Convolution of two random variables with common pdf of $e^{-t}$

Question

Let $X$ and $Y$ be two random variables with common pdf where $f_X(t) = f_Y (t) = e^{−t}$ for $t > 0$, and $0$ otherwise. Find the density function of $2X + Y$.

I attempted the following convolution:

$$\int_{-\infty}^{\infty}f_X(t-y)f_Y(y)dy$$

$$\int_{0}^{t}f_X(2x)f_Y(y)dy$$

$$e^{-t}\int_{0}^{t}f_X(t-y)dy$$

Now I can substitute $f_X(u)$ for $f_X(t-y)$ by converting the bounds of the integral, and since $t-0=t$ and $t-t=0$ and $du=-dy$ and I have

$$-e^{-t}\int_{t}^{0}f_X(u)du$$

$$-e^{-2t}\int_{t}^{0}du$$

This is obviously incorrect and I am sure there must be more than one mistake, I am just not sure what they are.

Answer 1

Let $Z=2X$, then $$F_{Z}(z)=\Pr[Z\le z]=\Pr[2X\le z]=\Pr[X\le \tfrac z2]=F_X(\tfrac z2)$$ Differentiating: $f_Z(z)=\frac12f_X(\tfrac z2)$.

The pdf of $T=Z+Y$ is therefore given by:

$$\begin{align} f_T(t)&=\int_{-\infty}^{+\infty}f_Z(\tau)f_Y(t-\tau)d\tau\\[8pt] &=\int_{-\infty}^{+\infty}\tfrac12f_X(\tfrac{\tau}2)f_Y(t-\tau)d\tau\\[8pt] &=\frac12\int_{0}^{t}e^{-\tau/2}e^{-(t-\tau)}d\tau\tag{t>0}\\[8pt] &=\frac12e^{-t}\int_{0}^{t}e^{\tau/2}d\tau\tag{t>0}\\[8pt] &=\frac12e^{-t}\,\cdot\,2(e^{t/2} - 1)\tag{t>0} \end{align}$$ and $f_T(t)=0, \forall t\le 0.$

Answer 2

Your mistake is when you say $f_X(2x) = e^{-2x}$. Normally the way you do it is as follows: $$F_{2X}(x) = Pr(2X <x ) = Pr(X < \frac{x}{2}) = \int\limits_{-\infty}^{\frac{x}{2}} f_X(t) \ dt $$ Take a change of variable of $u = 2t$ you get $$F_{2X}(x) =\int\limits_{-\infty}^{x} f_X(\frac{u}{2}) \frac{1}{2} du = \frac{1}{2} \int\limits_{-\infty}^{x} f_X(\frac{u}{2}) du$$ The PDF is the derivative of the CDF, i.e. $$f_{2X}(x) = F_{2X}'(x) = \frac{1}{2}f_X(\frac{u}{2}) = \frac{1}{2}e^{-\frac{x}{2}} \qquad x>0$$ Now, you can convolve $f_{2X}(x) = \frac{1}{2}e^{-\frac{x}{2}}$ with $f_Y(y) = e^{-y}$, as you were doing.