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For a sample, I know the sample size, and the mean and standard deviation of the values ​​transformed using the natural logarithm. I need to calculate the mean and standard deviation of the original data (which I don't have). Is this possible?

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3 Answers 3

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@Ben is correct that you don't have enough information. However, you can approximate the mean and standard deviation of a transformed variable using the delta method.

The delta method for the mean says that $\overline{f(x)} \approx f(\bar x) + \frac{1}{2} f^{''}(\bar x) \cdot \textrm{Var}(x)$ (see here); in your case $f(x) = \exp(x)$, so if $M$ and $S$ are the mean and standard deviation on the log scale, then the mean on the original scale should be $\exp(M) + \frac{1}{2} \exp(M) S^2 = (1+S^2/2) \exp(M)$.

The standard deviation on the transformed scale is approximately $f'(\bar x) \sigma(x)$, so your original-scale SD should be approximately $S \exp(M)$.

The information in @Glen_b's answer gives you some information about the likely accuracy of the approximations: for example, their expression translates to $\exp(M)(1 + S^2/2 + \textrm{skew}/6 + ...)$ (where "skew" is estimated as the mean of $(x - \bar x)^3$ on the log scale, not normalized in any way). If you have some guess about the magnitude of the skew of the log-scale data you can use this to decide whether the approximation for the mean is likely to be OK ...

I'd encourage you to do some experiments with made-up data (e.g. using rlnorm() in R) to confirm that I didn't make any mathematical errors.

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Unfortunately, since the logarithm is a nonlinear transformation, that information is insufficient to compute the mean and standard deviation of the original sample. There are many possible means and standard deviations for the original sample that would correspond to the known mean and standard deviation of the logarithms.

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If you treat the set of values as a discrete distribution with probability $\frac{1}{n}$ on each value, you can see that it depends on the collection of higher moments of the logs. Looking at just the mean to start:

\begin{eqnarray} E[e^X] &=& E[e^{\mu+(X-\mu)}] \\ &=& e^{\mu}\cdot E[e^{(X-\mu)}] \\ &=& e^{\mu}\cdot E[1 + (X-\mu) + (X-\mu)^2/2!\,+\,(X-\mu)^3/3!\,+\,\ldots\,] \\ &=& e^{\mu}\cdot [1 + 0 + E[(X-\mu)^2]/2!\,+\,E[(X-\mu)^3]/3!\,+\,\ldots\,] \\ &=& e^{\mu}\cdot [1 + \mu_2/2! + \mu_3/3!\,+\,\ldots\,] \end{eqnarray}

so the mean of the original values is not determined by just the first moment and the second central moment of the logs. (Convergence of the series expansion shouldn't be a problem in relation to a sample - the "support" of the sample is bounded.)

We see that the leading term in the expansion after accounting for a function of the first two moments involves the third moment on the log scale. This is helpful; it suggests strong left or right skew on the log scale will tend to push the expectation up or down away from what you could infer from the first two moments of the logs.

Bounds on sample moments should be possible though, if you know some things about the sample besides the sample mean and variance, like the sample max and min, say.

On the other hand, if you're trying to estimate population quantities and you know something (or are prepared to assume something) about the population shape you may be able to produce suitable estimates.

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