructed from these sets by performing the set operations of union, intersection and
negation any number of times.
(3-2)

4. Distribution Function
The role of the subscript X in (3-3) is only to identify the
actual r.v is said to the Probability Distribution
Function associated with the r.v X.
(3-3)

5. Properties of a PDF if g(x) is a distribution function, then (i)
(ii) if then
(iii) for all x.
(3-4)

6. Properties of a PDF Suppose defined in (3-3) (i) and
(ii) I f then the subset Consequently the event since implies As a result
=>the probability distribution function is nonneg ative and monotone nondecreasing.
(3-5)
(3-6)
(3-7)

7. Properties of a PDF (iii) Let and consider the event since (3-8)
using mutually exclusive property of events we get
But and hence
(3-8)
(3-9)
(3-10)
(3-11)

8. Properties of a PDF Thus But the right limit of x, and hence
i.e., is right-continuous, justifying all properties of
a distribution function.
(3-12)

9. Additional Properties of a PDF
(iv) If for some then
This follows, since implies
is the null set, and for any will be a
subset of the null set.
(v)
We have
(vi)
The events and are mutually
exclusive and their union represents the event
(3-13)
(3-14)
(3-15)

10. Additional Properties of a PDF
(vii)
Let and From (3-15) or
Thus the only discontinuities of a distribution function
occur at points where
is satisfied.
(3-16)
(3-17)
(3-18)
According to (3-14),Fx(x0+) the limit of Fx(x) asx->x0 from the right always exists and equals Fx(x0) However the left limit value F(x0-) need not equal ThusF(x0) Thus Fx(x) need not be continuous from the left. At a discontinuity point of the distribution, the left and right limits are different.
(3-19)

11. continuous-type & discrete-type r.v
X is said to be a continuous-type r.v if
And from
=>
If is constant except for a finite number of jump discontinuities (piece-wise constant; step-type), then X is said to be a discrete-type r.v. If is such a discontinuity point, then
(3-20)

12. Example
Example 3.1: X is a r.v such that Find Solution: For so that and for so that (Fig.3.2)
at a point of discontinuity we get
Fig. 3.2

13. Example Example 3.2: Toss a coin. Suppose the r.v X is such that Find
Solution: For so that
at a point of discontinuity we get
Fig.3.3

14. Example Example:3.3 A fair coin is tossed twice, and let the r.v X
represent the number of heads. Find
Solution: In this case

15. Example
From Fig.3.4,
Fig. 3.4

16. Probability density function (p.d.f)
(3-21)
Since
from the monotone-nondecreasing nature of
=> for all x

17. Probability density function (p.d.f)
will be a continuous function, if X is a continuous type r.v
if X is a discrete type r.v as in (3-20), then its p.d.f has the general form (Fig. 3.5)
represent the jump-discontinuity points in As Fig. 3.5 shows represents a collection of positive discrete masses, and it is known as the probability mass function (p.m.f ) in the discrete case.
Fig. 3.5

18. Probability density function (p.d.f)
From (3-23),
Since
=>
the area under in the interval represents the probability in (3-22).
(3-22)
(a)
(b)

19. Continuous-type random variables
Normal (Gaussian):
Where
the notation  will be used to represent (3-23).
(3-23)
Fig. 3.7

20. Continuous-type random variables
Uniform:  (Fig. 3.8)
(3.24)
Fig. 3.8

21. Continuous-type random variables
Exponential:  (Fig. 3.9)
(3.25)
Fig. 3.9

22. Exponential distribution
Assume the occurences of nonoverlapping intervals are independent, and assume:
q(t): the probability that in a time interval t no event has occurred.
x: the waiting time to the first arrival
Then we have: P(x>t)=q(t)
t1 and t2 : two consecutive nonoverlapping intervals,

23. Exponential distribution
Then we have: q(t1) q(t2) = q(t1+t2)
The only bounded solution is:
So the pdf is exponential.
If the occurrences of events over nonoverlapping intervals are independent, the corresponding pdf has to be exponential.

24. Memoryless property of exponential distribution
Let Consider the events
and Then

25. Continuous-type random variables
4. Gamma:  if (Fig. 3.10)
If an integer
(3-26)
Fig. 3.10

26. Continuous-type random variables
The exponential random variable is a special case of gamma distribution with
The (chi-square) random variable with n degrees of freedom is a special case of gamma distribution with

27. Continuous-type random variables
5. Beta:  if (Fig. 3.11)
where the Beta function is defined as
Beta distribution with a=b=1 is the uniform distribution on (0,1).
Fig. 3.11
(3-27)
(3-28)

28. Note that is the same as Gamma 7. Rayleigh:  if (Fig. 3.13)
6. Chi-Square:  if (Fig. 3.12)
Note that is the same as Gamma
7. Rayleigh:  if (Fig. 3.13)
8. Nakagami – m distribution:
(3-29)
(3-30)
Fig. 3.12
Fig. 3.13
(3-31)

29. 11. Student’s t-distribution with n degrees of freedom (Fig 3.16)
9. Cauchy:  if (Fig. 3.14)
10. Laplace: (Fig. 3.15)
11. Student’s t-distribution with n degrees of freedom (Fig 3.16)
Fig. 3.14
(3-34)
(3-33)
(3-32)
Fig. 3.15
Fig. 3.16

30. 12. Fisher’s F-distribution
(3-35)

31. Discrete-type random variables
1. Bernoulli: X takes the values (0,1), and
2. Binomial:  if (Fig. 3.17)
(3-36)
(3-37)
Fig. 3.17
The probability of k successes in n experiments with replacement (in ball drawing)

32. Discrete-type random variables
3. Poisson:  if (Fig. 3.18)
(3-38)
Fig. 3.18

33. Discrete-type random variables
Poisson distribution represents the number of occurrences of a rare event in a large number of trials.
Pk Increasing with k from 0 to λ and decreasing after that.

34. 4. Hypergeometric: 5. Geometric:  if
The probability of k successes in n experiments without
replacement (ball drawing)
5. Geometric:  if
(3-39)
(3-40)

35. 6. Negative Binomial: ~ if
7. Discrete-Uniform:
(3-41)
(3-42)

36. A box contains a white balls and b black balls. A ball is
Polya’s distribution:
A box contains a white balls and b black balls. A ball is
drawn at random, and it is replaced along with c balls of
the same color. If X represents the number of white balls
drawn in n such draws, find the probability
mass function of X.

37. :probability of drawing k white balls ,followed by n – k black balls
Solution:
: probability of drawing k successive white balls
:probability of drawing k white balls ,followed by
n – k black balls
(3-43)
(3-44)
Interestingly, pk in (3-51) also represents the probability of drawing k white balls and (n – k) black balls in any other
specific order (i.e., The same set of numerator and denominator terms in (3-51) contribute to all other sequences
as well.) But there are (n,k) such distinct mutually exclusive sequences and summing over all of them,

38. if draws are done with replacement, then c = 0 and (3-45)
Polya distribution: probability of getting k white balls
in n draws.
if draws are done with replacement, then c = 0 and
(3-45)
simplifies to the binomial distribution
(3-45)

39. if the draws are conducted without replacement,
Then c = – 1 in (3-45), and it gives
which represents the hypergeometric distribution.
(3-46)

40. Finally c = +1 gives (replacements are doubled)
(3-47) is Polya’s +1 distribution.
(3-47)