The document discusses discrete and continuous random variables and their probability mass functions (pmf), probability density functions (pdf), and cumulative distribution functions (cdf). It provides examples of discrete and continuous random variables. It defines pmf as P(X=x) and pdf as f(x)=P[(x-a/2)≤X≤(x+a/2)]/a for discrete and continuous variables respectively. The cdf is defined as F(x)=P(X≤x). It also discusses mathematical expectation (mean) as E(X)=ΣxP(x) for discrete variables and ∫xf(x)dx for continuous variables.

1. Discrete and Continuous , pmf, pdf and cdf

2. Random Variable:
 In the applications of theory of probability. It is more
convenient to work, numerical outcomes rather than
non-numerical outcomes. Because of this we introduce
the idea of random variables.

3. Definition of Random Variable:
A random variable is a function on a sample space
associated with a random experiment assuming any value
from R and assigning a real no. to each and every sample
point of the random experiment. A random variable is
denoted by a capital letter.
Ex: If a coin is tossed three times and if x denotes the no. of
heads then x is a random variable. In this case, the sample
space is given by:
S = { HHH, HHT, HTH, HTT, THH, THT, TTH, TTT} = Ω
and we fin out that X=0 if the sample point is TTT
X=1( If sample point is HTT, THT or TTT)
X=2( If sample point is HHT, HTH or THH)

4. R.V: X: Ω → R
Where Ω IS domain of X and R is the co-domain (or) range of
X.
HHH
HHT
HTH
HTT
THH
THT
TTH
TTT
3
2
2
1
2
1
1
0

5. A one dimensional R.V ‘X’ is said to be discrete if it takes at most
countable no. of values i.e., integral values,
For example
1. The size of a family.
2. No. of accidents on road.
3. No. of customers in a bank.
4. No. of cars manufactured by the company.
5. The sales of commodities in a departmental store.
6. The no. of transactions in a bank per day.
7. The no. of customers enter an ATM counter during a time
8. No. of apples in a basket
9. no. of flats in an apartment etc
Discrete R.V:

6. Continuous R.V:
A one dimensional R.V ‘X’ is said to be continuous if it
assumes all possible values between the limits.
For example:
1. The temperature at a location.
2. The life time of electronic component.
3. The reaction temperature at chemical laboratory.
4. Current in a semi-conductor diode
5. Height, weight, etc.

7. Probability Mass Function (pmf):
Let X be a one dimensional discrete random variable.
The function p(x)= P(X=x) is said to be a probability
mass function if it satisfies the properties of
1. Non-negativity: p(x)≥0, for all x
2. Normality: ∑x p(x) =1

8. Probability Density Function (pdf):
Let X be a one dimensional continuous random
variable. The function
f(x)=Lt a→0 pr{[(x-a/2)≤X≤(x+a/2)]÷a} where
a=dx
to be a probability density function if it satisfies the
properties of non-negativity i.e., f(x)≥0, for all x.
∫x f(x)dx=1

9. Cumulative Distribution Function
(cdf) (or) Distribution function (df):
The distribution function of a one dimensional R.V ‘X’ is
given by
F(x)=P(X≤x), where x∈R
According to the classification, the cdf,
Cdf, F(x)= ∑x≤xp(x), discrete R.V
∫X≤xf(x)dx, continuous R.V
Properties:
1. F(x) lies between 0 and 1
2. F(+∞)=1
3. F(-∞)=0
4. F(x) is a non-decreasing and right continuous function.

10. Mathematical Expectation (or)
Expectation (E(X)):
Let p(x)/f(x) be the pmf/pdf of a one dimensional
random variable ‘X’, according to the classification of
discrete/continuous. Then the expected value of R.V is
given by
E(X)= ∑x xp(x), discrete R.V
∫x xf(x)dx, continuous R.V
Provided the summation and integration are
absolutely convergent.

11. Note: 1
The mathematical expectation, E(X) is nothing but the
mean of the distribution.
Let x1, x2…….xn be a set of ‘n’ independently, identically
distributed (iid) observations.
By definition, µ(mean, average, A.M)= sum of the
observations÷no.of observations
=
(x1+x2+x3+………………..xn)/n
=x1.1/n+x2.1/n+x3.1/n+…………….xn.1/n
=x1P(x1)+x2.P(x2)+x3.P(x3)+………xn.P(xn) =n∑i=1xi.P(xi)
∴µ=∑x x.p(x)=E(X)

12. Note 2
The E(X) plays a vital role in data analysis for the
computation of the measures of central tendency
parameters (mean, median, mode, G.M, H.M) and
measures of dispersion (range, mean deviation,
quartile deviation and standard deviation).

