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While skipping through the class notes I noticed one exercice that I couldn't solve:

Suppose we have $\mu$ - probability distribution in $\mathbb{R}$.

Recalling that $\mu_k \rightarrow \mu$ iff $\int_{\mathbb{R}} f(x)\mu_k(dx) \rightarrow \int_{\mathbb{R}} f(x) \mu(dx)$, $\forall f \in C(\mathbb{R})$, $f$ bounded $k\rightarrow \infty$. We have to show:

  1. $\exists \mu_k$ - series of continuous distributions that $\mu_k \rightarrow \mu$.
  2. $\exists \mu_k$ - series of discrete distributions that $\mu_k \rightarrow \mu$.
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Hint: Look at the cdfs. If $\mu_k, \mu$ are probability distributions with cdfs $F_k, F$, recall $\mu_k \to \mu$ weakly iff $F_k(x) \to F(x)$ at all points $x$ where $F$ is continuous. Can you approximate $F$ in this sense by continuous cdfs? By cdfs which are step functions? –  Nate Eldredge Jun 8 '12 at 14:31
    
We didn't actually had the weak convergence defined in our classes, so I guess the only thing we can use here is that $\mu_k \rightarrow \mu$ iff $\int_{\mathbb{R}} f(x)\mu_k(dx) \rightarrow \int_{\mathbb{R}} f(x) \mu(dx)$, $\forall f \in C(\mathbb{R})$, $k\rightarrow \infty$. –  user974514 Jun 9 '12 at 3:42
    
Which properties of weak convergence do you know? –  Davide Giraudo Jun 9 '12 at 8:30
    
The only thing we know is the fact that was mentioned above. –  user974514 Jun 9 '12 at 11:35
    
Your definition should say "...for all bounded, continuous functions $f$..." –  Byron Schmuland Jun 13 '12 at 22:17
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1 Answer

up vote 0 down vote accepted

Hints:

  • Every distribution is a barycenter $p\mu_d+(1-p)\mu_c$ where $0\leqslant p\leqslant1$, $\mu_d$ is discrete and $\mu_c$ is continuous.
  • As a consequence, it suffices to show that every discrete distribution is a limit of continuous ones, and vive versa.
  • In one direction, show that, for every $x$, the (continuous) uniform distribution on the interval $[x-\varepsilon,x+\varepsilon]$ converges to the (discrete) Dirac distribution at $x$ when $\varepsilon\to0^+$.
  • In the other direction, show that, for every continuous distribution $\mu_c$, the (discrete) distribution $\sum\limits_n\mu_c([n\varepsilon,(n+1)\varepsilon))\,\delta_{n\varepsilon}$, where the sum runs over every integer $n$, converges to $\mu_c$ when $\varepsilon\to0^+$.
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