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Let $D_\theta$ denote a distribution with parameter $\theta$, that has density wrt. Lebesgue measure. Let $X$ have distribution $D_\theta$ and let $\hat{\theta}_n$ be a sequence of estimates of $\theta$, converging to $\theta$ in probability (or almost surely if that is required).

Let $Q_\theta(p)$ denote the quantile function of $D_\theta$. Do we have

$$ P(X \in [Q_{\hat{\theta}_n}(p_1), Q_{\hat{\theta}_n}(p_2)]) \to P(X \in [Q_\theta(p_1), Q_\theta(p_2)]) $$

for every $p_1 \leq p_2$?

If not, what further assumptions do we need? Continuity of $Q$ in $\theta$ seems like it might do the trick but can we infer that from simply knowing that $D_\theta$ is absolutely continuous wrt. Lebesgue measure? Any ideas?

Lundborg
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1 Answers1

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Continuity will work, since the cumulative distribution function $F$ associated to $X$ is absolutely continuous, and therefore if $Q_{\hat{\theta}_n}(p_1) \to Q_\theta(p_1)$ and $Q_{\hat{\theta}_n}(p_2) \to Q_\theta(p_2)$, it follows that

$$ F(Q_{\hat{\theta}_n}(p_2)) - F(Q_{\hat{\theta}_n}(p_1)) \to F(Q_{\hat{\theta}}(p_2)) - F(Q_{\hat{\theta}}(p_1)) $$ which is what you are looking for. However, the fact that $D_{\theta}$ is absolutely continuous does not imply this, since this says nothing about the regularity of $\theta \mapsto D_{\theta}$. For example, we might define a family of densities as $$ f_{\theta}(x) = \begin{cases} \mathbb{1}_{[0,1]}(x), \quad &\theta = 0,\\ \frac{1}{|\theta|}\mathbb{1}_{[0, |\theta|]}(x), \quad &|\theta| > 0. \end{cases}$$ Then we have $Q_{\theta}(1/2) = |\theta|/2 \to 0$ as $\theta \to 0$, but $Q_{0}(1/2) = 1/2$.

A sufficient condition would be to require that $\theta \to D_{\theta}$ is continuous wrt convergence in distribution, as this implies convergence of the quantile functions in continuity points (for a proof see here) along with $F$ being strictly increasing, as this implies that $Q$ is continuous.

user159517
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