Suppose that $X$ and $Y$ are independent continuous random variables. Find the density of $X + Y$. Suppose $X$ and $Y$ are independent continuous random variables. Show the density function of $X + Y$ is:
$$f_{x + y}(a) = \int\limits_{-\infty}^{\infty} f_x(a - y)f_y(y) dy$$
Let $f(x,y)$ be the joint PDF of $X$ and $Y$. Then since $X$ and $Y$ are independent, $f(x,y) = f_X(x)f_Y(y)$. Then I'm stumped. I'm also confused why the function is now a single variable in terms of $a$.
 A: let $Z = X + Y$
the distribution function for $Z$ conditioned on $X=x$
$P_{Z|X}(Z \le z | X = x) = P(X + Y \le z | X = x) = P(x + Y \le z | X = x) = [Y \text{ is independent of X}] = P(Y \le z - x)$
differentiating, we'll get conditional density: $f_{Z|X}(z|x) = f_Y(z-x)$
Finally, to get the marginal density for $Z$, let's integrate joint distribution $f_{Z,X}$ over $X$
$f_Z(z) = \int\limits_{-\infty}^{\infty} f_{X,Z}(x,z) dx = \int\limits_{-\infty}^{\infty} f_X(x) f_{Z|X}(z|x) dx = \int\limits_{-\infty}^{\infty} f_X(x) f_Y(z-x)$
A: Let $(\Omega,\mathcal{F},P)$ be the probability space under consideration.
Let $t\in\mathbb{R}$ and $A=\{(x,y)\in\mathbb{R}^{2}\mid x+y\leq t\}$.
Let $\mu_{XY}$ be the distribution measure induced by the random
vector $(X,Y)$ and $\mu_{X}$, $\mu_{Y}$ the distribution measures
induced by the random variables $X$ and $Y$ respectively. (i.e.,
$\mu_{XY}(B)=P\left(\left[(X,Y)\in B\right]\right)$, $B\in\mathcal{B}(\mathbb{R}^2)$,
etc...). Since $X,Y$ are independent, we have that $\mu_{XY}=\mu_{X}\times\mu_{Y}$.
Moreover, it is assumed that $X$, $Y$ admit pdf $f_{X}$ and $f_{Y}$
respectively, so $d\mu_{X}=f_{X}(x)dx$ and $d\mu_{Y}(y)=f_{Y}(y)dy$.
Note that for any $\omega\in\Omega$, $X(\omega)+Y(\omega)\leq t$
iff $1_{A}(X,Y)(\omega)=1$. Therefore, $1_{A}(X,Y)$ is precisely the indicator function of the set $[X+Y\leq t]$. Hence
\begin{eqnarray*}
P\left([X+Y\leq t]\right) & = & \int1_{A}(X,Y)dP\\
 & = & \int1_{A}(x,y)d\mu_{XY}(x,y)\\
 & = & \int1_{A}(x,y)d(\mu_{X}\times\mu_{Y})(x,y)\\
 & = & \int\left\{ \int1_{A}(x,y)d\mu_{X}(x)\right\} d\mu_{Y}(y)\\
 & = & \int\left\{ \int1_{A}(x,y)f_{X}(x)dx\right\} f_{Y}(y)dy.
\end{eqnarray*}
In the above, we have invoked Fubini-Tonelli Theorem. Moreover, the
second equality is a result due to Doob (i.e., $\int\phi(X)dP=\int\phi(x)d\mu(x)$
for any $\mu$-integrable Borel function $\phi:\mathbb{R}^{n}\rightarrow\mathbb{R}$,
where $X$ is a $n$-dimensional random vector and $\mu$ the distribution
measure associated with $X$). For each fixed $y$, $1_{A}(x,y)=1$
iff $x+y\leq t$ iff $x\in(-\infty,t-y]$. Therefore, $\int1_{A}(x,y)f_{X}(x)dx=\int_{-\infty}^{t-y}f_{X}(x)dx=\int_{-\infty}^{t}f_{X}(x'-y)dx'$,
by considering the substitution $x'=x+y$. Hence,
\begin{eqnarray*}
P\left([X+Y\leq t]\right) & = & \int_{-\infty}^{\infty}\int_{-\infty}^{t-y}f_{X}(x)dxf_{Y}(y)dy\\
 & = & \int_{-\infty}^{\infty}\int_{-\infty}^{t}f_{X}(x-y)f_{Y}(y)dxdy.\\
 & = & \int_{-\infty}^{t}\left\{ \int_{-\infty}^{\infty}f_{X}(x-y)f_{Y}(y)dy\right\} dx\\
 & = & \int_{-\infty}^{t}g(x)dx,
\end{eqnarray*}
where $g(x)=\int_{-\infty}^{\infty}f_{X}(x-y)f_{Y}(y)dy$.
Observe that $(x,y)\mapsto f_{X}(x-y)f_{Y}(y)$ is non-negative and
measurable with respect to the product $\sigma$-algebra $\mathcal{B}(\mathbb{R})\otimes\mathcal{B}(\mathbb{R})$.
By Tonelli theorem, $g$ defined in above is measurable. Moreover,
$\int g(x)dx=\int\left\{ \int f_{X}(x-y)f_{Y}(y)dy\right\} dx=\int\left\{ \int f_{X}(x-y)f_{Y}(y)dx\right\} dy=\int\left\{ \int f_{X}(x)f_{Y}(y)dx\right\} dy=1$.
That is, $g$ is integrable.
Denote the c.d.f. of $X+Y$ by $F_{X+Y}$. That $F_{X+Y}(t)=\int_{-\infty}^{t}g(x)dx$
indicates that $F_{X+Y}$ is absolutely continuous and $F_{X+Y}'(t)=g(t)=\int_{-\infty}^{\infty}f_{X}(t-y)f_{Y}(y)dy$
(a.e.).
In another word, $X+Y$ admits a p.d.f. and the p.d.f. is $g$ defined in above.
A: Just apply the fundamental theorem of calculus and the definition for probability density function::-
$\begin{align}\dfrac{\mathrm d~~}{\mathrm d a}\int_{-\infty}^\infty h(a,y)\,\mathrm d y&= \int_{-\infty}^\infty\dfrac{\partial h(a,y)}{\partial a\qquad}\,\mathrm dy\\[2ex]\dfrac{\mathrm d~~}{\mathrm d a}\int_{-\infty}^a g(x)\,\mathrm d x &= g(a)\\[2ex]&\text{so...}\\[3ex] f_{\small X+Y}(a) &=\dfrac{\mathrm d~~}{\mathrm d a}\mathsf P(X\leq a-Y)\\[1ex]&=\dfrac{\mathrm d~~}{\mathrm d a}\int_{-\infty}^\infty\int_{-\infty}^{a-y} f_{\small X, Y}(x,y)\,\mathrm d x\,\mathrm d y\\[1ex]&~~\vdots\end{align}$


I'm also confused why the function is now a single variable in terms of a.

$X+Y$ is a single random variable.   It's probability density function would therefore be monovariate.   The token $a$ was chose by the author to be the argument of this function.
