Let $X,Y,Z$ be random variables assuming values on $\mathbb{N}$ with $X \neq Y$. Let define
$$Z = \begin{cases} X, \quad if Y > X\\ Y, \quad if Y < X \end{cases}$$
Obtain $\mathbb{E}[Z-X]$.
My attempt was:
$$\mathbb{E}[Z-X] = \mathbb{E}[X-X]\cdot \mathbb{P}(Y>X) + \mathbb{E}[Y-X]\cdot \mathbb{P}(Y<X) = \mathbb{E}[Y-X]\cdot \mathbb{P}(Y<X)$$
Is this result correct? If it is, how can I proove?
No, the result is not correct because we know from the definition of $Z$ that $Z \le X$ for all possible outcomes of $X$ and $Y$. This is because if $Y > X$, then $Z = X$, and if $Y < X$, then $Z = Y < X$. So $Z$ is never greater than $X$, hence $Z - X \le 0$, with equality occurring if and only if $\Pr[Y < X] = 0$. So your expectation cannot be positive.
However, $\mathbb E[Y - X]$ could be positive. For instance, suppose $X$ and $Y$ are independent and $$\Pr[X = x] = \begin{cases} 1/2, & x = 1 \\ 1/2, & x = 2 \end{cases}, \quad \Pr[Y = y] = \begin{cases} 1/4, & y = 0 \\ 3/4, & y = 100. \end{cases}$$ Then $\mathbb E[X] = 1/2(1 + 2) = 3/2$, and $\mathbb E[Y] = 75$, hence $$\mathbb E[Y-X] = \mathbb E[Y] - \mathbb E[X] = 75 - 3/2 > 0,$$ and $\Pr[X > Y] = \Pr[Y = 0] = 1/4$, which would mean that for our example, $$\mathbb E[Y - X] \Pr[X > Y] = 147/8.$$ But as we stated earlier, $\mathbb E[Z - X] < 0$, and we can calculate this explicitly: $$\Pr[Z = z] = \begin{cases} 1/4, & z = 0 \\ 3/8, & z = 1 \\ 3/8, & z = 2. \end{cases}$$ Notice we cannot have any other outcomes of $Z = \min(X,Y)$. Then $$\mathbb E[Z - X] = \mathbb E[Z] - \mathbb E[X] = 9/8 - 3/2 = -3/8 < 0.$$ We can still use linearity of expectation even when $Z$ and $X$ are not independent.
The correct expression is $$\mathbb E[Z - X] = \mathbb E[Y - X \color{red}{\,\mid\, Y < X}] \Pr[Y < X].$$ This is because the law of total expectation for discrete variables is $$\mathbb E[B] = \sum_{a \in \Omega} \mathbb E[B \mid A = a] \Pr[A = a]$$ where $\Omega$ is the support of the discrete variable $A$.
If we apply the correct formula, we have $$\Pr[Y - X = k \mid Y < X] = \begin{cases} 1/2, & k = -1 \\ 1/2, & k = -2. \end{cases}$$ This is because the only way $Y < X$ is if $Y = 0$, and when this happens, $Y - X = -X$. Therefore $$\mathbb E[Y - X \mid Y < X] = (1/2)(-1 + (-2)) = -3/2,$$ and $$\mathbb E[Z - X] = \mathbb E[Y - X \mid Y < X] \Pr[Y < X] = (-3/2)(1/4) = -3/8,$$ which matches our earlier calculation.