Fubini-Tonelli theorem

Question

I have the following question with me: Let $X, Y$ be independent real valued random variables and define $Z:=X+Y$. Let $F_X, F_Y$ and $F_Z$ be the respective distribution functions of $X$, $Y$ and $Z$. Show that $$ F_Z(a)=\int F_X(a-y) P^Y(d y), \quad a \in \mathbb{R} . $$ If $X$ has continuous density $f_X$, show that $$ f_Z(a):=\int f_X(a-y) P^Y(d y), \quad a \in \mathbb{R} $$ is the density of $Z$. If $X$ has continuous density $f_X$ and $Y$ has continuous density $f_Y$, show that $$ f_Z(a):=\int_{-\infty}^{\infty} f_X(a-y) f_Y(y) d y, \quad a \in \mathbb{R} $$ is the density of $Z$.

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This is what I've tried: I can use the Fubini-Tonelli theorem to calculate $F_Z(a)$. Also, integrating is easier than taking derivative.

The following is the theorem: Let $(E, \mathcal{E}, P)$ and $(F, \mathcal{F}, Q)$ be probability spaces. a) Define $R(A \times B)=P(A) Q(B)$ for $A \in \mathcal{E}, B \in \mathcal{F}$. R extends uniquely to a probability measure $P \otimes Q$ on $\mathcal{E} \times \mathcal{F}$ (it is called the product measure). b) If $f$ is measurable and integrable (or positive) w.r.t. $\mathcal{E} \otimes \mathcal{F}$, then the functions $$ \begin{aligned} x & \mapsto \int f(x, y) Q(d y), \\ y & \mapsto \int f(x, y) P(d x) \end{aligned} $$ are measurable and $$ \begin{aligned} \int f d P \otimes Q & =\int\left\{\int f(x, y) Q(d y)\right\} P(d x) \\ & =\int\left\{\int f(x, y) P(d x)\right\} Q(d y) \end{aligned} $$

But I'm not still not able to fully figure this out, and I would appreciate help. Thank you!!

Answer 1

This result is just known as "Convolution Theorem", and you can find many proofs of it online. The main idea is best understood if $X,Y:\Omega\longrightarrow \mathbb{R}$ are absolutely continuous with densities $f_X$ and $f_Y$. Then their common distribution is also absolutely continuous and given by $$F_{(X,Y)}(x,y) := P(\{\omega\in\Omega : X(\omega)\leq x, Y(\omega)\leq y\}) = \int\limits_{-\infty}^y\int\limits_{-\infty}^x f_{(X,Y)}(\tilde x, \tilde y)d \tilde x d\tilde y.$$

The rest is just calculation: \begin{align*} \int\limits_{-\infty}^t f_{X+Y}(z)dz &= F_{X+Y}(t)\\ &= P(\{\omega\in\Omega : X(\omega)+Y(\omega) \leq t\}) \\ &= P(\{\omega\in\Omega: X(\omega)\leq t-y,Y(\omega)=y\in\mathbb{R}\})\\ & = \int\limits_\mathbb{R} \int\limits_{-\infty}^{t-y} f_{(X,Y)}(x,y)dx dy \\ &= \int\limits_{\mathbb{R}}\int\limits_{-\infty}^t f_{(X,Y)}(y-z,y)dzdy \text{ (here we substituted } z = x+y)\\ &= \int\limits_{-\infty}^t \int\limits_{\mathbb{R}}f_{(X,Y)}(y-z,y)dydz \text{ (by Fubini)}\\ & = \int\limits_{-\infty}^t\left( \int\limits_\mathbb{R}f_X(y-z)\cdot f_Y(y)dy \right)dz \text{ by independence of $X$ and $Y$.} \end{align*}

And since the density of a distribution is unique, this means that $f_{X+Y}(z) = \int\limits_\mathbb{R}f_X(y-z)\cdot f_Y(y)dy $.

In the case where $X,Y$ do not have densities, the proof is entire analogous, just a bit more difficult to write, since you have to work with the induced measures instead of the densities.