Convolution of Two i.i.d. Exponential Distributions I have $X,Y$ random variables, both equally distributed under the exponential distribution. However, for some reason the convolution seems to vanish:
$$Z = X+Y $$
Therefore,
$$f_Z := \int_{0}^{x}f_X(x-t)f_Y(t)dt  $$
$$f_Z = \alpha_1\alpha_2\frac{e^{-\alpha_1 x}-e^{-\alpha_2 x}}{\alpha_1-\alpha_2},\hspace{6mm} \alpha_1 = \alpha_2 $$
The Algebra behind the convolution says $Z$ vanishes. But I don't find any intuitive reason on why this happens.
 A: You have done your algebra incorrectly. In your answer you are dividing by zero (which I might add doesn't mean the convolution vanishes as you say). Also, you didn't mention it, but I'm also going to assume $X$ and $Y$ are independent (otherwise you can't just take the convolution of the respective pdfs). The exponential distribution has pdf $$f(x) = \begin{cases} 
\lambda e^{-\lambda x}, & x\geqslant 0 \\
0, & x<0 . 
\end{cases} $$ Hence, if $x>0$ then \begin{align*}
\int_{-\infty}^\infty f(y)f(x-y) d y &= \lambda^2\int_0^x e^{-\lambda y}e^{-\lambda(x-y)} d y \\
&= \lambda^2\int_0^x e^{-\lambda x} d y = \lambda^2 x e^{-\lambda x}. 
\end{align*} If $x<0$ then $$ \int_{-\infty}^\infty f(y)f(x-y) d y = 0.$$ Thus, $$ 
f_Z(x) = \begin{cases} 
\lambda^2 x e^{-\lambda x}, & x\geqslant 0 \\
0, & x<0. 
\end{cases}
$$ This is nonzero.
A: Another way is to "re-cycle" your result which, btw, should have been:
$$f_Z = \alpha_1\alpha_2\frac{e^{-\alpha_1 x}-e^{-\alpha_2 x}}{\alpha_2-\alpha_1}$$
(otherwise, you have a negative valued density).
Here is how: setting $\alpha_2=\alpha_1+\epsilon$ and making $\epsilon$ tend to $0$, for a fixed $x$, we get:
$$\lim_{\epsilon \to 0} f_Z(x) = \lim_{\epsilon \to 0}\alpha_1(\alpha_1 + \epsilon)\frac{e^{-\alpha_1 x}-e^{-(\alpha_1+\epsilon) x}}{(\alpha_1+\epsilon)-\alpha_1}=-\alpha_1^2\lim_{\epsilon \to 0}\frac{e^{-(\alpha_1+\epsilon) x}-e^{-\alpha_1 x}}{\epsilon}$$
We recognize in this limit the derivative of $e^{- \alpha_1 x}$ with respect to $\alpha_1$ (in particular, $x$ is viewed as a constant) giving back the desired density:
$$(-\alpha_1^2)(-xe^{- \alpha_1 x})=\alpha_1^2xe^{- \alpha_1 x}$$
