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I want to prove $[X|y] \rightarrow \delta_0(X)$, as $y$ goes to $\infty$, i.e. the distribution of $[X|y]$ converges to degenerate distribution at zero for large enough $y$. I know the density $f(x|y)$ upto a multiplicative constant and also the fact that $$ \frac{f(x|y)}{f(0|y)} \rightarrow 0 \quad \text{as } y \rightarrow \infty \quad \forall x \in \mathcal{D}(x)\setminus \{0\} $$

Using this fact, how do I prove $[X|y] \rightarrow \delta_0(X)$? (Or provide a counter-example).

VitalStatistix
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    Could you clarify one thing? If you "know a density...up to a multiplicative constant," then you know the density itself because it must integrate to unity. Why don't you then just evaluate the limit directly? – whuber Jul 18 '11 at 13:06
  • Hi Whuber ... I can't evaluate the limit directly because it's difficult to integrate this density and obtain the constant analytically. Numerically I have studied the behavior (by applying sampling techniques that don't require the constant) and it's indeed going to degenerate distribution at zero. Also, I have found a counter example - of distribution that satisfies the ratio property but is not asymptotically degenerate. I think I need a stronger condition than this ... – VitalStatistix Jul 18 '11 at 13:36

1 Answers1

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This is a counter example. Take $f(x|y) \propto 1_{[-1/y,1/y]}(x) + 1_{[y,y+1]}(x)$. Then
$$\frac{f(x|y)}{f(0|y)} = 1_{[-1/y,1/y]}(x) + 1_{[y,y+1]}(x) \to 0$$ for $y \to \infty$ when $x \neq 0$. However, this sequence of distributions does not converge weakly.

NRH
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  • Thanks .. I need a stronger result for proving the convergence. I took the ratio of two integrals - $\frac{\int_{\epsilon}^{1} f(x|y) dx}{\int_0^{\epsilon} f(x|y) dx} \rightarrow 0 $ as $ y \rightarrow 0$. From this, it is easy to show the convergence. Thanks for the counter-example. – VitalStatistix Jul 18 '11 at 13:44