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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.

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

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