Standard error of estimated sum or product mean Updated question: Given two sample means ($\bar X, \bar Y$) and sample standard deviations ($S_X, S_Y$) with different sample sizes ($n_X, n_Y$), I want to calculate the standard errors ($SE_\theta, SE_\rho$) of estimated sum ($\theta$) and product ($\rho$) means.
The given sample means and sample standard deviations are calculated as
$\bar X = \frac{\sum\limits_{i=1}^{n_X} X_{i}}{n_X}$, $S_X = \sqrt{\frac{\sum\limits_{i=1}^{n_X} (X_{i} - \bar X)^2}{n_X - 1}}$, $\bar Y = \frac{\sum\limits_{i=1}^{n_Y} Y_{i}}{n_Y}$ and $S_Y = \sqrt{\frac{\sum\limits_{i=1}^{n_Y} (Y_{i} - \bar Y)^2}{n_Y - 1}}$
where $n_X \neq n_Y$.
To clarify, I provide two made-up examples where the variables $X$ and $Y$ are assumed to be independent:
Sum: Imagine a population of sharks spending 5 ± 0.2 h d$^{–1}$ ($\bar X \pm S_X$) feeding on seagrass and 3 ± 0.1 h d$^{–1}$ ($\bar Y \pm S_Y$) feeding on fish. The first mean is based on 20 sampled sharks ($n_X$= 20), the second on 10 ($n_Y$= 10). If I want to estimate the sum mean, I use $\theta = \bar X + \bar Y$. In this example $\theta$ = 8 h d$^{–1}$.
Product: Imagine a forest with a mean density of 10 ± 3 plants m$^{-2}$ ($\bar X \pm S_X$) and a mean photosynthetic rate of 20 ± 0.5 $\mu$mol O$_2$ plant$^{-1}$ min$^{-1}$ ($\bar Y \pm S_Y$). The first mean is based on 15 replicate measurements ($n_X$= 15), the second on 40 ($n_Y$= 40). If I want to estimate the product mean, I use $\rho = \bar X\bar Y$. In this example $\rho$ = 200 $\mu$mol O$_2$ m$^{-2}$ min$^{-1}$.
I want to know how to calculate $SE_\theta$ and $SE_\rho$ using the provided information in each example.
The accepted answer to this question provides this equation for $SE_\theta$:
$SE_\theta = \sqrt{\frac{S_Y^2}{n_Y} + \frac{S_X^2}{n_X}}$
This very helpful paper provides this equation for $SE_\rho$:
$SE_\rho = \sqrt{\frac{\bar X^2S_Y^2}{n_Y} + \frac{\bar Y^2S_X^2}{n_X} + \frac{S_X^2S_Y^2}{n_Xn_Y}}$
Are these equations correct?
Answer: I have found my own answer in the meantime and thought it might be useful to share it here. The equations for the standard error of the sum mean ($SE_\theta$) and the standard error of the product mean ($SE_\rho$) are correct. They are derived from $Var(\bar X + \bar Y) = Var(\bar X) + Var(\bar Y)$ and $Var(\bar X\bar Y) = E(X)^2Var(\bar Y) + E(Y)^2Var(\bar X) + Var(\bar X)Var(\bar Y)$.
When the standard errors of $\bar X$ and $\bar Y$ are given, $SE_\theta$ and $SE_\rho$ can be calculated as
$SE_\theta = \sqrt{SE_\bar X^2 + SE_\bar Y^2}$ and $SE_\rho = \sqrt{\bar X^2SE_\bar Y^2 + \bar Y^2SE_\bar X^2 + SE_\bar X^2SE_\bar Y^2}$
where $SE_\bar X = \frac{S_X}{\sqrt{n_X}}$ and $SE_\bar Y = \frac{S_Y}{\sqrt{n_Y}}$.
 A: The 2nd equation can't be exactly right as it implies that when both sample means are 0, $\sigma_3$ must be 0. Looks like your sources for the 2nd equation are ignoring the $Var(X)Var(Y)$ term under the assumption that its size will be negligible compared to the other terms.
The 1st equation is a better approximation, as per the derivation given on the Cross Validated post you linked, assuming that $\sigma_1, \sigma_2$ are the standard deviations of $\bar{x}_1, \bar{x}_2$, not the sample standard deviations. (Intuition: as you increase your sample sizes, your sample standard deviations will converge to the population standard deviation, but the standard deviation in your estimates of $\bar{x}_1, \bar{x}_2$ should converge to 0, so the standard deviation in your estimate of $\bar{x}_3$ should converge to 0). To be exact, you would need to use the true population means rather than the sample means.
Edit: rereading your question it's unclear whether you are looking for the standard error in the estimate of the product mean (this is what "I want to calculate the standard deviation of the product of the first two sample means" means) or just for the standard deviation in the distribution of the product of the two independent variables (which I think is what your forest calculation is doing).
A: For your revised problem:
Given: $X$ and $Y$ are independent random variables with different sample sizes, say $n$ and $m$ respectively.
The variance of two independent random variables, in this case, being the sample means $\bar X$ and $\bar Y$, is:
$$ {\rm Var}(\bar X \space \bar Y) ={\rm Var}(\bar X){\rm Var}(\bar Y)+{\rm Var}(\bar X)(E(\bar Y))^2+{\rm Var}(\bar Y)(E(\bar X))^2$$
where:

*

*$E(\bar X) = \mu_X$

*$E(\bar Y) = \mu_Y$

*${\rm Var}(\bar X) = \frac1n {\rm Var}(X)$

*${\rm Var}(\bar Y) = \frac1m {\rm Var}(Y)$
That is all there is to it.
