please answer the question below:

Question

(a) Prove that the ? coefficient is equal to 1 if and only if f11 = f1+ = f+1.

(b) Show that if A and B are independent, then P(A, B) × P(A, B) =

P(A, B) × P(A, B).

(c) Show that Yule’s Q and Y coefficients

(d) Write a simplified expression for the value of each measure shown in

Tables 6.11 and 6.12 when the variables are statistically independent.

Answer 1

(a) Instead of proving f11 = f1+ = f+1, we will show that P(A, B) =

P(A) = P(B), where P(A, B) = f11/N, P(A) = f1+/N, and P(B) =

f+1/N. When the ?-coefficient equals to 1:

? = P(A, B) ? P(A)P(B)



P(A)P(B)



1 ? P(A)

1 ? P(B)



= 1

The preceding equation can be simplified as follows:



P(A, B) ? P(A)P(B)

2

= P(A)P(B)



1 ? P(A)

1 ? P(B)





P(A, B)

2 ? 2P(A, B)P(A)P(B) = P(A)P(B)



1 ? P(A) ? P(B)





P(A, B)

2 = P(A)P(B)



1 ? P(A) ? P(B)+2P(A, B)





We may rewrite the equation in terms of P(B) as follows:

P(A)P(B)

2 ? P(A)



1 ? P(A)+2P(A, B)



P(B) + P(A, B)

2 = 0

The solution to the quadratic equation in P(B) is:

P(B) = P(A)? ? P(A)2?2 ? 4P(A)P(A, B)2

2P(A) ,

where ? = 1 ? P(A)+2P(A, B). Note that the second solution, in

which the second term on the left hand side is positive, is not a feasible

solution because it corresponds to ? = ?1.