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In general, one can show that if $X_1, \ldots, X_n \text{ i.i.d. } \sim F$, then: $$\sqrt{n}(X_{(np)} - p)\overset{d}\rightarrow \mathrm{N}\left(0, \frac{p(1 - p)}{f^{2}(\xi_p)}\right)$$ where $f$ is the density of $F$. This result is classical, and can be proven in many ways, for one possible reference, see Example 2.4.9 in Lehmann's book: Elements of large-sample theory. In your case, based on this result, you can show that: $$\sqrt{n}(X_{(n/3)} - 1/3)\overset{d}\rightarrow \mathrm{N}\left(0, \frac{2}{9}\right)$$. The above statement is about convergence in distribution, using Slutsky's theorem, your educated case can be confirmed --- but you have to point out the convergence mode: it converges in probability. It is incorrect to just say "it converges to 1/3". To show it converges in probability to 1/3, it is more straightforward, you can just show $\mathrm{Var}(X_{(n/3)}) \rightarrow 0$ as $n \to \infty$.

In general, one can show that if $X_1, \ldots, X_n \text{ i.i.d. } \sim F$, then: $$\sqrt{n}(X_{(np)} - p)\overset{d}\rightarrow \mathrm{N}\left(0, \frac{p(1 - p)}{f^{2}(\xi_p)}\right)$$ where $f$ is the density of $F$. This result is classical, and can be proven in many ways, for one possible reference, see Example 2.4.9 in Lehmann's book: Elements of large-sample theory. In your case, based on this result, you can show that: $$\sqrt{n}(X_{(n/3)} - 1/3)\overset{d}\rightarrow \mathrm{N}\left(0, \frac{2}{9}\right).$$ The above statement is about convergence in distribution, using Slutsky's theorem, your educated case can be confirmed --- but you have to point out the convergence mode: it converges in probability. It is incorrect to just say "it converges to 1/3". To show it converges in probability to 1/3, it is more straightforward, you can just show $\mathrm{Var}(X_{(n/3)}) \rightarrow 0$ as $n \to \infty$.

In general, one can show that if $X_1, \ldots, X_n \text{ i.i.d. } \sim F$, then: $$\sqrt{n}(X_{(np)} - p)\overset{d}\rightarrow \mathrm{N}\left(0, \frac{p(1 - p)}{f(\xi_p)}\right)$$ where $f$ is the density of $F$. This result is classical, and can be proven in many ways, for one possible reference, see Example 2.4.9 in Lehmann's book: Elements of large-sample theory. In your case, based on this result, you can show that: $$\sqrt{n}(X_{(n/3)} - 1/3)\overset{d}\rightarrow \mathrm{N}\left(0, \frac{2}{9}\right)$$. The above statement is about convergence in distribution, using Slutsky's theorem, your educated case can be confirmed --- but you have to point out the convergence mode: it converges in probability. It is incorrect to just say "it converges to 1/3". To show it converges in probability to 1/3, it is more straightforward, you can just show $\mathrm{Var}(X_{(n/3)}) \rightarrow 0$ as $n \to \infty$.

In general, one can show that if $X_1, \ldots, X_n \text{ i.i.d. } \sim F$, then: $$\sqrt{n}(X_{(np)} - p)\overset{d}\rightarrow \mathrm{N}\left(0, \frac{p(1 - p)}{f^{2}(\xi_p)}\right)$$ where $f$ is the density of $F$. This result is classical, and can be proven in many ways, for one possible reference, see Example 2.4.9 in Lehmann's book: Elements of large-sample theory. In your case, based on this result, you can show that: $$\sqrt{n}(X_{(n/3)} - 1/3)\overset{d}\rightarrow \mathrm{N}\left(0, \frac{2}{9}\right)$$. The above statement is about convergence in distribution, using Slutsky's theorem, your educated case can be confirmed --- but you have to point out the convergence mode: it converges in probability. It is incorrect to just say "it converges to 1/3". To show it converges in probability to 1/3, it is more straightforward, you can just show $\mathrm{Var}(X_{(n/3)}) \rightarrow 0$ as $n \to \infty$.

In general, one can show that if $X_1, \ldots, X_n \text{ i.i.d. } \sim F$, then: $$\sqrt{n}(X_{(np)} - p)\overset{d}\rightarrow \mathrm{N}\left(0, \frac{p(1 - p)}{f(\xi_p)}\right)$$ where $f$ is the density of $F$. This result is classical, and can be proved in many ways, for one possible reference, see Example 2.4.9 in Lehmann's book: \emph{Elements of large-sample theory}. In your case, based on this result, you can show that: $$\sqrt{n}(X_{(n/3)} - 1/3)\overset{d}\rightarrow \mathrm{N}\left(0, \frac{2}{9}\right)$$. The above statement is about convergence in distribution, using Slutsky's theorem, your educated case can be confirmed --- but you have to point out the convergence mode: it converges in probability. It is incorrect to just say "it converges to 1/3". To show it converges in probability to 1/3, it is more straightforward, you can just show $\mathrm{Var}(X_{n/3)}) \rightarrow 0$ as $n \to \infty$.

In general, one can show that if $X_1, \ldots, X_n \text{ i.i.d. } \sim F$, then: $$\sqrt{n}(X_{(np)} - p)\overset{d}\rightarrow \mathrm{N}\left(0, \frac{p(1 - p)}{f(\xi_p)}\right)$$ where $f$ is the density of $F$. This result is classical, and can be proven in many ways, for one possible reference, see Example 2.4.9 in Lehmann's book: Elements of large-sample theory. In your case, based on this result, you can show that: $$\sqrt{n}(X_{(n/3)} - 1/3)\overset{d}\rightarrow \mathrm{N}\left(0, \frac{2}{9}\right)$$. The above statement is about convergence in distribution, using Slutsky's theorem, your educated case can be confirmed --- but you have to point out the convergence mode: it converges in probability. It is incorrect to just say "it converges to 1/3". To show it converges in probability to 1/3, it is more straightforward, you can just show $\mathrm{Var}(X_{(n/3)}) \rightarrow 0$ as $n \to \infty$.