0
$\begingroup$

Let $X$ have pdf $$f_X(x)=e^{-x} \qquad \text{ for } x \ge 0$$ and $Y$ have $$f_Y(y)=1 \qquad \text{ for } 0\le y \leq 1$$
$X, Y$ both independent. What is the pdf for $Z=X+Y$?

Using convolution formula I get $$f_Z(z)=\int_{-\infty}^{\infty}f_X(z-y)f_Y(y)dy$$

How do I figure out my limits of integration? Since $f_X(x)$ is from $0\leq x \lt \infty$ so should $f_X(z-y)$ be. That makes want to integrate from $0$ to $z$, which is apparently wrong.

$\endgroup$
  • $\begingroup$ In your integral, as $y$ sweeps from $-\infty$ to $\infty$, the integrand must be $0$ except when $y \in [0,1]$, right? So straightaway, the limits of the integral can be reduced. Next, let's pick a numerical value (say $3$) for $z$. As $y$ sweeps from $0$ to $1$, is $f_X(3-y)$ always nonzero? and if so, what is $f_X(3-y)$? What if we had chosen $z=5$ instead? or $z=9.382$? Just when you were ready to jump to conclusions, I suggest: try it for $z = 0.5$. $\endgroup$ – Dilip Sarwate Nov 3 '14 at 23:16
2
$\begingroup$

Write the densities \begin{align} f_Y(y) &= 1_{0<y<1} \\ f_X(x) &= 1_{x > 0} e^{-x} \end{align} so that the convolution becomes \begin{align} \int_{-\infty}^\infty 1_{0<z-x<1} 1_{x > 0} e^{-x} dx &= \int_{\max(0, z-1)}^z e^{-x} dx \\&= e^{-\max(0, z-1)} - e^{-z} \\ f_Z(z)&= e^{\min(0, 1-z)} - e^{-z} \end{align}

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.