Understanding the behaviour of order statistics of samples of uniform distribution Let $P$ denote any continuous distribution with density $p$ on $[0,1]$ and $Q$ the uniform distribution on $[0,1]$ whose density is $1$. 

Let $X_1,\ldots,X_n$ be $n$ i.i.d samples drawn from the distribution $Q$ and $X_{(1)},\ldots,X_{(n)}$ be their order statistics and let $Y$ be a random variable distributed acccording to $P$. Define weights as:
$$
w_i=\frac{p(X_{(i)})}{q(X_{(i)})}= p(X_{(i)}), 1 \leq i \leq n
$$
and 
$$
\tilde{w}_i = \mathbb{P}(Y \in [X_{(i)},X_{(i+1)}))= F_{P}(X_{(i+1)})-F_P(X_{(i)}).
$$

I am trying to understand if there exists any notion of similarity or convergence between the random variables $w_i$ and $\tilde{w}_i$ as $n \to \infty$.

Any help towards this would be appreciated. 

The reason why I encountered this is because of the observation that if I use a Gaussian Kernel Density estimator with weights $w_i$ and $\tilde{w}_i$ respectively, both the weights are able to fit the pdf $p(x)$ perfectly across a range of distributions $P$ and $Q$. I want to know if this is a manifestation of some underlying fact about the above defined weights. I am attaching the plots for completeness:
In short, the two weights sequences you define are asymptotically equivalent in probability. I put a sketch of proof, see below.

But I'd recommend to have a closer look at Chapter 7 of [David H.A., Nagaraja H.N. Order Statistics (2003)]. I believe the topics discussed there are pretty close to the current one.

Proposition. $\forall\rho&gt;0\;\;\mathbb{P}(|w_i-\tilde{w}_i|&gt;\rho)\to 0, \;n \to \infty$.

Proof. 

a) A basic fact of order statistics is that $X_{(i)}$ converges in probability to $\frac{(i-1)/n\;+\; i/n}{2}$ as $n\to\infty$. 

b) If we denote $\delta_{i+1}=X_{(i+1)}-X_{(i)}$, then it's possible to get $\delta_{i+1}=\textit{o}(1/n)$ and

$$ F_{P}(X_{(i+1)}) = F_P(X_{(i)}) + p(X_{(i)})\cdot\delta_{i+1} + \textit{o}(1/n^2)
$$
using the Taylor approximation; the notation here is $p(\bullet)\equiv F^{\prime}_P(\bullet)$.

c) Finally, 

$\mathbb{P}(|w_i-\tilde{w}_i|&gt;\rho) = \mathbb{P}(|p(X_{(i)}) - \left(F_{P}(X_{(i+1)}) - F_P(X_{(i)})\right) |&gt;\rho) = \mathbb{P}(|p(X_{(i)})\cdot\left(1-\delta_{i+1} \right) + \textit{o}(1/n^2)|&gt;\rho).
$ 

(We just used b) and then a).) As we see, the left-hand side of the inequality is of order $\textit{o}(1/n)$ (in probability, see a)). Thus, we can choose appropriate (big enough) $n^\ast$ given any positive $\rho$, which means that the probability of the event $\left\{\ldots &gt;\rho\right\}$ tends to zero. This concludes the proof.
In short, the two weights sequences you define are asymptotically equivalent in probability. I put a sketch of proof, see below.

But I'd recommend to have a closer look at Chapter 7 of [David H.A., Nagaraja H.N. Order Statistics (2003)]. I believe the topics discussed there are pretty close to the current one.

Proposition. $\forall\rho&gt;0\;\;\mathbb{P}(|w_i-\tilde{w}_i|&gt;\rho)\to 0, \;n \to \infty$.

Proof. 

a) A basic fact of order statistics is that $X_{(i)}$ converges in probability to $\frac{(i-1)/n\;+\; i/n}{2}$ as $n\to\infty$. 

b) If we denote $\delta_{i+1}=X_{(i+1)}-X_{(i)}$, then it's possible to get $\delta_{i+1}=\textit{o}(1/n)$ and

$$ F_{P}(X_{(i+1)}) = F_P(X_{(i)}) + p(X_{(i)})\cdot\delta_{i+1} + \textit{o}(1/n^2)
$$
using the Taylor approximation; the notation here is $p(\bullet)\equiv F^{\prime}_P(\bullet)$.

c) Finally, 

$\mathbb{P}(|w_i-\tilde{w}_i|&gt;\rho) = \mathbb{P}(|p(X_{(i)}) - \left(F_{P}(X_{(i+1)}) - F_P(X_{(i)})\right) |&gt;\rho) = \mathbb{P}(|p(X_{(i)})\cdot\left(1-\delta_{i+1} \right) + \textit{o}(1/n^2)|&gt;\rho).
$ 

(We just used b) and then a).) As we see, the left-hand side of the inequality is of order $\textit{o}(1/n)$ (in probability, see a)). Thus, we can choose appropriate (big enough) $n^\ast$ given any positive $\rho$, which means that the probability of the event $\left\{\ldots &gt;\rho\right\}$ tends to zero. This concludes the proof.