I have 2 variables, which I can't be sure to call them random variables. They are aggregates from multiple measurements of two distinct phenomena. I don't have access to the raw measurements data, just the aggregates: Var A with mean $\mu_1$ and standard deviation $\sigma_1$, and Var B with mean $\mu_2$ and standard deviation $\sigma_2$.

I want to calculate Var C which given by the following formula:


I'd imagine for Var C's mean $\mu_3$, this boils down to


How about Var C's standard deviation $\sigma_3$? Is it even possible to calculate Var C mean and standard deviation given only the means and standard deviations for Var A and B?


  • $\begingroup$ Perhaps in your case $μ_3≈μ_1/μ_2 + 2$ would be close to correct. Certainly not as a reliable general rule. For example, if $μ2\approx 0,$ then this method might give a useless result. (Same as previous Comment, but with typo in notation repaired.) $\endgroup$ – BruceET Dec 20 '17 at 7:41

Comment continued, with specific examples.

Suppose $X_1 \sim \mathsf{Norm}(\mu_1=200,\, \sigma_1=20)$ and $X_2 \sim \mathsf{Norm}(\mu_2=150,\, \sigma_1=20).$ Then, by simulation, $E(X_1/X_2) +2 \approx 3.36$ and $\mu_1/\mu_2 + 2 = 3.33.$ Not a horrible approximation.

x1 = rnorm(10^6, 200, 20)
x2 = rnorm(10^6, 150, 20)
mean(x1/x2) + 2
## 3.358685  # aprx E(X1/X2) + 2
200/150 + 2
## 3.333333

However, if $X_1 \sim \mathsf{Norm}(\mu_1=20,\, \sigma_1=10)$ and $X_2 \sim \mathsf{Norm}(\mu_2=15, \sigma_1=10).$ Then $E(X_1/X_2) +2$ is unstable because you are dividing by values very near 0 some of the time. (Theoretically speaking, $E(X_1/X_2)$ is undefined.) Here, $3.33$ is not a useful approximation.


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