Mediator. Full luxury bayes

bayesian
estadística
causal inference
R
2022
Author

jlcr

Published

February 12, 2022

Continuando con la serie sobre cosas de inferencia causal y full luxury bayes, antes de que empiece mi amigo Carlos Gil, y dónde seguramente se aprenderá más.

Este ejemplo viene motivado precisamente por una charla que tuve el otro día con él.

Sea el siguiente diagrama causal

  library(tidyverse)
  library(dagitty)
  library(ggdag)
  
  g <- dagitty("dag{ 
  x -> y ;
  z -> y ;
  x -> z
 }")
  
  
  ggdag(g)

Se tiene que z es un mediador entre x e y, y la teoría nos dice que si quiero obtener el efecto directo de x sobre y he de condicionar por z , y efectivamente, así nos lo dice el backdoor criterio. Mientras que si quiero el efecto total de x sobre y no he de condicionar por z.

  adjustmentSets(g, exposure = "x", outcome = "y", effect = "total"  )
#>  {}
  adjustmentSets(g, exposure = "x", outcome = "y", effect = "direct"  )
#> { z }

Datos simulados


set.seed(155)
N <- 1000 

x <- rnorm(N, 2, 1) 
z <- rnorm(N, 4+ 4*x , 2 ) # a z le ponemos más variabilidad, pero daría igual
y <- rnorm(N, 2+ 3*x + z, 1)

Efecto total de x sobre y

Tal y como hemos simulado los datos, se esperaría que el efecto total de x sobre y fuera

\[ \dfrac{cov(x,y)} {var(x)} \approx 7 \]

Y qué el efecto directo fuera de 3

Efectivamente

Efecto total

# efecto total
lm(y~x)
#> 
#> Call:
#> lm(formula = y ~ x)
#> 
#> Coefficients:
#> (Intercept)            x  
#>       5.881        7.112
# efecto directo
lm(y~x+z)
#> 
#> Call:
#> lm(formula = y ~ x + z)
#> 
#> Coefficients:
#> (Intercept)            x            z  
#>      2.0318       3.0339       0.9945

Full luxury bayes

Hagamos lo mismo pero estimando el dag completo


library(cmdstanr)
library(rethinking)
set_cmdstan_path("~/Descargas/cmdstan/")

dat <- list(
  N = N,
  x = x,
  y = y,
  z = z
)
set.seed(1908)

flbi <- ulam(
  alist(
    # x model, si quiero estimar la media de x sino, no me hace falta
    x ~ normal(mux, k),
    mux <- a0,
    z ~ normal( mu , sigma ),
    
    mu <- a1 + bx * x,
   
    y ~ normal( nu , tau ),
    nu <- a2 + bx2 * x + bz * z,

    # priors
    
    c(a0,a1,a2,bx,bx2, bz) ~ normal( 0 , 0.5 ),
    c(sigma,tau, k) ~ exponential( 1 )
  ), data=dat , chains=10 , cores=10 , warmup = 500, iter=2000 , cmdstan=TRUE )
#> Running MCMC with 10 parallel chains, with 1 thread(s) per chain...
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Y recuperamos los coeficientes y varianzas


precis(flbi)
#>            mean         sd      5.5%    94.5%     n_eff     Rhat4
#> bz    1.0131931 0.01557219 0.9883466 1.038241  8226.042 1.0005919
#> bx2   2.9610334 0.07098661 2.8447278 3.073501  9640.122 1.0004040
#> bx    4.1535508 0.06180592 4.0552767 4.252771  8753.661 1.0001456
#> a2    1.9440850 0.09315785 1.7952495 2.093027 11266.267 1.0001373
#> a1    3.7089914 0.13578637 3.4916007 3.926010  8709.448 1.0002893
#> a0    1.9857103 0.03108199 1.9357773 2.035712 16252.310 0.9997523
#> k     0.9719412 0.02166791 0.9378896 1.007410 15276.158 0.9999036
#> tau   0.9859183 0.02216068 0.9512944 1.021932 15735.140 1.0001540
#> sigma 1.9636543 0.04382554 1.8950489 2.034130 15855.098 1.0000898
# extraemos 10000 muestras de la posteriori 
post <- extract.samples(flbi, n = 10000)

Y el efecto directo de x sobre y sería

quantile(post$bx2, probs = c(0.025, 0.5, 0.975))
#>     2.5%      50%    97.5% 
#> 2.823659 2.962330 3.099855

En este ejemplo sencillo, podríamos estimar el efecto causal de x sobre y simplemente sumando las posterioris

quantile(post$bx + post$bx2, c(0.025, 0.5, 0.975))
#>     2.5%      50%    97.5% 
#> 6.928046 7.114180 7.299252

También podríamos obtener el efecto causal total de x sobre y simulando una intervención. En este caso, al ser la variable continua, lo que queremos obtener como de diferente es y cuando \(X = x_i\) versus cuando \(X = x_i+1\).

Siempre podríamos ajustar otro modelo bayesiano dónde no tuviéramos en cuenta a z y obtendríamos la estimación de ese efecto total de x sobre y, pero siguiendo a Rubin y Gelman, la idea es incluir en nuestro modelo toda la información disponible. Y tal y como dice Richard McElreath , Statistical Rethinking 2022, el efecto causal se puede estimar simulando la intervención.

Así que el objetivo es dado nuestro modelo que incluye la variable z, simular la intervención cuando \(X = x_i\) y cuando \(X = x_i+1\) y una estimación del efecto directo es la resta de ambas distribuciones a posteriori de y.

