How does a conditional expectation transform into a simple expectation?

Question

Suppose I have random variables $X$ and $N$.

It is given

$E[X|N] = N + cE[X] + d$

Why is the above statement equivalent to

$E[X] = E[N] + cE[X] + d$

where $c$ and $d$ are just constants in real numbers.

Answer 1

Given $\mathbb{E}_{X}[X\mid N] = N + c \cdot \mathbb{E}_{X}[X] + d$, $$\mathbb{E}_{X}[X] = \mathbb{E}_{N}\left[\mathbb{E}_{X}[X \mid N]\right] = \mathbb{E}_{N}\left[N + c \cdot \mathbb{E}_{X}[X] + d\right] = \mathbb{E}_{N}[N] + \underbrace{c\cdot \mathbb{E}_{X}[X]}_{\text{constant with respect to }N}+d\text{,}$$ by double expectation.

Answer 2

This is called the tower property: for every random variables $X$ and $Y$ such that $X$ is integrable, $$E(E(X\mid Y))=E(X).$$