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Consider a sample $X_1,X_2,...X_n\overset{\text{iid}}{\sim}F$. Let $T_n=F_n^{-1}(1/2)$ be the sample median where, $F^{-1}(x)=\inf\{t:F(t)\ge x\}$ and $F_n(y)=\frac{1}{n}\sum_{i=1}^n\mathbb{I}(X_i\le y)$. Assume that $N=2m-1$. Then, $T_n=X_{(m)}$.

Now, let $X^*_1,...X_n^*$ be an iid sample from $F_n$. Then, $$P_*(X_{(m)}^*=X_{(k)}|X_1,...X_n)=\sum_{j=0}^{m-1}{n\choose j}\frac{(k-1)^j(n-k+1)^{n-j}-k^j(n-k)^{n-j}}{n^n}.$$

[Ref. The Jackknife & Bootstrap (Jun Shao)]

I'm having trouble understanding how they arrived at this expression. Any hints on how to proceed?

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    $\begingroup$ To obtain $X_{(k)}$, it must be drawn a certain number of times $(1,2,..,n)$ with in total the same number of draws less than or equal to $X_{(k)}$ and more than or equal to $X_{(k)}$. $\endgroup$
    – Xi'an
    Commented Feb 11 at 10:32
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    $\begingroup$ That helped! thank you $\endgroup$
    – zaira
    Commented Feb 11 at 12:11
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    $\begingroup$ You can answer your own question, if you wish @zaira. That would be helpful for any future visitors (and you might be able to check your work too by compiling it). $\endgroup$ Commented Feb 11 at 12:21
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    $\begingroup$ See also stats.stackexchange.com/questions/533449/… $\endgroup$ Commented Feb 12 at 13:25

1 Answer 1

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Hint.

$$P_*\left(X_{(m)}^*=X_{(k)}\right)=P_*\left(X_{(m)}^*\ge X_{(k)}\right)-P_*\left(X_{(m)}^*>X_{(k)}\right).$$

To calculate $P_*\left(X_{(m)}^*\ge X_{(k)}\right),$ we observe that this only happens if $X_{(1)},\ldots,X_{(k-1)}$ occupy at most $(m-1)$ spots in the new (resampled) sample. Otherwise, it forces the sample median to be from $X_{(1)},\ldots, X_{(k-1)}$. Hence, $$P_*\left(X_{(m)}^*\ge X_{(k)}\right)=\sum_{j=0}^{m-1}\frac{{n\choose j}(k-1)^j(n-k+1)^{n-j}}{n^n},$$ where $n\choose j$ is for choosing $j$ spots in the unordered new sample of size $n$ (to fit in observations from $X_{(1)},\ldots, X_{(k-1)}$).

Similarly, for $P_*\left(X_{(m)}^*>X_{(k)}\right)$ it only happens if we $X_{(1)},\ldots,X_{(k-1)}$ and $X_{(k)}$ are limited to at most $(m-1)$ spots.

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    $\begingroup$ +1, zaira, for making a concise answer out of the hint given. $\endgroup$ Commented Feb 11 at 18:15

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