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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?

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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 $x$.

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

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

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