If $X$ is a random variable, then Chebyshev's inequality states that $$\Pr(|X-E(X)|\ge t)\le\frac{Var(X)}{t^2}.$$ If the "spread"/variance of the distribution $X$ is drawn from is very large, then for small $t$ this inequality gives no useful information. My question is if we get several draws from $X$, can we say something a bit stronger, something like this:
If $X_1,\ldots, X_N, X_{N+1}$ are iid, then can we bound $\Pr(\min_{1\le i\le N} |X_i - X_{N+1}| \ge t)?$ It seems like we should be able to say that this would be small with much higher confidence than in the vanilla Chebyshev, since the $N$ samples we draw before comparing the distance to $X_{N+1}$ intuitively seems to capture the spread of the distribution, and would potentially yield a better concentration than comparing to the mean. Any help is appreciated!
Since $X_{N+1}$ is independent of the vector $\left(X_1,\dots,X_N\right)$, it follows (from Fubini's theorem) that $$ \Pr\left(\min_{1\le i\le N} |X_i - X_{N+1}| \ge t\right)=\int_{\mathbb R}\Pr\left(\min_{1\le i\le N} |X_i - x| \ge t\right)\mathrm{d}{\Pr}_{X_{N+1}}(x). $$ Using independence of the vector $\left(X_1,\dots,X_N\right)$ and then the fact that all the components have the same distribution, we get that $$ \Pr\left(\min_{1\le i\le N} |X_i - x| \ge t\right)=\prod_{i=1}^n\Pr\left(\left\lvert X_i-x \right\rvert \geq t\right)=\Pr\left(\left\lvert X_1-x \right\rvert \geq t\right)^N $$ hence $$ \Pr\left(\min_{1\le i\le N} |X_i - X_{N+1}| \ge t\right)=\int_{\mathbb R}\Pr\left(\left\lvert X_1-x \right\rvert \geq t\right)^N\mathrm{d}{\Pr}_{X_{N+1}}(x). $$