Question

1. . Toss 4 fair coins and consider the following two r.v.'s:

X = number of Head's in the first 3 coin tosses, Y = number of Head's in the last 2 coin tosses, what is the covariance Cov(3+2X,4+Y )?

Answer 1

For all the outcomes, the value of X and Y are first computed here as:

Outcome	X	Y	3 + 2X	4 + Y	(3 + 2X)(4 + Y)
HHHH	3	2	9	6	54
HHHT	3	1	9	5	45
HHTH	2	1	7	5	35
HTHH	2	2	7	6	42
THHH	2	2	7	6	42
HHTT	2	0	7	4	28
HTHT	2	1	7	5	35
HTTH	1	1	5	5	25
THTH	1	1	5	5	25
TTHH	1	2	5	6	30
THHT	2	1	7	5	35
TTTH	0	1	3	5	15
TTHT	1	1	5	5	25
THTT	1	0	5	4	20
HTTT	1	0	5	4	20
TTTT	0	0	3	4	12

The expected values are first computed here as:

E(3 + 2X) = Average of the (3 + 2X) values

= (9*2 + 7*6 + 5*6 + 2*3) / 16

= 6

E(4 + Y) is computed here as: = average of the (4 + Y) column above

= (4*6 + 5*8 + 4*4)/16

= 5

E [ (3 + 2X)(4 + Y)] is computed as the avg of the last column in the table as:

= (54 + 45 + ... + 12)/16

= 488 / 16

= 30.5

Therefore now the covariance here is computed as:

Cov(3 + 2X, 4 + Y) = E [ (3 + 2X)(4 + Y)] - E(3 + 2X)E(4 + Y)= 30.5 - 6*5 = 0.5

Therefore 0.5 is the required variance here.