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I've a multiplicative model for which $Y = X*Z$, for which $Y$ and $Z$ is known. I want this model to be additive (therefore using logarithms) to figure out the variance of $X$. I have $$ln(Y) = ln(X) + ln(Z)$$ and so $$VAR(ln(Y)) = VAR(ln(X)) + VAR(ln(Z))$$ and finally $$VAR(ln(X) = VAR(ln(Y)) - VAR(ln(Z))$$.

I'm not looking for a way to calculate $X$, only the variance of $X$. I now know $VAR(ln(X))$, but is there I can "revert" this variance to get the variance for $X$?

develarist
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Srick
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  • is $X$ data unknown and the reason why the variance of it can't be estimated? – develarist Sep 18 '20 at 19:08
  • Exactly. I've a multiplicative model for which Y = X*Z, for which Y and Z is known. I want this model to be additive (therefore using logarithms) to figure out the variance of X. But I can't really figure it out. – Srick Sep 18 '20 at 19:11
  • is $Var(\log(X))$ just a given, or was it estimated somehow from $Y$ or $Z$? – develarist Sep 18 '20 at 19:13
  • Yes: Y and Z is known. I used logarithms to make Y=X*Z into ln(Y) = ln(X) + ln(Y). From here I figured that I can get VAR(ln(Y)) = VAR(ln(X)) + VAR(ln(Z)), and ultimately VAR(ln(X)) = VAR(ln(Y)) - VAR(ln(Z)). But now I'm looking for VAR(X) – Srick Sep 18 '20 at 19:15
  • separate question: if you knew $Var(X)$, how would you get $X$ from it? – develarist Sep 18 '20 at 19:17
  • I'm not really looking to estimate $X$, only the variance. This might sound weird, but I'm trying to estimate variance in a measurement, so knowing the actual values is unnecessary. – Srick Sep 18 '20 at 19:23
  • I'll get back to this since my answer is incomplete – develarist Sep 18 '20 at 19:26
  • Thank you very much! – Srick Sep 18 '20 at 19:28
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    Your formula for $\text{Var}(\text{ln}(Y))$ is incomplete, as it is missing the covariance between the 2 terms on the right. See the [Wikipedia page](https://en.wikipedia.org/wiki/Variance#Sum_of_correlated_variables). [This page](https://stats.stackexchange.com/a/5790/28500) shows how to get the approximate variance of a transformation of a variable having known mean and variance. – EdM Sep 18 '20 at 20:16
  • Your model is not fully specified. For instance, if it has an additive error term, making it $Y=XZ+\epsilon,$ then taking logs won't work at all. – whuber Sep 18 '20 at 20:57
  • hard to get the variance of a variable from its log's variance based on the variance of logs of other variables, $$\ln Y = \ln X + \ln Z$$ $$Var(\ln Y) = Var(\ln X) + Var(\ln Z)$$ $$Var(\ln X) = Var(\ln Y) - Var(\ln Z)$$ $$\dots ?$$ – develarist Sep 18 '20 at 19:22
  • I realize that I should have added this info in my question, but Y and Z is of different sizes (not the same amount of samples), so the above equation isn't suitable for me. Ie, I have 1000 samples of X (unknown values), 1000 samples of Y and 100 samples of Z (both known), and I'm trying to estimate the variance of X, not getting all the actual values of X. – Srick Sep 18 '20 at 19:27
  • ok your comments here and above should be edited into the question – develarist Sep 18 '20 at 19:29
  • Done! Thank you. – Srick Sep 18 '20 at 19:35

2 Answers2

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The answer by @ericperkerson is completely right, you cannot in general find $\text{Var}(X)$ using only $\text{Var}(\ln X)$, but I would like to bring out the delta method. If $X\sim\mathcal N(\mu,\sigma^2)$ and $\sigma$ is small enough*, we have the following approximation: $$f(X)\overset{\text{approx.}}\sim\mathcal N(f(\mu), f'(\mu)^2\cdot\sigma^2)$$

In the case of $\ln X$, this gives us:

$$\text{Var}(\ln X)\approx\frac{\text{Var}(X)}{\mathbb E(X)^2}$$

So, if besides the variance of $X$ you also know its expectation, you can find this simple approximation for $\text{Var}(\ln X)$. How good this approximation is likely to depend on the shape and standard deviation of your distribution.

* small enough here means $\sigma^2<<\frac{1}{f''(\mu)}$, so that our approximation can safely ignore any curvature of $f(x)$.

PedroSebe
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This is not possible in general.

This is because it's possible to find different random variables $X$ and $Y$ where $V(\log(X)) = V(\log(Y))$ but $V(X) \ne V(Y)$, so even if you know $V(\log(X))$ you cannot recover the original variance $V(X)$ (at least not without making some additional assumptions).

For example, suppose we know that $X \sim \text{Lognormal} (\mu, \sigma^2)$, and hence $\log (X) \sim \text{Normal}(\mu, \sigma^2)$. The variance of a log-normal random variable is

$$V(X) = (\exp(\sigma^2) - 1)\exp(2\mu + \sigma^2).$$

This means that even if we know $V(\log(X)) = \sigma^2 = \log(2) \approx 0.693$, for example, the variance of $X$ still depends on $\mu$ through the equation \begin{align} V(X) & = (\exp(\log(2)) - 1)\exp(2\mu + \log(2)) \\ & = (2 - 1)\exp(2\mu) \exp(\log(2)) \\ & = 2 \exp(2\mu). \end{align}

Hence, the variance of $X$ could be literally any positive number, even though we know $V(\log(X)) = \log(2)$. For example, it's possible that $\mu = 0$ which would give $V(X) = 2$, but it's also possible that $\mu = 1$ which would give $V(X) = 2 \exp(2) \approx 14.778$, and just knowing $V(\log(X))$ is not enough to distinguish between these possibilities.

Eric Perkerson
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