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Would a propensity score be valid as an instrumental variable in a quasi experimental context? I've seen papers that explore the question from the opposite direction: can an instrumental variable be used in the calculation of propensity scores. Thanks.

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No.

Let's say we have treatment $T$, outcome $Y$, and some other variables $X$. The propensity score is then defined as

$$p(x) = P(T = 1|X = x)$$

This is first and foremost a purely statistical construct which has no direct connection to causality. Causality comes into the picture if we assume "ignorability":

$$P(Y_{t}|T, X) = P(Y_{t}|X)$$

where $Y_{t}$ is the potential outcome of $Y$ when $T$ is set to $t$. This holds for example when the back-door criterion holds on the causal graph you assume (i.e., $X$ blocks all back-door paths from $T$ to $Y$).

Rubin and Rosenbaum argued that it will then hold that

$$P(Y_{t}|T, p(X)) = P(Y_{t}|p(X))$$

which makes estimation easier since one does not need to deal with the potentially high-dimensional $X$, but simply $p(x)$, which is a single number.

For instrumental variable analysis, you are looking for a $X$ such that at the very least it holds that ("instrument independence") $$P(Y_{t}|X) = P(Y_{t})$$

(usually, you assume more, but let's ignore this). However, typically, $X$ that fulfill the "ignorability" assumption above contains variables that influence both $T$ and $Y$. Influencing $Y$ would violate instrument independence. Since the propensity score is simply a function of these $X$, they also will violate instrument independence.

Literature: Pearl, Judea: Understanding propensity scores. In: Causality, 2nd ed. CUP.

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    $\begingroup$ Good answer. I wonder if you were to calculate a propensity score from several instruments and no true confounders if the resulting propensity score could be a used as a unidimensional summary of the instruments, which itself would be an instrument. $\endgroup$ – Noah May 27 '17 at 20:33
  • $\begingroup$ Yes, at least under linearity, that's what 2SLS and GMM do; they estimate E[T|X] in the first stage, where X are the instruments. This also happens to the be the propensity score. But this is not the IV estimator as such. Also, propensity scores are mostly (always?) discussed when it comes to adjustment for matching, not IV estimation. $\endgroup$ – Julian Schuessler Jun 1 '17 at 9:06
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3 questions need to pay attention when using propensity score (PS) as covariate in the regression model. In the practice, PS is got from logistic regression. For example: the linear model.

1: $Y=\beta_0 + \beta_1 T + \beta_2P +\epsilon$ where $T$ is treatment and $P$ is PS. Because $P=\frac {\exp(X\beta)}{1+\exp(X\beta)}$, we have $Y=\beta_0 + \beta_1 T + \beta_2\frac {\exp(X\beta)}{1+\exp(X\beta)} +\epsilon$. The question is: Why the response variable $Y$ has this strange relation with covariate $X$? Why we need this limitation?

2: Should we consider the variance of $P$, because $P$ is estimated from data, not observed true value?

3: If $X$ has high dimensional and/or co-linearity problem when the $X$ is used in linear model directly, $X$ has high dimensional and/or co-linearity problem in the logistic regression when you try to get PS. Is it possible to resolve this problem in logistic regression?

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  • $\begingroup$ I don't see how this addresses the question. OP asks about instruments. $\endgroup$ – Julian Schuessler Jun 1 '17 at 9:08

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