Mathematics Assignment - Desklib

Mathematics Assignment
Student Name:
Instructor Name:
Course Number:
2nd October 2019

Q1.
a) P(X=Y) = P(X=Y=1) or P (X=Y=2) or P(X=Y=3) or P(X=Y=4)
=0.04+0.04+0.04+0.06
P(X=Y) =0.18
P(X¿ 2 ,Y≤ 3 ¿) = P(X¿ 3 ,Y¿ 1 ¿or P(X¿ 3 ,Y¿ 2 ¿or P(X¿ 3 ,Y¿ 3 ¿or P(X¿ 4 ,Y¿ 1 ¿or P(X¿ 4 ,Y¿ 2 ¿or
P(X ¿ 3 ,Y¿ 3 ¿
P(X¿ 2 ,Y≤ 3 ¿=0.06+0.05+0.04+0.05+0.10+0.03
P(X¿ 2 ,Y≤ 3 ¿= 0.33
b) Marginal distributions f X(x ¿ of X
We need to write the row totals
Marginal distributions f Y ( y ¿ of Y
We need to write the column totals
c) Conditional probability for X given that Y=3
f X∨ y ( x ¿= f XY (x , y )
f X (x) = Joint probability
marginal probability
P ( X =x , Y =3)
P (Y =3) = P (X =1, Y =3)
P(Y =3) ∨ P(X =2 ,Y =3)
P(Y =3) ∨ P( X=3 , Y =3)
P(Y =3) ∨ P ( X=4 , Y =3)
P(Y =3)
P ( X =1, Y =3)
P(Y =3) = 0.01
0.1 =0.1
y=1 y=2 y=3 y=4 y=5 Total
x=1 0.04 0.11 0.01 0.07 0.07 0.3
x=2 0.05 0.04 0.02 0.05 0.04 0.2
x=3 0.06 0.05 0.04 0.02 0.03 0.2
x=4 0.05 0.1 0.03 0.06 0.06 0.3
Total 0.2 0.3 0.1 0.2 0.2 1
x 1 2 3 4
f X(x ¿ 0.3 0.2 0.2 0.3
x 1 2 3 4 5
f Y ( y ¿ 0.2 0.3 0.1 0.2 0.2

P ( X =2, Y =3)
P(Y =3)
0.02
0.1 =0.2
P ( X =3 ,Y =3)
P(Y =3)
0.04
0.1 =0.4
P ( X =4 , Y =3)
P(Y =3)
0.03
0.1 =0.3
Conditional probability of X given that Y=3
d) Independence
For discrete variables, independence means that the probability in a cell must be
the product of the marginal probability of its rows and columns.
In row 1 and column 1, the probability is 0.04
However the product of marginal probability is 0.2×0.3=0.06
0.04≠ 0.06
X and Y are dependent (not independent) since 0.04 ≠ 0.06
Q2.)
Suppose X and Y are independent.
Let 0≤ k ≤ m+n
P(X+Y=k) =∑
i=0
k
P( X=i , Y =k −i)
∑
i=0
k
P ( X =i ) P(Y =k −i)
But P ( X=i ) =(m
i ) p
i
(1− p)m −i and P(Y =k−i)=( n
k−i ) pk−i (1−p)n−k +i
∑
i=0
k
P ( X =i ) P(Y =k −i) becomes
=∑
i=0
K
(m
i ) pi (1−p)m−i(( n
k−i ) pk−i (1−p)n−k +i
= pk (1−p)n +m−k
∑
i=0
k
(m
i )( n
k −i )
x 1 2 3 4
f X∨ y=3(
x ¿
0.1 0.2 0.4 0.3

Taking ( n+ m
k )=∑
i=0
k
(m
i )( n
k −i )
= pk (1−p)n +m−k
∑
i=0
k
(m
i )( n
k −i ) is reduced to
¿ ( n+m
k ) pk (1−p)n +m−k
Thus we may conclude that X+Y have a binomial distribution with the
parameters m+n and p.
Q3).
a)
We need to draw the region D which corresponds to 0 ≤ y ≤ 1. This is the region
above y ≥0 and below the line y=1.
The line y=1 intersects the line x=0 and x=2 at points (0,1) and (2,1).
Thus we have g1 ( x ) =1∧f 2 ( x ) =1
∫∫ f ( x , y ) dA=∫
0
2
∫
0
1
C x2 y3 dydx
¿ C∫
0
2
[∫
0
2
1
4 x2 y 4
]0
1
dx
¿ C∫
0
2
1
4 x
2
=1
¿ 8
12 C=1
y=1
y=0
x=2
x=0