COMPLEX ANALYSIS,PROBABILITY AND STATISTICAL METHODS(18MAT41)
MODULE - 05
JOINT PROBABILITY DISTRIBUTION
INTRODUCTION
We have discussed probability distribution associated with a single random
variable. The same can be generalized for two or more random variables. We
discuss probability distributions associated with two random variables referred
to as a joint distribution.
JOINT DISTRIBUTION AND JOINT PROBABILITY DISTRIBUTION
If X & Y are two discrete random variables, we define the joint probability
function of X & Y by
P (X = x, Y = y) = f (x,y)
Where f(x,y) satisfy conditions
f(x,y) ≥0 and ∑𝑥 ∑𝑦 𝑓(𝑥, 𝑦) = 1
The second condition means that the sum over all the values of x and y is equal
to one.
Suppose X = {𝑥1 , 𝑥2 , … 𝑥𝑚 } and Y = {𝑦1 , 𝑦2 , … 𝑦𝑛 } then P ( X = xi , Y = yi)
denoted by J ij.
It should be observed that f is a function on the Cartesian product of the sets X
and Y as we have
X × Y = {(𝑥1, 𝑦1 ), (𝑥2 , 𝑦2 ) … (𝑥𝑚 , 𝑦𝑛 )}
f is also referred to as joint probability density function of X and Y in the
respective order. The set of values of this function f(xi, yi) = J ij for i = 1,2,...m, j
= 1,2,...n is called the joint probability distribution of X and Y.These values are
presented in the form of a two way table called the joint probability table.
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X Y 𝑦1 𝑦2 ... 𝑦𝑛 Sum
𝑥1 𝐽11 𝐽12 𝐽1𝑛 f(x1)
𝑥2 𝐽21 𝐽22 ... 𝐽2𝑛 f(x2)
... ... ... ... ...
𝑥𝑚 𝐽𝑚1 𝐽𝑚2 ... 𝐽𝑚𝑛 f(xm)
𝑠𝑢𝑚 g(y1) g(y2) g(yn) 1
MARGINAL PROBABILITY DISTRIBUTION
In the joint probability table {f(x1), f(x2). . . f(xm) }are the sum of horizontal
entries and {g(y1), g(y2), . . . g(yn)} are the sum of vertical entries in the joint
probability distribution table. These are called marginal probability distribution
of X and Y respectively.
INDEPENDENT RANDOM VARIABLES
The discrete random variable X and Y are said to be independent random
variables if P ( X = xi , Y = yj) = P ( X = xi ) ∙ P(Y = yi)
i.e f(xi) g(yj) = Jij
Expectation, Variance, Covariance and Correlation
Expectation
𝜇𝑋 = E(X) = ∑𝑥 ∑𝑦 𝑥 𝑓(𝑥, 𝑦) = ∑𝑖 𝑥𝑖 𝑓(𝑥𝑖 )
𝜇𝑌 = E(Y) = ∑𝑥 ∑𝑦 𝑦 𝑓(𝑥, 𝑦) = ∑𝑗 𝑦𝑗 𝑔(𝑦𝑗 )
𝜇𝑋𝑌 = E(XY) = ∑𝑖 ∑𝑗 𝑥𝑖 𝑦𝑗 𝐽𝑖𝑗
Variance
𝜎𝑋2 = E ( X2) – [𝐸(𝑋)]2
𝜎𝑌2 = E ( Y2) – [𝐸(𝑌)]2
Covariance
COV (X,Y) = E (XY) – E(X)∙ E(Y)
Correlation
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𝐶𝑂𝑉 (𝑋,𝑌)
Correlation of X and Y = 𝜌(𝑋, 𝑌) =
𝜎𝑋 𝜎𝑌
NOTE:
If X and Y are independents, E(X , Y) = E(X)∙ E(Y) and hence
COV (X , Y) = 0 = 𝜌(𝑋, 𝑌)
PROBLEMS
1. The joint probability distribution of two random variables X and Y ia as
follows.
X Y -4 2 7
1 1/8 1//4 1/8
5 1/4 1/8 1/8
Compute the following
(a) E(X) and E(Y) (b) E( XY) ( c) 𝜎𝑋 and 𝜎𝑦 (d) COV (X , Y)
( e) 𝜌(𝑋, 𝑌)
Solu: The distribution is obtained adding the all the respective row entries
and also the respective coloumn entries.
