Question

The relation between the Yule's coefficient Q and colligation coefficient Y is:

• A. \(Q = Y/(1-Y^2)\)
• B. \(Y = Q/(1+Q^2)\)
• C. \(Q = 2Y/(1+Y^2)\)
• D. \(Y = 2Q/(1+Q^2)\)

Hint:

Write Q and Y in terms of \(p=\sqrt{ad}\) and \(q=\sqrt{bc}\), then simplify \(2Y/(1+Y^2)\).

Answer 1

The Correct Option is C

Step 1: Let \(a=(AB)\), \(b=(A\beta)\), \(c=(\alpha B)\), \(d=(\alpha\beta)\) be the four ultimate class frequencies for two attributes.

Step 2: Yule's coefficient of association is \(Q = \dfrac{ad-bc}{ad+bc}\), and the coefficient of colligation is \(Y = \dfrac{\sqrt{ad}-\sqrt{bc}}{\sqrt{ad}+\sqrt{bc}}\).

Step 3: Let \(p=\sqrt{ad}\) and \(q=\sqrt{bc}\). Then \(Y=\dfrac{p-q}{p+q}\) and \(Q=\dfrac{p^2-q^2}{p^2+q^2}\).

Step 4: Compute \(1+Y^2 = 1+\dfrac{(p-q)^2}{(p+q)^2} = \dfrac{(p+q)^2+(p-q)^2}{(p+q)^2} = \dfrac{2p^2+2q^2}{(p+q)^2}\).

Step 5: So \(\dfrac{2Y}{1+Y^2} = \dfrac{2(p-q)/(p+q)}{[2p^2+2q^2]/(p+q)^2} = \dfrac{2(p-q)(p+q)}{2(p^2+q^2)} = \dfrac{p^2-q^2}{p^2+q^2} = Q\).

Step 6: Hence \(Q = \dfrac{2Y}{1+Y^2}\), matching option (C).

\[\boxed{Q=\frac{2Y}{1+Y^2}}\]