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Let $X$ and $Y$ be two independent binomial random variables where ๐‘‹โˆผ๐ต(๐พ,๐‘ž), ๐‘Œโˆผ๐ต(๐พ,๐‘) (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.

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  • $\begingroup$ Assuming neither of $q$ or $p$ equals $0$ or $1,$ just apply the Central Limit Theorem. $\endgroup$
    – whuber
    Mar 27, 2020 at 16:05

1 Answer 1

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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 $Y$ :)

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    $\begingroup$ +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. $\endgroup$
    – whuber
    Mar 27, 2020 at 16:47

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