Let $Y\sim\text{Exp}(1)$ be a random variable. I denote the random variable $X$ as $X=\min(Y,1)$. The task is to find the moment generating function of $X$.
By simply calculating the probability I managed to find the CDF of $X$ is: $$F_X(t) = \begin{cases} 1-e^{-t}, & 0 \le t <1\\ 1, & t \ge 1\\ 0, & \text{otherwise} \end{cases}$$
Here I got stuck. Since $X$ is not a continuous variable, it does not have a PDF, and without it I do not know how to calculate the moment generating function. ($X$ is also not discrete).
I will appreciate some help.
The definition of the moment generating function does not require a PDF. $$\phi_X(t) := E[e^{tX}].$$
From here, it may be helpful to write $$e^{tX} = e^{tX} \mathbf{1}\{Y > 1\} + e^{tX} \mathbf{1}\{Y \le 1\}$$ and compute the expectation of the two terms separately.
Edit for more details:
$$E[e^{tX} \mathbf{1}\{Y > 1\}] = E[e^{tY} \mathbf{1}\{Y > 1\}] = \int_1^\infty e^{ty} f_Y(y) \, dy = \cdots.$$ $$E[e^{tX} \mathbf{1}\{Y \le 1\}] = e^t E[\mathbf{1}\{Y \le 1\}] = e^t P(Y \le 1) = \cdots.$$