Convolutions: Let U~Unif(0,1) and X~Expo(1), independently. Find the PDF of U +X.

Let U~Unif(0,1) and X~Expo(1), independently. Find the PDF of U +X.

Solution:

$$f_T(t) = \int_{-\infty}^{\infty}f_Y(t-x)*f_X(x)dx$$

$$= \int_{-\infty}^{\infty}1*\lambda e^{-\lambda x}dx$$

Integrate over 0 to t. Can someone confirm this is correct, and why? I think because it is expo distribution, x and t must be >0 by some rule? My other guess would be integrate over 0 to infinity, but that would leave us with a numerical value, not a PDF with a variable, which is a clue that could be incorrect.

$$= \int_{0}^{t}\lambda e^{-\lambda x}dx$$

$$=1-e^{-\lambda x} |_0^t$$

$$= 1-e^{-\lambda t} - (1-e^0)$$

$$1-e^{-\lambda t}$$

for t>0

Can I read the answer as: In conclusion, this basically says adding the uniform distribution has no effect on the exponential PMF.

Be careful: if $$U\sim \mathcal U(0,1)$$, then $$f_U(t-x)=\left\{\begin{matrix}1 & 0
Since $$0 is the same as $$t-1, your integral becomes $$\int_{t-1}^t \lambda e^{-\lambda x}dx,$$ and actually this is true if $$t-1>0$$ (that is, $$t>1$$), since $$f_X(x)=\left\{\begin{matrix}\lambda e^{-\lambda x} & x>0\\ 0 & \text{otherwise}.\\ \end{matrix}\right.$$
If $$t\in [0,1]$$, then the integral is between $$0$$ and $$t$$.
Finally, the integral is zero if $$t<0$$.