Qt chapter 7 random variables

1. Random Variables
By Rohan Bhatkar

2. GROUP 1
Murtuza Indorewala
Ayushi Jain
Apeksha Mehta
Rohan Bhatkar
Yash Rawani
Mohammed Driver
Mayur Sancheti
Nandan Shah
01
05
08
11
21
24
32
33

3. Content-
Random Variables
Probability Mass Function
Discrete Random Variables
Probability Distribution
Distribution Function
Variance
Expectation
Continuous Random Variables

4. Random Variables
Consider an experiment of throwing two dice. We know that this
experiments has 36 outcomes .
Let x be the sum of numbers on the uppermost faces.
For ExampleThe value of x is 2 if outcome of the experiment is (1,1) &
probability of this outcome is 1/36.
In this case we can associate real number with each outcome of
random experiment or group of outcomes of random experiment.
Here X is called Random Variables.

5. Probability Mass Function
If x is discrete random variables values x1, x2, x3, …., xn then probability of each value is
described by a function called the Probability Mass Function. The probability that random variable X
takes values xi is denoted by p(xi).
Illustration Suppose fair coin is marked 1 & 2, dice numbered 1, 2, 3, 4, 5 & 6 are thrown simultaneously then
probability mass function of random variables X which is sum of numbers on coin & dice is
obtained as under.
SolutionThe sample space is:
S={(1,1), (1,2), (1,3), (1,4), (1,5), (1,6), (2,1), (2,2), (2,3), (2,4), (2,5), (2,6)}
Note that n(s) = 12
X: sum of numbers on coin & dice.
X = 2, 3, 4, 5, 6, 7, 8
P[x=2] = {(1,1)} = 1/12
P[x=3] = {(1,2)(2,1)} = 2/12
P[x=4] = {(1,3)(2,2)} = 2/12
P[x=5] = {(1,4)(2,3)} = 2/12
P[x=6] = {(1,5)(2,4)} = 2/12
P[x=7] = {(1,6)(2,5)} = 2/12
P[x=8] = {(2,6)} = 1/12
Thus probability distribution of random variable X is as under:
X=x
2
3
4
5
6
7
8
P[X=x]
1/12
2/12
2/12
2/12
2/12
2/12
1/12
Note that P[xi] ≥ 0 & ∑ (xi) = 1

6. Discrete Random Variable
A random variable which can assume only a countable number of real values & the values taken by
variables depends on outcome of random experiment is called Discrete Random Variables.
For
1.
2.
3.
ExampleNumber of misprints per page of book.
Number of heads in n tosses of fair coin.
Number of throws of dice to get first 6.
Illustration 1A bag contains 20 currency notes. 10 of Rs. 5, 5 of Rs. 10, 3 of Rs. 20 & 2 of Rs. 50. a note is
selected at random from the bag. Find the expected value of the domination of the note drawn.
SolutionDefine the discrete random variable
X = Denomination on the note drawn in Rs.
X
Rs. 5
Rs. 10
Rs. 20
Rs. 50
P[X = x]
1/2
1/4
3/20
1/10
xi. pi
5/2
5/2
3
5
E(X) = ∑xipi = 26/2 = 13 Rs.
Expected value of the denomination of the note drawn is Rs. 13.

7. Illustration 2The probability distribution of discrete random variable X is given below:
X
8
12
16
20
24
P[X=x]
1/8
1/6
3/8
1/4
1/12
Calculate mean & variance of X.
SolutionX
8
12
16
20
24
P[X=x]
1/8
1/6
3/8
1/4
1/12
xipi
1
2
6
5
2
xip2i
8
24
96
100
8
Mean = E(X) = ∑xipi
= 16
Var (X) = E (X2) – [E (X)]2
E (X2) = ∑xipi 2
Var (X) = 276 – 256
= 20

8. Continuous Random Variables
Consider the small interval (X, X + dx) of length dx round the point x.
Let f(x) be any continuous function of x so that f(x) represent the probability that falls in very
small interval (X, x + dx)
Symbolically P[x ≤ X ≤ x + dx] = f(x) dx.
y
f(x) dx
Y = f(x)
x
X – dx/2X + dx/2
In the figure f(x) dx represents the area bounded by the curve y = f(x); X-axis a& the ordinates x
and x + dx.
The function f(x) so defined is known as probability density function of random variable X & usually
abbreviated as p.d.f. The expression f(x) dx usually written as F (x), is known as probability
differential & f(x) is known as probability density curve. The probability that X lie in the interval dx
is f(x) dx. Thus the p.d.f. of random variable x is defined as
f(x) = lim P [x ≤ X ≤ x + δx]
δx → 0
δx

