I am trying to read Norris' book on Markov chains. In particular, at page 160 he states the central limit theorem :
Theorem 4.4.1 (Central limit theorem). Let $X_1, X_2,…$ be a sequence of independent and identically distributed real-valued random variables with mean 0 and variance $t ∈ (0,∞)$. Then, for all bounded continuous functions $f$, as $n → ∞$ we have
$\mathbb{E}[f(X_1+\ldots+X_n)/\sqrt{n}]\to \int_{\mathbb{R}}f(x)\phi_t(x)dx$
I thought until now that the central limit theorem was
Let $X_1, X_2,…$ be a sequence of independent and identically distributed real-valued random variables with mean 0 and variance $t ∈ (0,∞)$. Then, for all bounded continuous functions $f$, as $n → ∞$ we have
$(X_1+\ldots X_n)/\sqrt{n}\xrightarrow{d}\mathcal{N}(0,t)$, where "$\xrightarrow{d}$" denotes convergence in distribution.
I don't really understand if these are equivalent. In the first theorem, $f$ has to be bounded so we cannot take the identity $f(x)=x$ which is not bounded right?
And I don't understand how to see that these formulation are equivalent…
Best Answer
$Y_n \to Y$ in distribution if and only if $Ef(Y_n) \to Ef(Y)$ for every bounded continuous function $f$. [ For more information see https://en.wikipedia.org/wiki/Convergence_of_random_variables ]
If $Y$ has a density $\phi$ then we can write $Ef(Y)$ as $\int f(x)\phi (x) dx$. In our case the density is $\phi_t$ defined by $\phi_t(x)=\frac 1 {t\sqrt {2 \pi}}e^{-x^{2}/2t}$.