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Convolutions

Convolutions

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

\[ \begin{aligned} F_{X}^{n\*} = F_{X} * \cdots * F_{X} \text{ $n$ times}\end{aligned} \]

It has the following properties for positive random variable \(X_{i}\)s

  1. \begin{align} F_{X}^{n*}(x) \leq F_{X}^{n}(x) \end{align} This can be proven as
\[ \begin{aligned} P(X_{1} + \cdots + X_{n} \leq x) &\leq P(X_{1} \leq x, \ldots, X_{n} \leq x)\newline P(X_{1} + \cdots + X_{n} \leq x) &\leq \prod_{i=1}^{n} P(X \leq x) \text{ by independence}\newline \text{or, } F_{X}^{n\*}(x) &\leq F_{X}^{n}(x) \end{aligned} \]
  1. \begin{align} F_{X}^{n*}(x) \geq F_{X}^{n+1}(x) \end{align} which follows immediately from the fact that
\[ \begin{aligned} P(X_{1} + \cdots + X_{n} \leq x) &\geq P(X_{1} \leq x, \ldots, X_{n+1} \leq x)\newline \end{aligned} \]
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 $X_{1} + X_{2} <= 1$.