3
$\begingroup$

Suppose we have a multinomial distribution with support $(X_1,...,X_n)$ and $\sum_{i=1}^nX_i=N$. Consider the probability distribution of $X_1/X_2$, say. Does this distribution have an expected value and variance? If so, can they be calculated or approximated?

The reason for the question is that in the limit as $n\rightarrow{}\infty{}$, $X_1$ and $X_2$ approach normal distributions and the ratio of normal distributions has no expected value (see e.g. Wikipedia on ratio distributions). Simulations seem to suggest that the multinomial case is better behaved and that $E(\frac{X_1}{X_2})\cong{}\frac{E(X_1)}{E(X_2)}$.

The question arose in trying to use the delta method to calculate the expected value and variance of $X_1$ and $X_2$ (in the multinomial case). From the above, it would appear that the delta method is inapplicable in this case.

Improve this question
$\endgroup$
5
  • $\begingroup$ Since the probability of $X_2$ being zero is positive (assuming the multinomial distribution is non-deterministic) the quantity $X_1/X_2$ is ill-defined with positive probability. $\endgroup$
    – J. Virta
    Commented Feb 9, 2017 at 13:24
  • $\begingroup$ Thank you. This is a good point, as if $X_1$ and $X_2$ are normal, $\frac{X_1}{X_2}$ is only ill-defined with probability 0. However, the only related reference I could find (Koopman, P. A. R. "Confidence intervals for the ratio of two binomial proportions." Biometrics (1984): 513-517.) would surely have the same problem with the ratio of two independent binomial random variables? In fact, my original interest was in the distribution (in the multinomial case) of $\frac{X_1}{X_1+X_2}$ where defining this as 0 for $X_1+X_2=0$ seems more reasonable. Would this make the problem better posed? $\endgroup$
    – Helmut
    Commented Feb 9, 2017 at 21:59
  • $\begingroup$ Hi @Helmut I am also trying to evaluate the X1/(X1 + X2) case, did you ever manage to find a solution? I am planning on seeing if the results in the paper listed in the accepted answer below can be reapplied to that case. But thought I's ask incase you had a solution... Thanks :) $\endgroup$
    – EHH
    Commented Oct 3, 2023 at 12:36
  • $\begingroup$ @EHH To be honest, I can't really recall this now, and the Duris paper sounds like your best option. Good luck. $\endgroup$
    – Helmut
    Commented Oct 4, 2023 at 0:18
  • $\begingroup$ @Helmut Thanks for getting back to me on this. OK I will roll up my sleeves and see what I can get out of the Duris formulations :) cheers! $\endgroup$
    – EHH
    Commented Oct 5, 2023 at 8:22

1 Answer 1

Reset to default
2
$\begingroup$

See this paper: Mean and variance of ratios of proportions from categories of a multinomial distribution, it deals exactly with this question.

https://jsdajournal.springeropen.com/articles/10.1186/s40488-018-0083-x

Briefly, for reasons stated in the paper, it is enough to consider the case where n=3. Then, it is shown that

$E\left[\frac{X_1}{X_2}\right] \approx \frac{p_1}{p_2}\left(1 + \frac{1}{Np_2}\right)$,

$var\left[\frac{X_1}{X_2}\right] \approx \frac{1}{N}\left(\frac{p_1}{p_2}\right)^2\left(\frac{1}{p_1} + \frac{1}{p_2}\right)$,

where the considered multinomial distribution is given by $(N,p_1,p_2,p_3)$. These results are obtained through delta method. There is also a more precise answer, but the equations are bit more complicated.

Improve this answer
$\endgroup$
3
  • $\begingroup$ If the link goes dead, as they often do, your answer will lose its value. Can you summarise the contents of the linked article or give enough detail so others can find it in future? $\endgroup$
    – mdewey
    Commented Jan 21, 2018 at 14:43
  • $\begingroup$ Note that the given approximations must be very rough, since there is a positive probability that $X_2=0$, so the expectation (and hence variance) do not exist (we could say it is $\infty$). $\endgroup$
    – kjetil b halvorsen ♦
    Commented Mar 17, 2018 at 1:05
  • $\begingroup$ The paper also has another, more precise approximation for $E[X_1/(X_2+1)]$, which solves this problem of undefined values. $\endgroup$
    – user54038
    Commented May 31, 2018 at 21:07

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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