Distribution of X : Distribution of Y :
xi 1 5 yj -4 2 7
f(xi) 1/2 1/2 g(yj) 3/8 3/8 1/4
a) E(X) = ∑ 𝑥𝑖 𝑓(𝑥𝑖 ) = (1)(1/2) + 5 (1/2) = 3 = 𝜇𝑥
E(Y) = ∑ 𝑦𝑗 𝑔(𝑦𝑗 ) = (-4)(3/8) + 2( 3/8) + 7(1/4) = 1 = 𝜇𝑦
b) E (XY) = ∑ 𝑥𝑖 𝑦𝑗 𝐽𝑖𝑗 = (1)(-4)(1/8) + (1) (2) ( 1/4) + (1) (7) ( 1/8)
+(5) (-4) ( 1/4)+ (5) (2) ( 1/8)+ (5) (7) ( 1/8)
= 3/2
c) 𝜎𝑋2 = E ( X2) – [𝐸(𝑋)]2 and 𝜎𝑌2 = E ( Y2) – [𝐸(𝑌)]2
Now E ( X2) = ∑ 𝑥 2 𝑓(𝑥𝑖 ) = (1)(1/2)+25(1/2) = 13
E ( Y2) = ∑ 𝑦 2 𝑔(𝑦𝑗 ) = (16)(3/8)+(4)(3/8) + (48)(1/4) = 79/4
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Hence 𝜎𝑋2 = 13 – (3)2 = 4 and 𝜎𝑌2 = (79/4) – (1)2= 75/4
75
Thus 𝜎𝑋 = 2 and 𝜎𝑦 = √( ) = 4.33
4
d) COV (X,Y) = E (XY) – E(X)∙ E(Y)
= (3/2) – 3 (1) = - 3/2
3
𝐶𝑂𝑉 (𝑋,𝑌) (− )
2
e) 𝜌(𝑋, 𝑌) = = = - 0. 1732.
𝜎𝑋 𝜎𝑌 75
(2)√( )
4
2. The joint probability distribution table for two random variables X and Y is
as follows.
X Y -2 -1 4 5
1 0.1 0.2 0 0.3
2 0.2 0.1 0.`1 0
Determine the marginal probability distributions of X and Y. Also compute
(a) Expectations of X , Y and XY
(b) S.D’s of X,Y
(c) covariance of X and Y (d) Correlation of X and Y
Further verify that X and Y are dependent random variables
Solu: Marginal distributions of X and Y are got by adding all the respective
row entries and the respective column entries.
xi 1 2 yj -2 -1 4 5
f(xi) 0.6 0.4 g(yj) 0.3 0.3 0.1 0.3
(a)
= E(X) = ∑ 𝑥𝑖 𝑓(𝑥𝑖 ) = (1)(0.6) + (2)(0.4) = 1.4
𝜇𝑦 = E(Y) = ∑ 𝑦𝑗 𝑔(𝑦𝑗 ) = (-2) (0.3) + (-1)(0.3) + 4 (0.1) + 5 (0.3) = 1
E (XY) = ∑ 𝑥𝑖 𝑦𝑗 𝐽𝑖𝑗 = (1)(-2)(0.1) + (1) (-1) (0.2) + (1) (4) (0) + (1) (5) (0.3)
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+ (2) (-2) (0.2) + (2) (-1) ( 0.1) + (2) (4) (0.1) + (2)(5) (0)
= 0.9
b) 𝜎𝑋2 = E ( X2) – [𝐸(𝑋)]2 and 𝜎𝑌2 = E ( Y2) – [𝐸(𝑌)]2
Now E ( X2) = ∑ 𝑥 2 𝑓(𝑥𝑖 ) = (1)(0.6) + (4) (0.4) = 2.2
E ( Y2) = ∑ 𝑦 2 𝑔(𝑦𝑗 ) = (4)(0.3) + 1(0.3) + 16(0.1) + 25(0.3) = 10.6
Hence 𝜎𝑋2 = 2.2 – (1.4)2 = 0.245 and 𝜎𝑌2 = (10.6) – (1)2= 9.6
Thus 𝜎𝑋 = 0,49 and 𝜎𝑦 = 3.1
c) COV (X,Y) = E (XY) – E(X)∙ E(Y)
= 0.9 – 1.4(1) = - 0.5
𝐶𝑂𝑉 (𝑋,𝑌) (−0.5)
e) 𝜌(𝑋, 𝑌) = = (0.49)(3.1) = - 0.3.
𝜎𝑋 𝜎𝑌
If X and Yare independent random variables we must have
f(xi) g(yj) = Jij
It can be seen that f(x1)g(y1) = (0.6)(0.3) = 0.18 and J11 = 0.1
i.e f(x1)g(y1) ≠ J11
Hence we conclude that X and Y are dependent random variables.
3.The joint probability distribution of two discrete random variables X and Y is
given by f(x,y)= k(2x + y) where x and y are integers such that 0 ≤ x ≤ 2 , 0 ≤
y≤3.
(a) Find the value of the constant k
(b) Find the marginal probability distributions of X and Y
(C) Show that the random variable X and Y are dependent.
