Suppose that $X \sim \text{Exponential}(\lambda)$ and $Y \sim \text{Exponential}(2\lambda)$ are independent. I want to find the density of their sum $Z = X + Y$.

Now, I know how to solve this problem using the convolution.I was wondering if it can be solved in a different fashion. In particular:

$$ f_Z(Z = z) = f_Z(X + Y = z) = f_Z(X = x, Y= z - x) $$

By independence we obtain:

$$ f_Z(X = x, Y= z - x) = f_X(X=x)f_Y(Y= z- x) = \lambda e^{-\lambda x}\mathbb{1}_{\{x > 0\}}2\lambda e^{-2\lambda (z -x) }\mathbb{1}_{\{z -x > 0\}} $$

After this point, the rest is algebra:

$$ \lambda e^{-\lambda x}\mathbb{1}_{\{x > 0\}}2\lambda e^{-2\lambda (z -x) }\mathbb{1}_{\{z -x > 0\}} = 2\lambda^2e^{\lambda x - 2\lambda z}\mathbb{1}_{\{z -x > 0\}} $$

This does not agree with the result I get from using the convolution formula. I am wondering why. Can I transform what I have here to match that or is there something fundamentally wrong in this procedure?


You state $f_Z(z) =\hspace{-1.5ex}?~~ f_{X}(x)~f_Y(z-x)$

However, just where did the $x$ come from? It's some value realised by $X$, but... which.

What you have should be the joint density function of $X,Z$: $~f_{X,Z}(x,z) = f_X(x)~f_Y(z-x)~$ by change of variables, and independence of $X,Y$.

Indeed, as $X$ can realise any supported value that allows $X+Y=z$, that is the purpose of the convolution formula.   We 'integrate out' the $x$.

$$\begin{align}f_Z(z) ~=~& \int_\Bbb R f_{X}(x)~f_Y(z-x)\operatorname d x \\ =~& \int_{\Bbb R} 2\lambda^2 e^{-\lambda x-2\lambda(z-x)}\mathbf 1_{x>0, z-x>0, z>0}\operatorname d x \\=~& 2\lambda e^{-2\lambda z}\int_0^z \lambda e^{\lambda x}\operatorname d x ~\mathbf 1_{z>0}\\=~& 2\lambda e^{-2\lambda z}(e^{\lambda z}-e^0) ~\mathbf 1_{z>0} \\=~& 2\lambda (e^{-\lambda z}-e^{-2\lambda z}) ~\mathbf 1_{z>0} \end{align}$$

  • $\begingroup$ Now it's perfectly clear! Thanks! $\endgroup$ – Orest Xherija Oct 28 '16 at 2:22

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.