Creamos función para calcular el efecto de la intervención y_do_x

get_total_effect <- function(x_value = 0, incremento = 0.5) {
  n <- length(post$bx)
  with(post, {
    # simulate z, y  for x= x_value
    z_x0 <- rnorm(n, a1 + bx * x_value  , sigma)
    y_x0 <- rnorm(n, a2  + bz * z_x0 + bx * x_value , tau)
    
    # simulate z,y for x= x_value +1 
    z_x1 <- rnorm(n, a1 + bx * (x_value + incremento)  , sigma)
    y_x1 <- rnorm(n, a2  + bz * z_x1 + bx2 * (x_value + incremento) , tau)
    
    
    # compute contrast
    y_do_x <- (y_x1 - y_x0) / incremento
    return(y_do_x)
  })
  
}

Dado un valor de x, podemos calcular el efecto total

y_do_x_0_2 <- get_total_effect(x_value = 0.2) 

quantile(y_do_x_0_2)
#>         0%        25%        50%        75%       100% 
#> -15.324628   2.395551   6.702379  10.987520  28.909002

Podríamos hacer lo mismo para varios valores de x

x_seq <- seq(-0.5, 0.5, length = 1000)
y_do_x <-
  mclapply(x_seq,  function(lambda)
    get_total_effect(x_value = lambda))

Y para cada uno de estos 1000 valores tendría 10000 valores de su efecto total de x sobre y.

length(y_do_x[[500]])
#> [1] 10000

head(y_do_x[[500]])
#> [1]  3.055107  4.988595  5.030397 12.469616 14.944735 22.881773

Calculamos los intervalos de credibilidad del efecto total de x sobre y para cada valor de x.

# lo hacemos simplemente por cuantiles, aunque podríamos calcular el hdi también, 

intervalos_credibilidad <-  mclapply( y_do_x, function(x) quantile(x, probs = c(0.025, 0.5, 0.975)))

# Media e intervalor de credibilidad para el valor de x_seq en la posición 500 
mean(y_do_x[[500]])
#> [1] 7.278184
intervalos_credibilidad[[500]]
#>      2.5%       50%     97.5% 
#> -4.867513  7.337732 19.761850

intervalo de predicción clásico con el lm

Habría que calcular la predicción de y para un valor de x y para el de x + 1, podemos calcular los intervalos de predicción clásicos parea cada valor de x, pero no para la diferencia ( al menos con la función predict)


mt_lm <- lm(y~x)
predict(mt_lm, newdata = list(x= x_seq[[500]]), interval  = "prediction")
#>        fit      lwr     upr
#> 1 5.877439 1.578777 10.1761
predict(mt_lm, newdata = list(x= x_seq[[500]] +1), interval  = "prediction")
#>        fit      lwr     upr
#> 1 12.98993 8.698051 17.2818

Pero como sería obtener el intervalo de credibilidad para la media de los efectos totales?

Calculando para cada valor de x la media de la posteriori del efecto global y juntando todas las medias.

slopes_mean <- lapply(y_do_x, mean)

quantile(unlist(slopes_mean), c(0.025, 0.5, 0.975))
#>     2.5%      50%    97.5% 
#> 6.023455 7.180465 8.326128

Que tiene mucha menos variabilidad que el efecto global en un valor concreto de x, o si juntamos todas las estimaciones

quantile(unlist(y_do_x),  c(0.025, 0.5, 0.975))
#>      2.5%       50%     97.5% 
#> -5.219611  7.168985 19.562729

Evidentemente, podríamos simplemente no haber tenido en cuenta la variable z en nuestro modelo bayesiano.

flbi_2 <- ulam(
  alist(
    # x model, si quiero estimar la media de x sino, no me hace falta
    x ~ normal(mux, k),
    mux <- a1,
    
    y ~ normal( nu , tau ),
    nu <- a2 + bx * x ,
    
    # priors
    
    c(a1,a2,bx) ~ normal( 0 , 0.5 ),
    c(tau, k) ~ exponential( 1 )
  ), data=dat , chains=10 , cores=10 , warmup = 500, iter=2000 , cmdstan=TRUE )
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Y obtenemos directamente el efecto total de x sobre y.

precis(flbi_2)
#>          mean         sd      5.5%    94.5%    n_eff     Rhat4
#> bx  7.1948083 0.06787458 7.0863389 7.302933 10446.14 1.0001211
#> a2  5.6085615 0.14948341 5.3685467 5.848723 10598.69 1.0001680
#> a1  1.9860307 0.03058033 1.9368689 2.034631 14116.13 1.0004122
#> k   0.9721054 0.02159773 0.9382407 1.006761 15044.78 0.9995902
#> tau 2.1898100 0.04953736 2.1120600 2.269831 14995.33 0.9999084
post2 <- extract.samples(flbi_2,  n = 10000)

quantile(post2$bx, probs = c(0.025, 0.5, 0.975))
#>     2.5%      50%    97.5% 
#> 7.061389 7.194525 7.327361

Pensamientos finales

  • Parece que no es tan mala idea incluir en tu modelo bayesiano toda la información disponible.

  • Ser pluralista es una buena idea, usando el backdoor criterio dado que nuestro dag sea correcto, nos puede llevar a un modelo más simple y fácil de estimar.

  • Como dije en el último post, estimar todo el dag de forma conjunta puede ser útil en varias situaciones.

  • Ya en 2009 se hablaba de esto aquí