What does a shift-in-location-only entail?

Question

A number of questions on this site describe the location shift assumption for interpreting the Mann-Whitney U as a test of the equality of medians (e.g. here, here, and here). I'm trying to figure out what this actually entails.

This seems pretty straightforward for some distributions, such as the normal distribution. For others, such as log-normal and beta distributions, changing one of the parameters can affect the mean, median and variance. What counts as a shift in location only? If, for example, I increase the location parameter in a log-normal distribution without changing the scale parameter, does this count as a change in location only, even though the variance will have increased? Similarly, for something like the Beta distribution, altering one of the shape parameters will affect the mean, median, and variance.

For clarity, I'm hoping to understand the concept, not decide on whether I'm justified in using the Mann-Whitney U for a specific set of data.

Answer 1

You're not changing the parameters of the distribution, at least not in the typical sense.

Let's consider an exponential distribution, $X\sim exp(\beta)$ with PDF $f_X(x\vert \beta) = \dfrac{e^{\frac{-x}{\beta}}}{\beta}$.

If we have a shift in mean of 1 to get $Y\sim exp(\beta+1)$, then, we change the variance, as you note. However, let's change the PDF itself to $ f_Y(y\vert \beta) = \dfrac{e^{\frac{-(y-1)}{\beta}}}{\beta}$; then we get the same sort of shape but shifted. This is no longer an exponential distribution, but it has the same shape.

That is a location-only shift.

You can play this same sort of game with other PDFs by remembering that, for functions $f(x)$ in general, $f(x-a)$ is $f(x)$ shifted to the right by $a$ (so a left shift if $a<0$).

EDIT: I say that you're not changing parameters in the traditional sense because the exponential distribution traditionally only has the one parameter, but let's work with the $TPM(\beta,a)$ distribution with PDF $ f_X(x\vert \beta,a) = \dfrac{e^{\frac{-(x-a)}{\beta}}}{\beta}$. Then you get a location shift (and only a location shift) by changing the $a$ parameter and not touching $\beta$.