Let $X$ be a uniform random variable on $[0,1]$, let Y be an exponential random variable with parameter $\lambda$ independent of $X$. Find the probability density function of $X + Y$
I know that this question has already an answer but I would like to know whether my solution is equivalent to the answers in this site,
My solution is: $$ (f_X * f_Y)(x) = \begin{cases} \displaystyle \int_{0}^1 f_X(y) f_Y(x-y)\,dy & \text{if }x\ge1, \\[10pt] \displaystyle\int_0^x f_X(y) f_Y(x-y)\,dy & \text{if }0 < x < 1, \\[10pt] 0 & \text{otherwise}, \end{cases} $$
But I always find written: $$ (f_X * f_Y)(x) = \begin{cases} \displaystyle \int_{x-1}^x f_X(y) f_Y(x-y)\,dy & \text{if }x\ge1, \\[10pt] \displaystyle\int_0^x f_X(y) f_Y(x-y)\,dy & \text{if }0 < x < 1, \\[10pt] 0 & \text{otherwise}, \end{cases} $$
Are they equivalent? I think so, but can you help me to make sure this is the case?
This has nothing to do with the question mentioned, I am not referring to any change of variable.