The question is
What is the expectation of $f(X)$? where $X \sim \mathcal N(0, 1)$ and $f$ is the PDF of $\mathcal N(0, 1)$
I think in this case we have
$$ E(X) = \int_{-\infty}^{\infty} f(x) \times f(x) dx $$ $$ E(X) = \int_{-\infty}^{\infty} \frac{1}{2\pi} (e^{-x^2/2})^2 dx $$
$$ E(X) = \dfrac{1}{2\sqrt{{\pi}}} $$
Am I right there? I don't have solutions for this question so validation of my approach (either way) would be appreciated. I could not find a similar question when searching (probably because I found wording the question hard). Thanks in advance.
Looks right to me. Another way of doing it is:
\begin{align} \mathbb E\left[f(a + X)\right] &= \frac{\mathrm d}{\mathrm d a}\mathbb E\left[\Phi\left(a + X\right)\right]\\ &= \frac{\mathrm d}{\mathrm d a}\mathbb E\left[\mathbb P\left[Y \le a+X \Big | X\right]\right]\\ &= \frac{\mathrm d}{\mathrm d a}\mathbb P\left[Y - X \le a\right]\\ &= \frac{\mathrm d}{\mathrm d a} \Phi\left(\frac{a - 0}{\sqrt{2}}\right)\\ &= \frac1{\sqrt 2}f\left(\frac{a}{\sqrt 2}\right) \end{align}
For $a = 0$ you have your equality.