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.
 A: 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 $\int \mathrm{d}F(z) = 1.$
