Convolution of Two i.i.d. Exponential Distributions [duplicate]

Question

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.

Answer 1

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.

Answer 2

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:

$$(-\alpha_1^2)(-xe^{- \alpha_1 x})=\alpha_1^2xe^{- \alpha_1 x}$$