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In a discussion on this forum lognormal distribution, standard-deviation and (physical) units the cumulative distribution function (PDF) of the lognormal distribution was analysed. The conclusion was that `$\ln(x)$, and hence, $\mu$ and $\sigma$, are unit-free'.

While this seems plausible to me, I have an issue with the unit of the probability density function, given by

$$f(x) = \dfrac1{x \sigma\sqrt{2\pi}} \exp{\dfrac{-(\log x-\mu)^2}{ 2\sigma^2}}$$

$\log x$ is dimensionless, by taking the ratio of $x$ with its unit, $[x]$, i.e. $\log(x/[x])$ [1]. If $\mu$ and $\sigma$ are dimensionless it means that they are calculated based on dimensionless $x$.

Question 1

Is $x$ in the denominator of $\dfrac1{x \sigma\sqrt{2\pi}}$ also dimensionless? If it is, this would mean that $f(x)$ is dimensionless. Is this correct?

Question 2

Because in general the PDF has units of inverse random variable units, as discussed in [2], I deduce that the units of a PDF are dependent on the distribution. Is that correct?

References

[1] Matta, Massa, Gubskaya & Knoll, (2011), Can One Take the Logarithm or the Sine of a Dimensioned Quantity or a Unit? Dimensional Analysis Involving Transcendental Functions, Journal of Chemical Education, Vol 88, No. 1

[2] https://www.phy.ornl.gov/csep/mc/node13.html - Introduction to Monte Carlo Methods by the Computational Science Education Project (Oak Ridge National Laboratory)

Update

Taking into account @whuber comments, the PDF reads in fact $$f(x)=\dfrac1{x\sigma\sqrt{2\pi}} \exp\dfrac{-(\log \dfrac{x}{[x]} - \mu)^2}{2\sigma^2}$$ with $[x]$ the dimention of $x$. $\mu$ and $\sigma$ being dimensionless. Only in this form $f(x)dx$ is dimentionless probability. Is this sensible?

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    $\begingroup$ I believe you will immediately see the answers to your questions if you consider the probability element $f(x)\mathrm{d}x,$ in which $x$ is in the units of the random variable and the entire element must be a probability. $\endgroup$
    – whuber
    Mar 18, 2019 at 16:55
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    $\begingroup$ Yes, I know that, just thought that having in the PDF formula one $x$ with dimention and one without would be somewhat weired, see update $\endgroup$
    – mjs
    Mar 19, 2019 at 9:26
  • $\begingroup$ @whuber would you agree with the new form of the PDF as in the 'Update'? $\endgroup$
    – mjs
    Mar 19, 2019 at 12:37
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    $\begingroup$ It helps to understand that $\log(x)$ is always dimensionless, at least if it is to make sense--which means $x$ lies in a set of positive values with $0$ as a natural lower bound. The only permissible changes of unit will multiply $x$ by a constant. That changes $x$ by an additive constant, which is taken up in the value of $\mu.$ Consequently $\sigma$ must be unitless, too, leaving the units of $f(x)$ to be those of $1/x,$ which exactly cancels $\mathrm{d}x,$ as it must. $\endgroup$
    – whuber
    Mar 19, 2019 at 13:46
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    $\begingroup$ I could try algebra. The key part of the original model is the numerator $\log(x)-\mu.$ Changing the units of $x$ multiplies it by some positive number $\lambda.$ This adds $\log\lambda$ to $\log x$ (and therefore does not alter the log standard deviation $\sigma$). Consequently, if you set $\mu^\prime=\mu+\log\lambda,$ the new model involving $$\log(x\lambda)-\mu^\prime = \log(x)+\log\lambda - (\mu+\log\lambda) = \log(x)+\mu$$ is identical to the original one. Thus, the choice of units doesn't matter (it only affects the interpretation of the log mean $\mu$). $\endgroup$
    – whuber
    Mar 19, 2019 at 22:28

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