For $X\sim \exp(\lambda)$ find $E\left( \min\left(\lfloor X\rfloor, m\right)\right)$

Question

Let $X$ be a random variable so $X\sim \exp(0.41)$ Find $E\left( \min\left(\lfloor X\rfloor, 4\right)\right)$.

I'm trying to figure out how to solve it and determine a general solution for $\lambda=0.41$ and $m=4$. As I understand $P\left(\lfloor X\rfloor=x\right)=P\left(x\leq X<x+1\right)$. Also in order to find $E(Y)$ where $Y=\min\left(\lfloor X\rfloor, 4\right)$ I tried to go with the following equation: $$ E(Y)=\int_0^\infty (1-P_Y(Y\leq y))dy $$ But how do I really connect between $E(Y)$ and $\lfloor X\rfloor$?

Answer 1

Hint: $Y$ takes only $5$ vales $0,1,2,3$ and $4$. $P(Y=0)=P(X<1)=1-e^{-0.41}, P(Y=1)=P(1 <X<2)=e^{-0.41}-e^{-0.82}$ etc. Of course $EY=(0)P(Y=0)+(1)P(Y=1)+(2)P(Y=2)+(3)P(Y=3)+(4)P(Y=4)$. Can you finish?