# Comparison of two scaled binomial or normal random variables

Let $$X$$ and $$Y$$ be two independent binomial random variables where 𝑋∼𝐵(𝐾,𝑞), 𝑌∼𝐵(𝐾,𝑝) (suppose $$p>q$$), $$a$$ and $$b$$ two real numbers. What conditions can we impose on $$a$$ and $$b$$ such that the probability Pr($$aX>bY$$) is approaching

• 1
• 1/2
• 0

as $$K$$ goes to infinity.

• Assuming neither of $q$ or $p$ equals $0$ or $1,$ just apply the Central Limit Theorem.
– whuber
Mar 27 '20 at 16:05

Let $$Z_x = \frac{X - Kq}{\sqrt{K}}, Z_y = \frac{Y - Kp}{\sqrt{K}}$$

By central limit theorem, $$Z_x \rightarrow_d N(0,q(1-q)), Z_y \rightarrow_d N(0,p(1-p))$$.

Then

$$P(aX > bY) = P(\frac{aX}{\sqrt{K}} > \frac{bY}{\sqrt{K}})$$

$$= P(a[Z_x + \sqrt{K}q] > b[Z_y + \sqrt{K}p])$$

$$= P(aZ_x - bZ_y > \sqrt{K}(bp - aq))$$

Note that, as K approaches infinity, $$aZ_x - bZ_y$$ converges to a $$N(0, a^2q(1-q) + b^2p(1-p))$$. The behavior of the probability depends on $$\sqrt{K}(bp - aq)$$:

• when $$bp - aq > 0$$, the probability approaches 0
• when $$bp - aq = 0$$, the probability approaches 1/2
• when $$bp - aq < 0$$, the probability approaches 1

BTW, it would feel much better to use $$p$$ for $$X$$ and $$q$$ for $$Y$$ :)

• +1. You could avoid the complications of dealing with two variables simply by applying the CLT to $aU-bV$ where $U$ and $V$ independently have Bernoulli$(q)$ and Bernoulli$(p)$ distributions.
– whuber
Mar 27 '20 at 16:47