Problem
Consider $X_1, X_2 \sim \varepsilon(1)$ independent and define $T = X_1 - 2X_2$.
I want to calculate the probability density function (pdf) of $T$, denoted by $f_T$, which can be obtained by convolving the pdfs of $X_1$ with $-2X_2$.
My result is \begin{equation} f_T(t) =\begin{cases} \frac{2}{3} \, e^{t/2} & t < 0 \\ \frac{2}{3} \, e^{-t} & t \geq 0 \end{cases} \end{equation}
Unfortunately, this pdf is too large by a factor of 2. Below, I show my calculations that led here. Please help me identify the error.
Calculations
Let $f_1$ denote the pdf of $X_1$ and $f_2$ denote the pdf of $-2X_2$. Using this notation, we calculate $f_2$ by deriving its respective cumulative distribution function (cdf), denoted by $F_2$.
For $F_2$ we obtain:
\begin{align} F_2(x) &= P(-2X_2 \leq x) \\ &= P \left (X_2 \geq -\frac{x}{2} \right) \\ &= \int_{ -x/2}^{\infty} \mathbf{1}_{[0, \infty)}(y) \cdot e^{-y} dy \\ &=\begin{cases} 1 & x > 0 \\ e^{x/2} & x \leq 0 \end{cases} \end{align}
Deriving $F_2$ yields: \begin{equation} f_2(x) = \frac{\partial F_2(x)}{\partial x} =\mathbf{1}_{(-\infty, 0]}(x) \cdot e^{x/2} \end{equation}
We now convolve $f_1$ with $f_2$: \begin{align} f_T(t) &= \int_{-\infty}^{\infty} f_1(x) f_2(t - x) dx \\ &= \int_{-\infty}^{\infty} \mathbf{1}_{[0, \infty)}(x) \cdot e^{-x} \cdot \mathbf{1}_{(-\infty, 0]}(t - x) \cdot e^{(t - x)/2} \, dx \\ &= \int_{0}^{\infty} \mathbf{1}_{[t, \infty)}(x) \cdot e^{(t - 3x)/2} \, dx \\ &= \int_{\text{max}(0, \, t)}^{\infty} e^{(t - 3x)/2} dx \\ &= \frac{2}{3} \, e^{\, t/2 \, - \, 3 \cdot \text{max}(0, \, t) /2} \\ &=\begin{cases} \frac{2}{3} \, e^{t/2} & t < 0 \\ \frac{2}{3} \, e^{-t} & t \geq 0 \end{cases} \end{align}
You are missing a factor of $1/2$ in the density $f_2(x)$: differentiating $$F_2(x) = \begin{cases}1, & x > 0 \\ e^{x/2}, & x \le 0\end{cases}$$ yields $$f_2(x) = \color{red}{\frac{1}{2}} e^{x/2}, \quad x \le 0.$$