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I have two normally-distributed independent random variables X and Y and I need to calculate its division Z.

As far as I understand the mean of Z is $\mu_Z = \frac{\mu_X}{\mu_Y}$, but I don't know how to calculate the Standard Deviation $\sigma_Z$.

Is $\sigma_Z = \frac{\sigma_X}{\sigma_Y}$?

VGonPa
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    The mean of $Z = \frac{X}{Y}$ is **not** $\frac{\mu_X}{\mu_Y}$. Whatever gave you that idea? Even for one random variable, $E[g(X)] \neq g(E[X])$ _in general_. This is a fundamental notion that you would do well to learn well. Engrave it on your heart in letters of gold.... – Dilip Sarwate May 12 '13 at 21:14
  • Answers can be found in several places here by searching our site for "normal ratio Cauchy". – whuber May 13 '13 at 15:06
  • You can always reduce this problem to finding the mean or expected value of the reciprocal of a random variable. However, [this is not necessarily the reciprocal of the mean of the random variable!](http://www.insight-things.com/dividing-by-means-and-expected-values) – Jan Rothkegel Jul 07 '16 at 05:47

1 Answers1

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The ratio of two standard normal random variables ($\mu = 0, \sigma = 1$) is a Cauchy distribution. The Cauchy has an undefined variance (and hence undefined standard deviation). For other normals, the distribution is complex, indeed.

See http://en.wikipedia.org/wiki/Ratio_distribution

Of course, for any two samples from random variables, you can compute whatever you like. E.g.,

set.seed(20181)
x1 <- rnorm(100)
x2 <- rnorm(100)

ratio <- x1/x2
sd(x1)
sd(x2)
sd(ratio)

The SD of the ratio is 5.35.

But with a different seed, the SD can be very different. I ran this with 3 different seeds and got values as high as 11.21.

iNyar
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Peter Flom
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  • @Royi - the issue is that indefinite integral is undefined. (Both its positive and its negative parts integrate to infinity.) See https://stats.stackexchange.com/questions/36027/why-does-the-cauchy-distribution-have-no-mean – James Feb 17 '21 at 12:14