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


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

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

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 $$f_U(t-x)=\left\{\begin{matrix}1 & 0<t-x<1\\0 & \text{otherwise}.\\ \end{matrix}\right.$$

Since $0<t-x<1$ is the same as $t-1<x<t$, 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 $t<0$.


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