9. Illustration 1A random variable X has following p.d.f.
f(x) =
K
-∞<x<∞
1 + x2
=0
Otherwise
Find k:
SolutionSince X is continuous random variable with density function f(x),
∞
⌠ f(x) dx
=1
-∞
∞
⌠
-∞
k[tan-1
K
1 + x2
x]
dx
∞
=1
=1
-∞
k[tan-1 ∞ - tan-1 ∞ (- ∞)] = 1
k[π/2 + (π/2)]
k
k
π
=1
=1
= 1/ π

10. Illustration 2State, with reason following are probability distribution or not.
X
0
1
2
P
0.3
0.5
0.2
SolutionAll pi’s are > 0 and 0.3 + 0.5 + 0.2 = 1
..It is Probability distribution
Illustration 3State, with reason following are probability distribution or not.
X
0
1
2
3
P
0.1
0.1
0.1
0.1
SolutionNot All pi’s are > 0 but 0.1 + 0.1 + 0.1 + 0.1 – 0.4 ≠ 1
..It is not Probability distribution

11. Probability Distribution
Illustration 1Obtain the probability distribution of the number of sixes in two losses of a die.
SolutionWhen we tosses a two die possible outcomes are
{
(1,1),
(2,1),
(3,1),
(4,1),
(5,1),
(6,1),
(1,2),
(2,2),
(3,2),
(4,2),
(5,2),
(6,2),
(1,3),
(2,3),
(3,3),
(4,3),
(5,3),
(6,3),
(1,4),
(2,4),
(3,4),
(4,4),
(5,4),
(6,4),
(1,5),
(2,5),
(3,5),
(4,5),
(5,5),
(6,5),
(1,6),
(2,6),
(3,6),
(4,6),
(5,6),
(6,6)
}
Total 36. out of this 10 containing one ‘6’. One containing two ‘6’ and ‘25’ does not containing ‘6’.
Suppose ‘x’ is random variable that represent number of ‘6’.
x
0
1
2
P
25/36
10/36
1/36

12. Illustration 2Obtain the probability distribution of the number of heads in two losses of a coin.
SolutionWhen we tosses a two die possible outcomes are
(H,H), (H,T), (T,H), (T,T) Total 4.
Suppose ‘x’ is random variable that represent number of head H.
X
0
1
2
P
1/4
2/4
1/4
(T,T)
(HT,TH)
(H,H)

13. Distribution Function
Let X be a random variable. The function F defined for all real values x by F(x) = P [X = x], For all
real x is called Distribution Function.
A distribution function is also called as Cumulative Probability Distribution Function.
Properties OF Distribution Function:

If F is the D.F. of random variable X & if a <b then
P[a < X ≤ b] = F(b) – F(a).

Values of all distribution functions lie between 0 & 1
i.e. 0 ≤ F(x) ≤ 1 for all x.

All distribution functions are monotonically non-decreasing
i.e. 0 < y then F(x) < F(y).

F(- ∞) = lim F(x) = 0
x-∞
F(+ ∞) = lim F(x) = 1
x→∞

If X is discrete random variable then

If values of discrete random variable X are like x1 < x2 < x3 < x4 … then P(Xn+1) = F(Xn+1) –
F(Xn)

If x is discrete random variable then D.E. is step function.
F(x) =
xi ≤ x
∑ P(xi)

14. Illustration 1Consider probability distribution of random variable x
X=x
1
2
3
4
5
6
P[X=x]
0.1
0.2
0.3
0.2
0.1
0.1
If F(x) is distribution function of random variable x then
F(1) = P[x ≤ 1] = P(1) = 0.1
F(2) = P[x ≤ 1] = P(1) + P(2)
= 0.1 + 0.2 = 0.3
F(3) = 0.1 + 0.2 + 0.3 = 0.6
F(4) = 0.1 + 0.2 + 0.3 + 0.2 = 0.8
F(5) = 0.1 + 0.2 + 0.3 + 0.2 + 0.1 = 0.9
F(6) = 1
Thus values of x & corresponding cumulative probability distribution function is as under:
X=x
1
2
3
4
5
6
P[X=x]
0.1
0.3
0.6
0.8
0.9
1.0
We will have graphical representation of random variable X, & the Graph of D.E.
1.0
Probability Distribution
P(x)
3
0.8
f(x)
2
Cumulative
Probability Distribution
0.6
0.4
1
0.2
1
2
3
X→
4
5
6
1
2
3
X→
4
5
6

15. Variance
If X is discrete random variable then variance of x is given by
Var (X) = E[X – X (X)]2
Note that Var (X) = µ2
Var (X) = µ2` - µ1`2
Var (X) = E (X2) – [E(X)] 2
The positive square root of the variance is known as standard deviation
S D (X) = + √Var (X)
IllustrationFor a random variable X, E (X) = 10 and Var (X) = 5. Find Var (3X + 5), Var(X – 2), Var (4X).
Also find E(5X – 4), E(4X + 3).
Solution-
Var (3X + 5) = Var (3X)
= 9 Var (X) = 9 x 5 = 45
Var (x - 2) = Var (X) = 5
Var (4X) = 16 Var (X) = 16 x 5 = 80
Var (5X - 4) = 5 E(X) – 4
Var (5X - 4) = 5 x 10 – 4 = 46
Var (4X + 3) = 4 E(X) + 3
Var (4X + 3) = 4 x 10 + 3 = 43

