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What impact should unit conversion have on the relationship between standard deviation and variance?

For example, when the units are kilograms, the sd is 3.156 kg , and the variance is 9.96 kg which is the sd^2.

However, when we convert this to grams, the sd is 3,156 g (the equivalent of 3.156 kg) but the variance is 9,963,829 g (the equivalent of 9,963.829 kg).

What is the correct way of doing this? For reference, I am working with a dataset that has been converted to kg, but the original units are grams.

set.seed(1)
in_kg <- runif(10, 0, 10)
sd(in_kg)
var(in_kg)

in_grams <- in_kg*1000
sd(in_grams)
var(in_grams)

Output:

> sd(in_kg)
[1] 3.156553
> var(in_kg)
[1] 9.963829

> sd(in_grams)
[1] 3156.553
> var(in_grams)
[1] 9963829
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    $\begingroup$ If the sd is in g, the variance is in g squared. If it is in kg the variance is in kg squared. The units get squared along with the numbers. Sure, in this case and many others the units are unusual and don't have a simple explanation, but that's just what it is mathematically and part of why using sd helps interpretation. $\endgroup$
    – Nick Cox
    Aug 17 '20 at 8:46
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If your unit conversion can be written as an affine linear mapping like $f(x) = mx + b$ where $m > 0$ (e.g. grams to kilograms is given by $f(x) = 1000x$, Celsius to Fahrenheit is given by $f(C) = 1.8C + 32$), and if $X$ is a random variable with $V(X) = \sigma_X^2$, then the random variable $Y = f(X)$ will have \begin{align} V(Y) & = E[(Y - E(Y))^2] \\ & = E[(mX + b - E(mX + b))^2] \\ & = E[(mX + b - (mE(X) + b))^2] \\ & = E[(mX - mE(X))^2] \\ & = E[m^2(X - E(X))^2] \\ & = m^2 E[(X - E(X))^2] \\ & = m^2 V(X) \\ \end{align} and hence the standard deviations have the relation $$ \text{Standard deviation of $Y$} = \sigma_Y = \sqrt{V(Y)} = \sqrt{m^2 V(X)} = |m|\sigma = m \sigma . $$ So the variance will scale like $\sigma_Y^2 = m^2 \sigma_X^2$ but the standard deviations will scale like $\sigma_Y = m \sigma_X$

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