Joint probability density function problem
Let X1 and X2 denote the proportion of two different chemicals found in a sample
mixture of chemicals and used as an insecticide. Suppose X1 and X2 have the joint
probability density function given by:
f(x) = { 2 0 <= x1 <= 1, 0 <= x2 <= 1, 0 <= x1+x2 <= 1
0 otherwise}
Let Y = X1 + X2, that is the total proportion of the two chemicals found in a sample
mixture.
a) Find E(Y) and V(Y)
Solution.
1)
E (Y )  E ( X 1  X 2 )


 dx  ( x

1

1
 x 2 ) f ( x1 , x 2 )dx 2

1
1 x1
0
0
  dx1  ( x1  x 2 ) * 2dx 2
1
  (1  x1 )dx1
2
0

2
3
2)
E (Y 2 )  E ( X 1  X 2 ) 2



2
 dx1  ( x1  x2 ) f ( x1 , x2 )dx2


1
1 x1
0
0
  dx1  ( x1  x 2 ) 2 * 2dx 2
1
  23 (1  x1 )dx1
3
0

1
2
Hence,
V (Y )  E (Y 2 )  [ E (Y )] 2
 12  ( 23 ) 2
 181
b) Find an interval in which values of Y should lie for at least 50% of the samples of
insecticide.
Solution.
P (Y  50%)
 P ( X 1  X 2  0.5)
 1  P ( X 1  X 2  0.5)
0.5
0.5  x1
0
0
 1   dx1
 2dx
2
0.5
 1   2 * (0.5  x1 )dx1
0
 1 
1
4
3
4
c) Find the correlation between X1 and X2 and interpret its meaning.
Solution. We can compute the following.




E( X 1 ) 
1
1 x1
0
0
  dx1
1
 dx1  x1 f ( x1 , x2 )dx2
x
1
* 2dx 2
  2 x1 (1  x1 )dx1
0

1
3


 dx1  x1 f ( x1 , x2 )dx2
E( X 1 ) 
2
2

1
  dx1

1 x1
x
2
* 2dx 2
1
0
0
1
  2 x1 (1  x1 )dx1
2
0

1
6
1
1 x1
0
0
  dx1




 dx1  x2 f ( x1 , x2 )dx2
E( X 2 ) 
x
2
* 2dx 2
1
  (1  x1 ) 2 dx1
0

1
3


 dx1  x2 f ( x1 , x2 )dx2
E( X 2 ) 
2
2

1
  dx1
0

1 x1
x
2
* 2dx 2
2
0
1
  23 (1  x1 ) 3 dx1
0

1
6




E( X 1 X 2 ) 
1
1 x1
0
0
  dx1
1
x x
 dx1  x1 x2 f ( x1 , x2 )dx2
1 2
* 2dx 2
  x1 (1  x1 ) 2 dx1
0
 121
So,
Cov( X 1 , X 2 )  E ( X 1 X 2 )  E ( X 1 ) E ( X 2 )
 121  13 * 13
  361
and
V ( X 1 )  E ( X 1 )  [ E ( X 1 )] 2
2
 16  ( 13 ) 2  181 ,
V ( X 2 )  E ( X 2 )  [ E ( X 2 )] 2
2
 16  ( 13 ) 2  181
Hence, the coefficient of correlation is
( X1 , X 2 )
  Cov
 11//1836   12
V ( X )V ( X )
1
2
As the coefficient of correlation is
( X1 , X 2 )
  Cov
 11//1836   12
V ( X )V ( X )
1
2
which is less than 0, we conclude that there is a negative relationship between X1 and X2.