Let $X$ and $Y$ be two independent random variable. Suppose $X$ is a continuous random variable (or equivalently $X$ has density function). Then $X+Y$ also is a continuous random variable.
I have already figured out a solution as follow:
$\forall A\in \mathscr{B}(\mathbb{R})$ with $\int_A dx=0$, we have $\mu_{X+Y}(A)=\mathbb{P}(X+Y\in A)=\iint\limits_{\{(x,y);x+y\in A\}}\mu_{(X,Y)}(dx,dy)=\iint\limits_{\{(x,y);x+y\in A\}}\mu_X(dx)\mu_Y(dy)=\int_\mathbb{R}\left(\int_{A-y}\mu_X(dx)\right)\mu_Y(dy)$.
Notice $\int_{A-y}=\int_Adx=0,$ we get $\mu_{X+Y}(A)=0,$ which implies that $\mu_{X+Y}(dx)$ is absolutely continuous with respect to $dx$.
Now my teacher asks me to find another way to prove this proposition by using characteristic function. My idea is to use the conclusion that if $X$ and $Y$ are independent, then $f_{X+Y}(t)=f_X(t)f_Y(t)$. However, I do not how to go further.
Any hint or solution is highly appreciated!
Best Answer
Well I think I've worked out a solution which in some sense computes the density function of $X+Y$ explicitly.
Solution:Lets denote $\xi=X+Y$. Suppose the characteristic function and distribution function of $X+Y$ is $f(t)$ and $F(x)$ respectively.
Here we need to use inversion formula of characteristic function, that is $F(x_2)-F(x_1)= \lim\limits_{T\to\infty}{\frac{1}{2\pi}}\int_{-T}^T\frac{e^{-itx_1}-e^{-itx_2}}{it}f(t)dt \\=\lim\limits_{T\to\infty}{\frac{1}{2\pi}}\int_{-T}^T\frac{e^{-itx_1}-e^{-itx_2}}{it}\iint_{\mathbb{R}^2}e^{i(x+y)t}p(x,y)dxdydt \\=\lim\limits_{T\to\infty}{\frac{1}{2\pi}}\int_{-T}^T\frac{e^{-itx_1}-e^{-itx_2}}{it}\iint_{\mathbb{R}^2}e^{i(x+y)t}p_X(x)dxdF_Y(y)dt \\=\lim\limits_{T\to\infty}{\frac{1}{2\pi}}\iint_{\mathbb{R}^2}e^{i(x+y)t}p_X(x)dxdF_Y(y)\int_{-T}^T\frac{e^{-itx_1}-e^{-itx_2}}{it}dt \\=\lim\limits_{T\to\infty}{\frac{1}{2\pi}}\iint_{\mathbb{R}^2}p_X(x)dxdF_Y(y)\int_{-T}^T\frac{e^{i(x+y-x_1)t}-e^{i(x+y-x_2)t}}{it}dt \\=\iint_{x_1\leq x+y\leq x_2}p_X(t-y)F_Y(y)dt.$
Let $x_1\to -\infty$, then we get
$F(x_2)=\iint_{x+y\leq x_2}p(x)dF_Y(y)dx\\=\int_{-\infty}^\infty d F_Y(y)\int_{-\infty}^{x_2-y}p(x)dx \\=\int_{-\infty}^\infty d{F_Y(y)}\int_{-\infty}^{X_2}p_X(t-y)dt\\=\int_{-\infty}^{x_2}\int_{-\infty}^\infty p_X(t-y)dF_Y(y)dt$
By definition, the density function of $X+Y$ is $\int_{-\infty}^\infty p_X(t-y)dF_Y(y)$