Solu: X = {𝑥𝑖 } = {0,1,2} and Y = {𝑦𝑗 } = {0,1,2,3}
f( x,y) = k(2x+y) and the joint probability distribution table is
formed as follows.
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X 0 1 2 3 Sum
Y
0 0 k 2k 3k 6k
1 2k 3k 4k 5k 14k
2 4k 5k 6k 7k 22
Sum 6k 9k 12k 15k 42k
a) We must have 42k = 1
∴ k = 1/42
b) Marginal probability distribution is as follows.
xi 0 1 2 yj 0 1 2 3
f(xi) 6/42 4/42 22/42 g(yj) 6/42 9/42 12/4 15/42
= 1/7 =1/3 =11/21 =1/7 =3/14 =2/7 =5/14
c) It can be easily seen that f(xi) g(yj) ≠ Jij
Hence the random variables ade dependent.
4. A fair coin is tossed thrice. The random variables X and Y are defined as
follows. X = 0 or 1 according as head or tail occurs on the first toss.
Y = Number of heads
(a) Determine the distribution of X and Y
(b)Determine the joint distribution of X and Y
(C) Obtain the expectations of X,Y and XY. Also find S.Ds of
X and Y
(d) Compute Covariance and Correlation of X and Y.
Solu. The sample space S and the association of random variables X
and Y is given by the following table
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S HHH HHT HTH HTT THH THT TTH TTT
X 0 0 0 0 1 1 1 1
Y 3 2 2 1 2 1 1 0
(a) The probability distribution of X and Y is found as follows.
X = {𝑥𝑖 } = {0,1} and Y = {𝑦𝑗 } = {0,1,2,3}
P(X=0) is 4/8 = 1/2, P( X = 1) is 4/8 = 1/2
P(Y=0) is 1/8 , P( Y = 1) is 3/8
P(Y=2) is 3/8 , P( Y = 3) is 1/8
Thus we have the following probability distribution of X and Y
xi 0 `1 yj 0 1 2 3
f(xi) 1/2 1/2 g(yj) 1/8 3/8 3/8 1/8
(b)The joint distribution of X and Y is found by computing
Jij = P( X = xi , Y = yj ) where we have
X1 = 0, X2 = 1 and y1 = 0, y2 = 1 , y3 = 2 , y4 = 3
J 11 = P ( X = 0, Y = 0) = 0
(X = 0 implies that their is a head turn out and Y the total
number of heads 0 is impossible)
J 12 = P ( X = 0, Y = 1) = 1/8 correspondings to the outcome HTT
J 13 = P ( X = 0, Y = 2) = 2/8=1/4; outcomes are HHT and HTH
J 14 = P ( X = 0, Y = 3) = 1/8;outcome is HHH
J 21 = P ( X = 1, Y = 0) = 1/8,outcome is TTT
J 22 = P ( X = 1, Y = 1) = 2/8=1/4; outcomes are THT ,TTH
J 23 = P ( X = 1, Y = 2) = 1/8,outcome is THH
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J 24 = P ( X = 1, Y = 3) = 0 since the outcome is impossible.
(These values can be written quickly by looking at the table of S ,
X,Y)
The required joint probability distribution of X and Y is as follows.
X 0 1 2 3 Sum
Y
0 0 1/8 1/4 1/8 1/2
1 1/8 1/4 1/8 0 1/2
Sum 1/8 3/8 3/8 1/8 1
(c) 𝜇𝑥 = E(X) = ∑ 𝑥𝑖 𝑓(𝑥𝑖 ) = (0) (1/2) + (1)(1/2) = 1/2
𝜇𝑦 = E(Y) = ∑ 𝑦𝑗 𝑔(𝑦𝑗 ) = (0) (1/8) + (1)(3/8) + 2 (3/8) + 3 (1/8) = 12/8 =3/2
E (XY) = ∑ 𝑥𝑖 𝑦𝑗 𝐽𝑖𝑗 = 0 + (0+ ¼+ 2/8 + 0) = 1/2
𝜎𝑋2 = E ( X2) – [𝐸(𝑋)]2 and 𝜎𝑌2 = E ( Y2) – [𝐸(𝑌)]2
𝜎𝑋2 = (0 + ½ ) – ¼ = ¼ 𝜎𝑌2 = (0+3/8 + 3/2 + 9/8) – ( 9/4) = 3- (9/4) = 3/4
Thus 𝜎𝑋 = 1/2 and 𝜎𝑦 = √3/2
c) COV (X,Y) = E (XY) – E(X)∙ E(Y)
= ½ - ¾ = - 1/4
𝐶𝑂𝑉 (𝑋,𝑌) (−1/4) 1
𝜌(𝑋, 𝑌) = = =-
𝜎𝑋 𝜎𝑌 √3/4 √3
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SAMPLING THEORY
INTRODUCTION
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Test for significance for large samples
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