# Convolution of measures and stochastic dominance

Consider two probability measures on $$\mathbb{R}$$, namely $$\mu$$ and $$\nu$$. Suppose that $$\mu$$ stochastically dominates $$\nu$$ so that $$\mu \ge \nu$$. Here I am considering the first-order stochastic dominance, i.e. $$\mu \ge \nu \iff \int_{\mathbb{R}} f(x) \: \mathrm{d}\mu(x) \ge \int_{\mathbb{R}} f(x) \: \mathrm{d}\nu(x)$$ for all non decreasing measurable functions $$f$$.

My question is: does the convolution between measures preserve this ordering? In particular, is the following statement true? $$\mu \ge \nu \implies \mu^{*n} \ge \nu^{*n}$$ where $$\mu^{*n}=\underbrace{\mu*\mu*\dots*\mu}_{n \text{-times}}$$.

I think this is trivial but I don't see how to prove it rigorously.

• Instead of convolution do you actually mean product measure ? Sep 5 '19 at 9:44
• Nope, I do mean convolution of measures. Sep 5 '19 at 14:03

Let $$f$$ is a non decreasing measurable functions, then
\begin{align} &\int f(z)\,(\mu\ast\mu)(\mathrm{d}z)=\mathsf{E}[f(X+Y)]\quad \text{where X,Y are i.i.d. and X\overset{d}{=}\mu}\\ &\quad=\int \Biggl[\int f(x+y)\,\mu(\mathrm{d}x)\Biggr]\mu(\mathrm{d}y) \qquad \text{For fixed y , f(x+y) is non decreasing in x }\\ &\quad\ge \int \Biggl[\int f(x+y)\,\nu(\mathrm{d}x)\Biggr]\mu(\mathrm{d}y) \qquad\text{ g(y)=\int f(x+y)\,\nu(\mathrm{d}x) is non decreasing in y }\\ &\quad\ge \int \Biggl[\int f(x+y)\,\nu(\mathrm{d}x)\Biggr]\nu(\mathrm{d}y)\\ &\quad=\int f(z)\,(\nu\ast\nu)(\mathrm{d}z) \end{align}