13. Properties of expectation:
1.Expectation of function of a R.V:
Let G(X) be a function of a one dimensional R.V ‘X’ then the expectation of the
function of R.V is given by
E{G(X)} =∑x g(x).p(x), discrete R.V.
=∫x g(x).f(x), continuous R.V.
Provided the summation and integration are absolutely convergent.
2. Expectation of linear function:
Let G(X)= α+βX is a linear function in X then E(α+ΒX)= α+βE(X)
By definition,
E(G(X))=E{α+βX}
=∑x(α+βX).p(x)
=∑x [α.p(x)+βx.p(x)]
=∑x α.p(x)+∑x β.x.p(x)
=α.∑xp(x)+β.∑xx.p(x)
E(α+ΒX)= α.1+ β.E(X) ⇒ α+βE(X)
⇒ E(α+ΒX)= α+βE(X)

14. Note:
1. If β=0 (slope=0, provided line parallel to X-axis)
then,
E(α+0.X)= α+0.E(X)
⇒ E(α)=α
Expectation is an operator is applicable to R.Vs and
function of R.Vs only but not constants.
2. If α=0, β=-1, then
E(α+ΒX)= α+βE(X)
E(0+(-1)X) = 0+(-1) E(X)
⇒ E(-X)= -E(X)

16. A car which is brought for loan, the
customer has to pay Emi which is a
Random Variable having the pmf is
RV X 1 2 3 4 5 6 7
PM
F
P(x) a 2a 3a 2a a2 2a2 (7a2
+a)

17. Find the following:
1. a
2. P(x ≥1)
3. P(x>3)
4. P(2≤X<6)
5. Mean, E(X)
6. Variance, V(X)
7. SD< Standard deviation
8. E{2X+2}
9. V{5-2X}
10. 10)E{G(x)}, G(x)=5X2-7X+10
11. CDF, F(X)

18. SOLUTION
Given that,
P(x)=P(X=x) is a pmf
So, it has the properties of
Non Negativity P(x)≥1 for all x
Normality ∑xP(x)=1
by using these conditions we can determine all constants
of “a”.

19.  1) Therefore, by normality ∑n
x P(x)=1
 ∑7
x=1 P(x)=1
 a+2a+3a+2a+a2+2a2+(7a2+a)=1
 10a2+9a-1=0
 10a2+10a=a-1=0
 10a(a+1)-1(a+1)=0
 (a+1)(10a-1)=0
 a=-1 or = 1÷10
 a= 1÷10 because of the condition of non negativity.

20. 2) P(x≤1)
P(1)= 1/10 i.e. 10% of amount.
3) P(X>3)
1-P(X≤3)1-{P(1)+P(2)+P(3)} = 1-
{(1/10)+(2/10)+)(3/10)} = 1-{6/10} = 4/10
4) P(2≤X<6)
= P(2)+P(3)+P(4)+P(5) =

21. 5) Mean; E(X)
By definition, µ=E(X)= ∑n
x P(x)
∑7
x=1 xP(x)=1P(1)+2P(2)+3P(3)+…..7P(7)
=
=
= 3.56

22. 6) By definition, V(X)=E{[X-E(X)]2}=E(X)2-mean2 eqn1
Now, E(X)2=∑n
x
2 P(x)= ∑7
x=1 x2
P(x)=12P(1)+22P(2)+…….72P(7)
E(X)2=12 +22 +…….72
E(X)2=16.1
But; according to eqn1
V(X)=E(X2)-Mean2
V(X)=16.1-(3.56)2=3.43
V(X)=3.43
7) SD(X)= = =1.859

23. 8) From the properties of expectation and variance
E{2x+2}=2E(X)+2
=2(3.56)+2
=9.12
9) V{5-2X}=(-2)2V(X)=4(3.42)=13.68
10)E{G(X)=E{G(X)}=E{5X2-&X+10}=5(16.1)-
7(3.56)+10=65.5

24. 11) CDF
By definition the cdf is F(X)=P(X=x); x € R;
These values are given in the following table
RV X 1 2 3 4 5 6 7
PMF P(x)
PDF F(x) 1

25. Bibliography
1. Quantitative aptitude-CA-CPT BOS ICAI
2. Class notes
3. Case study - Self