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Assignment 3 Probability and Statistics I: a binary problem
Deadline for delivery: Friday March 27th 2009, 3.30
Binary numbers consist of digits 0 and 1. Suppose we have an arbitrary binary number of 5 of
these digits (the probabilities of 0 and 1 are equal). A sequence of identical digits is called a
run. Example: 01110 consists of three runs and the length of the largest run is 3.
Consider an arbitrary 5-digit binary number: X is the length of the largest run and Y is the
number of runs.
a. Determine the table of the joint distribution of X and Y. (First list all possible 5-digit
binary number and their value of X and Y.)
b. Determine the distribution of the length of the longest run, its expected value and its
variance.
c. Compute the correlation coefficient and give an interpretation of its value.
d. Determine E(X | X +Y = 5) and var( X | X+Y = 5)
Solutions: a.
rij
X Y
11111
5 1
11110
4 2
11101
3 3
11011
2 3
10111
3 3
01111
4 2
11100
3 2
11010
2 4
10110
2 4
01110
3 3
11001
2 3
10101
1 5
01101
2 4
10011
2 3
01011
2 4
00111
3 2
P(X=x en Y=y) :
rij
00000
00001
00010
00100
01000
10000
00011
00101
01001
10001
00110
01010
10010
01100
10100
11000
x \ y
1
2
3
4
1
0
0
0
0
2
0
0
3
0
2
16
4
0
2
16
5
1
16
0
P(Y=y)
1
16
4
16
X
5
4
3
2
3
4
3
2
2
3
2
1
2
2
2
3
5
1
16
Y
1
2
3
3
3
2
2
4
4
3
3
5
4
3
4
2
P(X=x) xP(X=x)
2
x P( X  x )
1
16
1
16
1
16
0
7
16
14
16
28
16
0
0
5
16
15
16
45
16
0
0
0
2
16
8
16
32
16
0
0
0
1
16
5
16
25
16
3
16
3
16
6
16
4
16
4
16
1
16
1
43
16
 E( X )
131
16
 E ( X 2)
2
b. See the computation in the table above. var( X )  E ( X 2)  ( EX ) 
c. E(Y)=3 and var(Y)=1, analogue X
247
256
, so:
E ( XY )   xyP( X  x  Y  y)  5  1 161  ....  115
16
115
43
14
cov(X,Y)= E(XY) - E(X)E(Y)= 16  16  3   16
(X ,Y ) 
cov(X , Y )
 X Y

 14
16   14  0.8908
247 1
247
256
There is a quite strong negative linear relation between the number of runs and the largest run
length (large run length coincide with just a few runs)
d. If X+Y = 5, then X can only take on the values 2 0r 3:
P(X=2|X+Y=5) =
P( X  2 en X  Y  5) P( X  2 en Y  3)
3 / 16


 0.6
P( X  Y  5)
P( X  Y  5)
3 / 16  2 / 16
And: P(X=3|X+Y=5) = 0.4
So E( X | X  Y  5)   xP( X  x | X  Y  5)  2 0.6  3  0.4  2.4
E( X 2 | X  Y  5)   x 2 P( X  x | X  Y  5)  4 0.6  9  0.4  2.4
var( X | X+Y = 5)= E(X2 | X +Y = 5) - E(X | X +Y = 5) 2 = 6 – 2.42 = 0.24
Marks:
l
a
b
c
d
points
2
2
3
3
Comment
Up to EX: 1 pt ,
varX: 1 pt
E(XY): 1 pt,
: 1 pt,
interpretation 1 pt
Conditional probs 1 pt , E(X|X+Y=5) 1 pt, var(X|X+Y=5) 1 pt