Why do discretised predictors have lower statistical power than continuous predictors? In designing an analysis, I'd like to decide between using discretised variables versus using the original, continuous variable (the reason being that in this particular case, collecting discretised data per se would be cheaper and easier).
However, my understanding is that discretised predictors lack statistical power in comparison with the continuous case. I.e. using a discretised predictor, I would have to collect substantially more data to identify an effect on a dependent variable.
My question: Is this true and what is the theoretical basis for the lack of statistical power of discretised variables?
I have found some evidence in a 1983 article "The Cost of Dichotomization" (link), but was wondering if there are alternative lines of argumentation that can be applied?
 A: Smaller Variance Leads To Smaller Power
Discretizing predictors is a bad idea as Frank Harrell explains here.  As to your titular question, let's examine a simple scenario.
Let's say I have a predictor which has standard normal distribution.  I'd like test the effect of this predictor on an outcome which is also normally distributed conditional on the predictor, with unit variance.  Were I to discretize the predictor, I would have a T test on my hands.  Leaving the predictor as is would just be a regression.  Further more, let's say I collect 25 observations and am interested in a smallest effect of 1 unit of whatever I am measuring.
The power for each test is given by
$$\gamma=1-\Phi\left[z_{1-\alpha / 2}-\left|\beta_{j}^{a}\right| \sigma_{x_{j}} \sqrt{n\left(1-\rho_{j}^{2}\right)} / \sigma_{y \mid \mathbf{x}}\right]$$
Where $\Phi$ is the normal CDF, $\alpha$ is the FPR, $
\beta_j^\alpha$ is smallest meaningful effect size, $\sigma_{x_j}$ is the variance of the predictor, $n$ is the sample size, $1/(1-\rho^2)$ is the variance inflation factor, and $\sigma_{y \vert x}$ is the residual standard deviation.
Let's plug in some of the terms I've mentioned here for economy of thought
$$\gamma=1-\Phi\left[1.96- 5 \sigma_{x_{j}}\right]$$
Because I've picked nice numbers, the power depends only on the variance of the predictor.  It isn't hard to see that this is a monotonic function in $\sigma$, hence larger variability in the predictor means greater power.  When you dichotomize, the largest $\sigma$ can be is 0.5.  In this example, dichotomizing would reduce my power from 99% to nearly 70%.
This fact is true across many GLMs, as the power equations are more or less similar.  Smaller variance in the predictor leads to smaller power for the associated test.  This is rather intuitive.  If you are interested in testing for the slope of a continuous predictor, you'd want that predictor to be as spread out as possible.  You can't estimate the slope of you have a little band in which to see the variation in $y$ due to $x$.
Additional Pitfalls
Dichotomizing is also a bad idea because it leads to residual confounding and discontinuity in effects.  For example, I'm very confident a woman who is 1 month pregnant is very different than a woman who is 9 months pregnant, but using a variable like is_pregnant treats them as if they were the same.  So even if it is the case that you can collect more data by discretizing, the quality of the estimate suffers.
