Define $Y = \min(X, c)$, for a given constant $c \in \mathbb{R}^+$, and nonzero real-valued random variable $X \sim F_X$, $F_X$ the cdf of $X$. I'm trying to find $\mathbb{E} \left(g(Y)\right)$, where $g(X) \sim \exp(1)$.
My attempt
First I have to find the distribution of $Y$. Since
$$ \begin{aligned} f_Y(y) &= f_X(y | X > c)P(X > c) + P(y = c | X \le c) P(X \le c) \end{aligned} $$
thus
$$ \begin{aligned} \mathbb{E} \left(g(Y)\right) &= \int_\mathbb{R} g(y) f_Y(y) dy \\ &= \int_\mathbb{R} g(x) f_Y(x) dx \\ &= \int_{0}^c \exp(-x) F_X(c)dx + \int_c^{\infty} \exp(-x)f_X(x)(1-F_X(c))dx \end{aligned} $$
But I think I'm lost. Am I on the right track?
$$E(g(Y)) =P(X\leq c) E(g(Y)|X\leq c) + P(X>c) E(g(Y)|X > c) =F(c) \cdot 1+(1-F(c) \cdot c.$$ Because when $X\leq c$ the mean of $g(Y)$ is an expontential cause $Y =X$ as the minimum of $X$ and $c$ and in the other cause the expected value of $c$ is $c$.