distribution of Z=X+Y

Question

If $X~Uni(0, 1)$ and $Y~Exp(1)$ what is the distribution of $ Z = X+Y$ considering $X$ and $Y$ are independent?

I used the moment generating function method and obtain $M_z (t) = \frac{\lambda e^t - \lambda}{\lambda t - t^2}$

I can't recall what that distribution is... Could it be gamma distribution it $\lambda = 1$?

Thanks for the help in advance!!!

Answer 1

$$Pr(Z<z)=Pr(X+Y<z)=Pr(Y<z-X)=\int_{0}^{1}Pr(Y<z-x)f_X(x)dx=\int_{0}^{1}Pr(Y<z-x)dx=\int_{z-1}^{z}Pr(Y<u)du$$for $z\le 1$ we have: $$\int_{z-1}^{z}Pr(Y<u)du=\int_{0}^{z}Pr(Y<u)du=\int_{0}^{z}1-e^{-u}du=z-1+e^{-z}$$and for $z\ge 1$:$$\int_{z-1}^{z}Pr(Y<u)du=\int_{z-1}^{z}1-e^{-u}du=1+e^{-z}(1-e)$$by differentiating we arrive at$$f_Z(z)=1-e^{-z}\quad,\quad 0\le z\le 1 \\(e-1)e^{-z}\quad , \quad z\ge1$$