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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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mjs
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    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. – whuber Mar 18 '19 at 16:55
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    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 – mjs Mar 19 '19 at 09:26
  • @whuber would you agree with the new form of the PDF as in the 'Update'? – mjs Mar 19 '19 at 12:37
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    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. – whuber Mar 19 '19 at 13:46
  • Somewhat convoluted answer but I take as yes. Any idea why no textbooks talk about probability distributions, their dimensions and units? – mjs Mar 19 '19 at 15:11
  • That sounds like a counterfactual claim to me. I'm sure some textbooks don't talk about dimensions and units, but many either do or explicitly tell you in their introduction what you're expected to know about basic science and math in order to read them. Just about all of them discuss pdfs. BTW, my previous comment was careful *not* to say yes, because the division by "$[x]$" is superfluous and may be misleading. – whuber Mar 19 '19 at 16:33
  • But to make $log(x)$ dimensionless you need to devide it by its unit. It turns out I don't understand the comments starting with "The only permissible...". Is there a way to express differently? – mjs Mar 19 '19 at 21:28
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    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 $\mu$). – whuber Mar 19 '19 at 22:28

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