Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

X and Y are 2 independent binomial random variables with parameters (n,p) and (m,q) respectively. (trials, probability parameter)

share|improve this question
3  
Isn't there positive probability that this ratio is undefined? –  guy Aug 27 '11 at 4:30

3 Answers 3

If $n$ and $m$ go to infinity while $p$ and $q$ are fixed, then the ratio $R=X/Y$ is well defined on an event of probability $1-o(1)$.

On this event (or conditionally on this event, since the two asymptotically equivalent), Edgeworth expansions of $X$ and $Y$ show that the expectation of $R$ behaves like $\dfrac{np}{mq}$ and that its variance behaves like $$ \left(\frac{np}{mq}\right)^2\left(\frac{1-p}{np}+\frac{1-q}{nq}\right). $$

share|improve this answer

There probably isn't a closed-form formula for this.

But $X$ has mean $np$ and standard deviation $\sqrt{np(1-p)}$, and $Y$ has mean $mq$ and standard deviation $mq(1-q)$.

Now you need a simple fact: if $X$ has mean $\mu$ and standard deviation $\sigma$, then $\log X$ has mean approximately $\log \mu$ and standard deviation approximately $\sigma/\mu$. This can be derived by Taylor expansion. Intuitively, $X$ "usually" falls in $[\mu-\sigma, \mu+\sigma]$ and so $\log X$ "usually" falls in $[\log (\mu-\sigma), \log (\mu+\sigma)]$. But we have

$$ \log (\mu \pm \sigma) = \log (\mu(1 \pm \sigma/mu)) = \log \mu \pm \log(1 \pm \sigma/mu) \approx \log \mu \pm \sigma/mu $$

where the approximation is the first-order Taylor expansion of $\log (1+x)$.

Therefore $\log X$ has mean approximately $\log np$ and standard deviation approximately $\sqrt{(1-p)/np}$; similarly $\log Y$ has mean approximately $\log mq$ and standard deviation approximately $\sqrt{(1-q)/mq}$.

So $\log X - \log Y = \log X/Y$ has mean approximately $\log(np/mq)$ and standard deviation approximately $$ \sqrt{{(1-p) \over np} + {(1-q) \over mq}}. $$

But you asked about $X/Y$. Inverting the earlier fact, if $Z$ has mean $\mu$ and standard deviation $\sigma$, then $e^Z$ has mean approximately $e^{\mu}$ and standard deviation approximately $\sigma e^\mu$. Therefore $X/Y$ has mean approximately $np/mq$ (not surprising!) and standard deviation approximately

$$ \left( \sqrt{{(1-p) \over np} + {(1-q) \over mq}} \right) {np \over mq}. $$

share|improve this answer

In essence you are asking for the distribution of the variable Z = X/Y, where Z can take values in $\left\{ 0, \frac{1}{m}, \ldots, \frac{n}{1}, \infty \right\}$, where the fractions are of the form $\frac{k}{l}$ with $0 \leq k \leq n$, and $1 \leq l \leq m$, and I'm assuming the convention $\frac{k}{0} = \infty$, which as Guy pointed out means this will be undefined, and the standard deviation will be infinite. So instead condition that this cannot happen $$ \mathbb{P}\left[Z = k/l | Y \neq 0 \right] = \frac{\mathbb{P}[Xl = YK \cap Y \neq 0]}{\mathbb{P}[Y \neq 0]} $$ So then this gives, $$ \begin{align} \mathbb{P}\left[Z = k/l | Y \neq 0 \right] & = (1- (1-q)^m) \mathbb{P}[ Xl = Yk \cap Y \neq 0]\\ & = (1-(1-q)^m) \sum p^i(1-p)^{(n-i)}q^j(1-q)^{(m-j)} \end{align} $$ where the sum is over all pairs $(i,j)$ with $0 \leq i \leq n$, $0 < j \leq m$ such that $i/j = k/l$... All in all it looks like its going to be a messy task to compute this sum, and to find the standard deviation.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.