Convolution operations are defined for both CDF and PDF/PMFs. Let $X$ and $Y$ be random independent variables, then \begin{alignat}{2} F_{X+Y}(x) &= F_{X} * F_{Y} &&= \int_{\mathbb{R}} F_{X}(x-y) dF_{Y}(y)\newline p_{X+Y}(x) &= p_{X} * p_{Y} &&= \int_{\mathbb{R}} p_{X}(x-y) p_{Y}(y) dy \end{align}
We can extend the idea to $n$ independent variables as \begin{align} F_{X}^{n*} = F_{X} * \cdots * F_{X} \text{ $n$ times}\end{align} It has the following properties for positive random variable $X_{i}$s
\begin{align} F_{X}^{n*}(x) \leq F_{X}^{n}(x) \end{align} This can be proven as \begin{align} P(X_{1} + \cdots + X_{n} \leq x) &\leq P(X_{1} \leq x, \ldots, X_{n} \leq x)\newline P(X_{1} + \cdots + X_{n} \leq x) &\leq \prod_{i=1}^{n} P(X \leq x) \text{ by independence}\newline \text{or, } F_{X}^{n*}(x) &\leq F_{X}^{n}(x) \end{align}
\begin{align} F_{X}^{n*}(x) \geq F_{X}^{n+1}(x) \end{align} which follows immediately from the fact that \begin{align} P(X_{1} + \cdots + X_{n} \leq x) &\geq P(X_{1} \leq x, \ldots, X_{n+1} \leq x)\newline \end{align} since the volume of the regions denoting the sums will be lower in the higer dimensions. This can be quickly verified by considering $X_{1} \leq 1$ and $X_{1} + X_{2} <= 1$.