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 $SE_\bar Y = \frac{S_Y}{\sqrt{n_Y}}$.

  • $\begingroup$ I think your notation and question needs some refinement. You refer to two means $\bar x_1$ and $\bar x_2$ of two random variables, but presumably your intention is to refer to two SAMPLE MEANS (not population means). The product of two population means is not a random variable (and hence the variance is zero); the product of two sample means is a random variable. Then you need to define whether your standard deviations are population standard deviations, or the square root of sample variance, and if the latter, what measure of sample variance. $\endgroup$
    – wolfies
    Nov 27, 2020 at 1:16
  • $\begingroup$ FINALLY, you need to define $\bar x_3$ mathematically: on the one hand, you describe $\bar x_3 = \bar x_1 \bar x_2$; on the other hand, you describe $\bar x_3$ as a sample mean. It is not clear what you intend. $\endgroup$
    – wolfies
    Nov 27, 2020 at 1:19
  • $\begingroup$ I have updated my question accordingly. Do you have an answer to my question? $\endgroup$ Nov 27, 2020 at 7:26
  • $\begingroup$ Your definition of the sample mean is incorrect, and should not have the square root. $\endgroup$
    – wolfies
    Nov 28, 2020 at 16:27

2 Answers 2


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).


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:

$$ {\rm Var}(\bar X \space \bar Y) ={\rm Var}(\bar X){\rm Var}(\bar Y)+{\rm Var}(\bar X)(E(\bar Y))^2+{\rm Var}(\bar Y)(E(\bar X))^2$$


  • $E(\bar X) = \mu_X$
  • $E(\bar Y) = \mu_Y$
  • ${\rm Var}(\bar X) = \frac1n {\rm Var}(X)$
  • ${\rm Var}(\bar Y) = \frac1m {\rm Var}(Y)$

That is all there is to it.

  • $\begingroup$ The solution is expressed in terms of central and raw moments of the population, the definitions are provided for those terms, and is completely general. $\endgroup$
    – wolfies
    Nov 28, 2020 at 13:07
  • $\begingroup$ Ok thanks, I see that you have defined the output, but of what variable is the sample size $n$. I want to be able to calculate the standard deviation for the product of two sample means with different sample sizes, so I need to differentiate between $n_X$ and $n_Y$. $\endgroup$ Nov 28, 2020 at 15:13
  • $\begingroup$ So you agree that the first equation I provided yields the best estimate? $\endgroup$ Nov 28, 2020 at 22:00
  • $\begingroup$ No - your first equation is incorrect: (i) it uses sample estimators, whereas the correct solution requires the population moments; (ii) the solution is not in any event an estimate, but the calculation of the exact variance of the product of two independent sample means $\endgroup$
    – wolfies
    Nov 29, 2020 at 2:42

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