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Considering two continuous random variables $X$ and $Y$ with $d.f \; F_X, F_Y$ I want to fin the distribution and distribution function of the sum $Z=X+Y$.

\begin{align} P\{Z \leq z\} &= P\{X+Y \leq z\} \\ &=P\{X \leq z - Y\} \\ &=F_X(z - Y) \end{align}

I know here that I have to use somehow the joint distribution but I can't figure out how. We know that if $T(\omega) = (X(\omega), Y(\omega))$:

\begin{align} P\{T \leq (a,b)\} &= P\{X\leq a, Y \leq b\} = \int \int_{x \leq a, y \leq b} f_{XY}(a,b)\;dxdy \end{align}

Using the Radon-Nikodym Theorem if and only $T$ is absolutely continuous w.r.t the two dimensional lebesgue measure.

How to connect these two to find that the distribution function of the sum of the convolution of the densities ? Can somebody give me a formal derivation of this ? My text book and my online research provide poor explanations.

EDIT : Tentative of formulation :

Let $T(\omega) = (X(\omega),Y(\omega))$ be a random element, and $\phi(x,y) = x+y$ we have :

\begin{align} P\{Z \in B\} &= P\{X+Y \in B\} \\ &=P\{\phi(X,Y) \in B\} \\ &=P\{(X,Y) \in \phi^{-1}(B)\} \\ &=P\{T \in \phi^{-1}(B)\} \\ &= \int_{\phi^{-1}(B)} f_{XY}(x,y)d(x,y) \quad \text{If $PT^{-1}$ is absolutly continuous w.r.t 2D LM} \\ &= \int\int_{\phi^{-1}(B)} f_{XY}(x,y)dx dy \quad \text{Radon-Nikodym} \\ &= \int_{\mathbb{R}}\int_{-\infty}^{B - y} f_{XY}(x,y)dx dy \quad \text{because $(x,y) \in \phi^{-1}(B) \implies x \in B - y\quad \forall y$} \end{align}

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In general $$ \eqalign{P(X + Y \le z) &= \iint_{\{(x,y): x+y \le z\}} dx\; dy\; f_{XY}(x,y)\cr &= \int_{-\infty}^\infty dx \int_{-\infty}^{z-x} dy \; f_{XY}(x,y) }$$

If $X$ and $Y$ are independent, $f_{XY}(x,y) = f_X(x) f_Y(y)$ so this becomes $$ \int_{-\infty}^\infty dx \int_{-\infty}^{z-x} dy\;f_X(x) f_Y(y) = \int_{-\infty}^\infty dx\; \int_{-\infty}^z dt\; f_X(x) f_Y(t-x)$$

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  • $\begingroup$ I hope the OP is familiar with this notation. $\endgroup$ – drhab Aug 27 '14 at 17:27
  • $\begingroup$ Thank you for you comment ! Can you explain me how do you derive the very first line ? The $X+Y$ random variable takes value to $\mathbb{R}$ so how can we end up with a double integral ? What measure are we considering here ? I think whatever measure it is, it looks like it needs to be absolutly continuous w.r.t a 2 dimensional lebegue measure which would derive the integral. Then are you using the Fubini theorem to split the two integrals ? I am sorry for the bugging but I need more information on how to derive your result ... $\endgroup$ – user149705 Aug 27 '14 at 17:45
  • $\begingroup$ The event $X + Y \le z$ is the same as $(X,Y) \in \{(x,y): x + y \le z\}$. The probability measure corresponds to the joint distribution of $X$ and $Y$. That it is absolutely continuous with respect to two-dimensional Lebesgue measure is exactly (thanks to Radon-Nikodym) the statement that $X$ and $Y$ have a joint density $f_{XY}$ (which I am assuming because you mentioned $f_{XY}$). $\endgroup$ – Robert Israel Aug 27 '14 at 20:33
  • $\begingroup$ Of course if it is not absolutely continuous, there is no such formula involving densities. $\endgroup$ – Robert Israel Aug 27 '14 at 20:36
  • $\begingroup$ And yes, of course Fubini is used to make a double integral into an iterated integral. $\endgroup$ – Robert Israel Aug 27 '14 at 20:38

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