I've encountered this problem and I'm not sure whether my logic is correct. Suppose I have a random sample of customer spending and I want to estimate the market share of a given store in a given quarter (which I define to be the proportion of sales made in that store against the sum of sales in all other competing stores in the locality).

I calculate this statistic by taking the ratio of total spend in the store of interest against the sum of total spend. In equation form the statistic of interest is:

$$ Market \ share = \frac{\sum_{c} X_c}{\sum_c Y_c} $$ where $c$ is a customer in the sample, $X_c$ is total spending by customer c on store X and $Y_c$ is total spending by customer c on all stores of the same type as store $X$.

Suppose I want to estimate a confidence interval for this estimate. I know by the Central Limit Theorem that the total sum spent, $\sum_{c} X_c$ and $\sum_c Y_c$, are both going to be normally distributed asymptotically.

However, since the market share is the ratio of two sums my intuition suggests that this statistic will not be asymptotically normally distributed (as the ratio of two normal distribution is not normally distributed in general). Due to my hesitance about the asymptotic distribution of this statistic I decided to construct the confidence interval by bootstrapping instead.

My questions on this issue would be, am I ignorant of some nice asymptotic result that I could use and is my logic correct towards why I should bootstrap this problem?

  • 8
    $\begingroup$ "Is the ratio of two sums normally distributed?" No. But in some cases a normal approximation may be reasonable; indeed the normal may well be a better approximation to this than to the numerator. $\endgroup$
    – Glen_b
    Sep 9, 2021 at 4:18
  • $\begingroup$ I am not too sure what is the problem. The sum of X and Y are both individually normally distributed. Assume they are all idd it will then become a ratio distribution of two normal distributions (Cauchy distribution). $\endgroup$ Sep 9, 2021 at 5:00
  • $\begingroup$ Hi @Glen_b can you please clarify what you mean by 'a better approximation to this than to the numerator'? I ran some simple Monte Carlo simulations taking the ratio of two normally distributed random variables and when the coefficient of variation is large for even one of them, the length of the confidence interval tends to be underestimated by the normal distribution. $\endgroup$ Sep 9, 2021 at 5:01
  • 2
    $\begingroup$ @PrestonLui from memory I believe that the Cauchy distribution only arises if the mean of both normal distributions are equal to zero. However, given these sums are total spend which is a non-negative quantity it is unlikely that they will be centered at zero. $\endgroup$ Sep 9, 2021 at 5:12
  • $\begingroup$ So the CLT applies to the average not the sums ( ie average spend per customer), and for this the delta method would work asymptotically en.wikipedia.org/wiki/Delta_method $\endgroup$
    – seanv507
    Sep 9, 2021 at 6:08

1 Answer 1


Experiment. The ratio is Cauchy, as noted, if numerator and denominator are both normal distributions centered at zero (not illustrated here).

However, the ratio is also often nowhere near normal if numerator and denominator are both centered several standard deviations above zero. Using R:

x1 = rnorm(100, 50, 7)
x2 = rnorm(100, 70, 8)
ratio = x1/x2

        Shapiro-Wilk normality test

data:  ratio
W = 0.94094, p-value = 0.0002201

qqnorm(ratio); qqline(ratio, col="blue")

enter image description here

hist(ratio, prob=T, col="skyblue2")

enter image description here

  • 7
    $\begingroup$ "Experiment". Indeed. The most useful sentence I was taught very early in my career was "I don't know. Let's go see...." $\endgroup$
    – Jason
    Sep 9, 2021 at 13:29
  • 1
    $\begingroup$ Nit: shouldn't market share be in $[0, 1]$? Using the notation in the question it seems that $Y \geq X \geq 0$. $\endgroup$
    – Adrian
    Sep 12, 2021 at 22:10
  • $\begingroup$ @Adrian. Point taken. Much of the discussion ignored that restriction. However, in my simulation the numerator has mean smaller than does denominator, so my ratio is always positive. Furthermore, using the restriction you mention, still gives a ratio strongly rejected by S-W as normal. Q-Q plot is similar to the one shown. (And, of course, histogram doesn't go above 1.) $\endgroup$
    – BruceET
    Sep 13, 2021 at 1:06

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.