# Causal Inference: Calculate Expectation

This question is from Kang&Schafer(2007), shown in the following picture:

where $$\pi_i$$ is the propensity score function, i.e, $$\mathrm{P}(T_i=1|Z_i=z_i)$$. I am very confused how I can derive $$r^{(1)}=\mathrm{P}(T_i=1)=\mathrm{E}_{Z_i}[\pi_i]=0.5$$ and $$\mu^{(1)}=\mathrm{E}[Y_i|T_i=1]=200$$. This question has bothered me for too long. Thanks a lot.

• Routine textbook questions need the self-study tag. – Carl Jul 27 '20 at 13:09
• Oh, it is actually from a paper in Statistical Science. – Morck Jul 27 '20 at 13:13

## 1 Answer

This one's a little tricky because you don't really need to calculate much at all: the result follows from the symmetries of the expit function and the Normal distribution.

The "expit" function is

$$\operatorname{expit}(x) = \frac{1}{1 + e^{-x}} = \frac{e^{x}}{e^x+1}= 1 - \frac{1}{1 + e^x} = 1 - \operatorname{expit}(-x),$$

demonstrating that for all numbers $$x,$$

$$\operatorname{expit}(x) + \operatorname{expit}(-x) = 1.\tag{*}$$

Geometrically, this symmetry means the graph of expit is invariant under a 180 degree rotation about $$(0,1/2):$$

The only other thing we need to know is that the assumptions on the $$z_{ij}$$ imply the distribution of $$Z = -z_{i1} + 0.5z_{i2} -0.25 z_{i3} -0.1 z_{i4}$$ is symmetric with zero mean: this follows immediately from the fact that all the individual means are zero and that each of the $$z_{ij}$$ has a distribution symmetric about its mean.

Let the distribution function of this linear combination be

$$F(z) = \Pr(Z \le z),$$

whence its symmetry can be expressed as

$$1 = F(z) + F(-z)$$

and therefore

$$\mathrm{d}(1) = 0 = \mathrm{d}\left(F(z)+F(-z)\right) = \mathrm{d}F(z) - \mathrm{d}F(-z),$$

showing that

$$\mathrm{d}F(-z) = -\mathrm{d}F(z).\tag{**}$$

This will be used in the change of variable $$z\to -z$$ below.

Compute expectations by splitting the integral into negative and positive halves and then substituting $$z=-z$$ in the negative half:

\begin{aligned} E[\pi_i] &= E[\operatorname{expit}(Z)] = \int \operatorname{expit}(z)\,\mathrm{d}F(z)\\ &=\int_{-\infty}^0 \operatorname{expit}(z)\,\mathrm{d}F(z) + \int_0 ^\infty\operatorname{expit}(z)\,\mathrm{d}F(z)\\ &=\int_\infty^0 \operatorname{expit}(-z)\,\mathrm{d}F(-z) + \int_0 ^\infty\operatorname{expit}(z)\,\mathrm{d}F(z)&\text{Change of variable}\\ &=\int_0^\infty \operatorname{expit}(-z)\,\mathrm{d}F(z) + \int_0 ^\infty\operatorname{expit}(z)\,\mathrm{d}F(z)&\text{From (**)}\\ &=\int_0^\infty \left(\operatorname{expit}(-z)+\operatorname{expit}(z)\right)\,\mathrm{d}F(z)\\ &=\int_0^\infty \mathrm{d}F(z)&\text{From (*).} \end{aligned}

That was the crux of the matter. I leave it to you to construct a similar demonstration that the latter integral is exactly one-half of $$\int \mathrm{d}F(z) = 1.$$

• Thx a lot for your answer! I believe I get it. But it only accounts for why $E[\pi_i]=0.5$. How can I solve for $\mu^{(1)}=200$? – Morck Aug 5 '20 at 9:33
• What does $\mu^{(1)}$ represent? Your formula gives it as a conditional expectation based on $T_i,$ but I don't see any information about $T_i$ in your post (or even about what $i$ is). – whuber Aug 5 '20 at 13:19
• $T_i$ is the indicator variable of the i-th unit being assigned to a treat or a control group. In my question, the only information about $T_i$ is $\pi_i=\mathrm{P}(T_i=1|Z_i=z_i)=\expit(z_i^\tau\beta)$, the probability of a unit being assigned to the treat given $z_i$. Since $Y_i=f(Z_i)$, you can see $Y_i$ is related to $T_i$ by $z_i$. Therefore, $\mathrm{E}[Y_i|T_i=1]\neq \mathrm{E}[Y_i|T_i=0]$ due to different conditional distribution of $Z_i$ given different value of $T_i$. – Morck Aug 6 '20 at 2:12
• So $T_i$ represents which group a unit was assigned to(equals to 1 if being assigned to the treat group and 0 otherwise), and $\mu^{(1)}$ is actually the mean outcome of the treat group. – Morck Aug 6 '20 at 2:19
• And you can regard $Z_i$ as the covariates, or baseline information of a patient, such as age, sex, income... – Morck Aug 6 '20 at 2:22