Sharp regression discontinuity with pymc
models#
import causalpy as cp
%load_ext autoreload
%autoreload 2
%config InlineBackend.figure_format = 'retina'
seed = 42
df = cp.load_data("rd")
Linear, main-effects, and interaction model#
Note
The random_seed
keyword argument for the PyMC sampler is not necessary. We use it here so that the results are reproducible.
result = cp.RegressionDiscontinuity(
df,
formula="y ~ 1 + x + treated + x:treated",
model=cp.pymc_models.LinearRegression(sample_kwargs={"random_seed": seed}),
treatment_threshold=0.5,
)
fig, ax = result.plot()
Auto-assigning NUTS sampler...
Initializing NUTS using jitter+adapt_diag...
Multiprocess sampling (4 chains in 4 jobs)
NUTS: [beta, sigma]
Sampling 4 chains for 1_000 tune and 1_000 draw iterations (4_000 + 4_000 draws total) took 2 seconds.
Sampling: [beta, sigma, y_hat]
Sampling: [y_hat]
Sampling: [y_hat]
Sampling: [y_hat]
Sampling: [y_hat]

Though we can see that this does not give a good fit of the data almost certainly overestimates the discontinuity at threshold.
Using a bandwidth#
One way how we could deal with this is to use the bandwidth
kwarg. This will only fit the model to data within a certain bandwidth of the threshold. If \(x\) is the running variable, then the model will only be fitted to data where \(threshold - bandwidth \le x \le threshold + bandwidth\).
result = cp.RegressionDiscontinuity(
df,
formula="y ~ 1 + x + treated + x:treated",
model=cp.pymc_models.LinearRegression(sample_kwargs={"random_seed": seed}),
treatment_threshold=0.5,
bandwidth=0.3,
)
fig, ax = result.plot()
Auto-assigning NUTS sampler...
Initializing NUTS using jitter+adapt_diag...
Multiprocess sampling (4 chains in 4 jobs)
NUTS: [beta, sigma]
Sampling 4 chains for 1_000 tune and 1_000 draw iterations (4_000 + 4_000 draws total) took 4 seconds.
There were 21 divergences after tuning. Increase `target_accept` or reparameterize.
Sampling: [beta, sigma, y_hat]
Sampling: [y_hat]
Sampling: [y_hat]
Sampling: [y_hat]
Sampling: [y_hat]

We could even go crazy and just fit intercepts for the data close to the threshold. But clearly this will involve more estimation error as we are using less data.
result = cp.RegressionDiscontinuity(
df,
formula="y ~ 1 + treated",
model=cp.pymc_models.LinearRegression(sample_kwargs={"random_seed": seed}),
treatment_threshold=0.5,
bandwidth=0.2,
)
fig, ax = result.plot()
Auto-assigning NUTS sampler...
Initializing NUTS using jitter+adapt_diag...
Multiprocess sampling (4 chains in 4 jobs)
NUTS: [beta, sigma]
Sampling 4 chains for 1_000 tune and 1_000 draw iterations (4_000 + 4_000 draws total) took 1 seconds.
Sampling: [beta, sigma, y_hat]
Sampling: [y_hat]
Sampling: [y_hat]
Sampling: [y_hat]
Sampling: [y_hat]

Using basis splines#
Though it could arguably be better to fit with a more complex model, fit example a spline. This allows us to use all of the data, and (depending on the situation) maybe give a better fit.
result = cp.RegressionDiscontinuity(
df,
formula="y ~ 1 + bs(x, df=6) + treated",
model=cp.pymc_models.LinearRegression(sample_kwargs={"random_seed": seed}),
treatment_threshold=0.5,
)
fig, ax = result.plot()
Auto-assigning NUTS sampler...
Initializing NUTS using jitter+adapt_diag...
Multiprocess sampling (4 chains in 4 jobs)
NUTS: [beta, sigma]
Sampling 4 chains for 1_000 tune and 1_000 draw iterations (4_000 + 4_000 draws total) took 2 seconds.
Sampling: [beta, sigma, y_hat]
Sampling: [y_hat]
Sampling: [y_hat]
Sampling: [y_hat]
Sampling: [y_hat]

As with all of the models in this notebook, we can ask for a summary of the model coefficients.
result.summary()
Difference in Differences experiment
Formula: y ~ 1 + bs(x, df=6) + treated
Running variable: x
Threshold on running variable: 0.5
Results:
Discontinuity at threshold = 0.41$CI_{94\%}$[0.23, 0.59]
Model coefficients:
Intercept -0.23, 94% HDI [-0.32, -0.15]
treated[T.True] 0.41, 94% HDI [0.23, 0.59]
bs(x, df=6)[0] -0.59, 94% HDI [-0.78, -0.41]
bs(x, df=6)[1] -1.1, 94% HDI [-1.2, -0.94]
bs(x, df=6)[2] 0.28, 94% HDI [0.13, 0.43]
bs(x, df=6)[3] 1.7, 94% HDI [1.5, 1.8]
bs(x, df=6)[4] 1, 94% HDI [0.66, 1.4]
bs(x, df=6)[5] 0.57, 94% HDI [0.37, 0.76]
sigma 0.1, 94% HDI [0.089, 0.12]