PDF of `not k-th smallest value' Let $X_1,\cdots,X_n$ be i.i.d. random variables with a distribution $F$ and pdf $f$.  The $k$-th smallest value among $(X_1,\cdots,X_n)$ is an order statistic $X_{(k;n)}$. Let the pdf of $X_{(k;n)}$ be $f_{k;n}$. 
Fix an index $i$. What would be the pdf or cdf of $X_i$ given that it is not the $k$-th smallest value? Could it be written in terms of $f$ and $f_{k;n}$? Would it be
$$\frac{1}{n-1}\sum_{i\neq k}f_{i;n} ?$$
 A: Let's work out the random variables involved and their joint distribution so we can apply a formula for conditional probabilities.
The key is to note that IID implies the $X_i$ are exchangeable: the distribution $(X_1, \ldots, X_n)$ is the distribution of $(X_{\sigma(1)},\ldots,X_{\sigma(n)})$ for any permutation $\sigma:\{1,\ldots,n\}\to\{1,\ldots,n\}$.  Therefore


*

*We may without any loss of generality focus on $i=1$, because the answer we obtain for it will be the answer for all other $i=2,\ldots,n$; and

*The chance that $X_i$ has rank $k$ is the chance that each other $X_j$ has rank $k$, $j\ne i$.  Since these $n$ equal chances must sum to unity, each of them will equal $1/n$.
For convenience, write $X$ for $X_1$.
Define the random variable $R$ to be the rank of $X$. By $(2)$, $R$ is uniformly distributed on the integers $1,2,\ldots, n$.  In particular, for any $k\in\{1,2,\ldots, n\}$,
$$\Pr(R \ne k) = \sum_{i\ne k}\Pr(R=i) = (n-1)\frac{1}{n}.$$
Writing the joint distribution of $(X,R)$ is a little delicate because the marginal distribution of $X$ is continuous while the marginal distribution of $R$ is discrete.  Let's address this by breaking all events into unions over the $n$ distinct possible values of $R$.  Thus, fixing any $k\in\{1,2,\ldots, n\}$,
$$\eqalign{
\Pr(X \le x\mid R \ne k) &= \frac{\Pr(X \le x,\ R \ne k)}{\Pr(R \ne k)} 
= \frac{1}{(n-1)/n}\Pr(X \le x,\ R \ne k) \\
&= \frac{n}{n-1} \sum_{i\ne k}\Pr(X \le x,\ R = i) \\
&= \frac{n}{n-1} \sum_{i\ne k}\Pr(X \le x \mid R = i)\Pr(R = i) \\
&= \frac{n}{n-1} \sum_{i\ne k}\Pr(X \le x \mid R = i)\frac{1}{n} \\
&= \frac{1}{n-1} \sum_{i\ne k} F_{i;n}(x).
}$$
The conditional density $f_{X\mid -k}$, for $X$ given it is not of rank $R=k$, is obtained by differentiating this with respect to $x$,
$$f_{X\mid -k}(x) = \frac{\mathrm d}{\mathrm{d}x} \Pr(X \le x\mid R \ne k) = \frac{1}{n-1} \sum_{i\ne k} f_{i;n}(x).\tag{1}$$
Since exchangeability implies the mean of the order statistic densities $f_{i;n}$ is the underlying density $f$, this may also be expressed in the closed form
$$f_{X\mid -k}(x) = \frac{1}{n-1}\left(n f(x) - f_{k;n}(x)\right).\tag{2}$$
To verify this result, here are the results of a simulation with $n=7$ from an underlying standard Normal distribution.  The conditional histograms of $X_1$ for each $k$ are shown, on which are superimposed plots of formula $(2)$. The match is excellent.  They are ordered to emphasize the symmetry between the distributions conditional on $k$ and conditional on $n+1-k$: the symmetry of the standard Normal distribution implies the graphs of the corresponding densities will be mirror images.
 
For those who would like to experiment, here is the R code to produce and modify this simulation.
n <- 7       # Number of variables
n.sim <- 1e5 # Size of (unconditional) simulation
#
# Conduct the simulation.
# The result is a list `y` of draws from each conditional simulation,
# indexed by the order statistic `k` that is removed from the conditioning
# events.
#
set.seed(17)
x <- matrix(rnorm(n*n.sim), nrow=n)
r <- apply(x, 2, order)
y <- sapply(1:n, function(k) x[1, r[k,] != 1])
#
# Define the reference distributions for a standard
# normal distribution according to formula (2).
#
f <- function(x, k, n, cdf=pnorm, pdf=dnorm) {
  f.k.log <- cdf(x, log.p=TRUE)*(k-1) + 
    cdf(x, log.p=TRUE, lower.tail=FALSE) * (n-k) + 
    pdf(x, log=TRUE) +
    (lfactorial(n) - lfactorial(k-1) - lfactorial(n-k))
  return((n * pdf(x) - exp(f.k.log)) / (n-1))
}
#
# Prepare to plot the results.
#
n.breaks <- 50

n.2 <- ceiling(n/2)
k <- c(1:n.2, n+1 - 1:n.2)
b.range <- range(unlist(y))
breaks <- seq(b.range[1], b.range[2], length.out=n.breaks)
colors <- rainbow(n)
#
# Plot the results.
#
mar <- par("mar")
par(mfrow=c(2, n.2), mar=c(5.1, 0.1, 2.1, 0.1))
invisible(sapply(k, function(k) {
  s <- expression(X[1])
  m <- substitute(k != i, list(i=k))
  hist(y[[k]], freq=FALSE, breaks=breaks, yaxt="n", border="Gray",
       xlab=s, ylab="", cex.lab=1.5, main=m)
  abline(curve(f(x,k,n), add=TRUE, col=colors[k], lwd=2))
}))
par(mfrow=c(1,1), mar=mar)

