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I have the following problem. Let $f: \mathbb{R} \to ]0,+ \infty[$ continuous, integrable function and $X_n$ a sequence of real random variables. If $\int_{\mathbb{R}}|F_n(x)-F(x)|f(x) dx \to 0$ then $X_n \to X$ in distribution.

I have proved by paradox. Indeed exists $x_0$ (point of continuity of $F$), $\varepsilon_0>0$ and $F_{n_k}$ such that $|F_{n_k}(x_0)-F(x_0|> \varepsilon_0 \quad \forall k$.

Than this produce an absurd since $0< \int_{\mathbb{R}}|F_{n_k}(x)-F(x)|f(x) dx \to 0$.

I would like to know if there is a another way to prove this fact.

Thanks

  • 0
    In your proof, why is the "absurd" statement absurd? It can very well be positive and tending to 0..2012-12-20
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    I'm starting to doubt this statement. If you choose $X_n$ to be uniform in, say $[0, 1/n]$ and let $X$ be identically $0$, then your statement can hold for a suitable $f$ and still not yield convergence in distribution of $X_n$ to $X$.2012-12-20
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    I don't get your point, if you want I can expand my proof; in you example $X_n \to 0$ in distribution.2012-12-20
  • 0
    Ah yes, sorry. It does indeed.2012-12-20

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