16. Properties OF Variance:
1. Variance is independent of change of origin. It means that if X is random variable & ‘a’ is
constant then variances of X and new variable X + a are same i.e.
Var (X) + Var (X + a)
consider
Var (X + a) = E [X + a – (E (X) + a)]2
Var (X + a) = E [X + a – E (X) + a]2
Var (X + a) = E [X –E (X)]2
Var (X + a) = Var (X).
2. Variance of random variable depends upon change of scale.
i.e. Var (a X) = a2 Var (X)
consider
Var (a X) = E [aX – E (aX)]2
= E [aX – a E (X)]2
= E a2 [X –E (X)]2
= a2 E [X –E (X)]2
= a2 Var (X).
3. From property 1 and 2 we can find Var (a X + b),
Var (a X + b) = Var (a X)
= a2 Var (X)
4. Variance of constant is 0. put a=0 in a x + b then
Var (a X + b)
= Var (b)
but
Var (a x + b) = a2 × Var (X)
Var (a x + b) = 0 × Var (X) = 0
Var (b) = 0
by property 1
by property 2

17. Expectation
Mathematical Expectation of Discrete Random Variable:
Once we have determined probability distribution function P(x) &
distribution function of discrete random variable X, we want to compute the mean or
variance of random variable X, the mean or expected value of X is nothing but
weighted average of value X where corresponding probabilities are taken as weights,
thus if X takes values x1, x2,.. With corresponding probabilities p(x1), p(x2),… then
mathematical expectation of X denoted by E(X) is given by E(X) = ∑ xi P(xi).
E(X) is also mean of random variable X.
E(X) exists if series on right hand side is absolutely convergent.

18. Illustration 1The p.m.f. of a random variable X is given below. Find E(X).
Hence E(2X + 5) & E(X - 5).
X=x
1
2
3
4
5
6
P[X=x]
0.1
0.15
0.2
0.3
0.15
0.1
SolutionExpected value of random variable is given by:
E(X) = ∑xi P[X = xi]
= 1(0.1) + 2(0.15) + 3(0.2) + 4(0.3) + 5(0.15) + 6(0.1)
= 0.1 + 0.3 + 0.6 + 1.2 + 0.75 + 0.6
E(X) = 3.55
E(2X + 5) = 2 E(X) + 5
= 2 E(3.55) + 5
= 12.1
E(X - 5) = E(X) - 5
= (3.55) - 5
= -1.45

19. Illustration 2Mr. A & Mr. B plays a game. The winner is paid Rs. 99. The winner is decided by throwing a dice &
the one who gets ‘6’ first time is winner. Find expected gain of A & B if first A throws dice.
Note that if A does not win then B throws & if B does not win then A again throws this
way they continue.
Solution-
Suppose getting ‘6’ is denoted by S and not getting ‘6’ is denoted by ‘F’ then the
probability of A winning the game is
= P(S) + P(FES) + P(FFFFS) + P(FFFFFFS) + …..
= 1/6 + (5/6) 1/6 + (5/6) 1/6 + (5/6) 1/6 + …..
= 1/6 [ 1 + (5/6) + (5/6) + …..]
= 1/6 x 1/ 1- 25/36
= 1/6 x 36/11 = 6/11
Thus expected gain of A = 99 x 6/11 = 54
Expected gain of
= 99 -54 = 45

20. Properties OF Expectation:

If X1 & X2 are two random variables then E(X1 + X2) = E(X1 ) + E(X2).
This result can be generalized for X1, X2, …, Xn i.e. n random variables
E(X1 + X2 + …. + Xn ) = E(X1) + E(X2) + …. + E(Xn)


If X & Y are independent random variable, E(XY) = E(X) E(Y)
If X is a random variable & a is constant then
E(a X) = a E(X) and
E(X + a) = E(X) + a
If X is a random variable and a & b are constants
E[a (X) + b] = a E(X) + b
If X is a random variable and a & b are constants & g (X) a function X is random variable then
E[a g (X) + b] = a E[g (X)] + b
If X1, X2, X3 ….,Xn are any n random variable and if a1, a2, ….,an are any n constants then
a1 X1 + a2 X2 + …. An Xn is called linear combination of n variables & expectation of linear
combination is given by
E(a1 X1 + a2 X2 + ….. + an Xn)
=
=
=
If X ≥ 0 then E(X) ≥ 0.
If X and Y are two random variables & if X ≤ Y then E(X) ≥ E(Y).
If X & Y are independent random variable then
E[a g (X) + h (Y)] = E[g (X)] E[h (Y)]
then g (X) is a function of X & its random variable. Also h (Y) is a function of Y & random
variable.
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21. Thank You
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