Expectation with order statistics Let $X_1,\ldots,X_n$ be non-negative i.i.d. random variables with cdf $F$ and let $X_{(n)}$ denote their maximum. For simplicity suppose that $X_1$ has a density $f$ and that $0\le X_1\le M$ almost surely.
Let $h: \mathbb R^n\to \mathbb R_{+}$ be some function that is strictly increasing in all components.
I am trying to calculate the conditional expectation
$$E(h(X_1,\ldots,X_n) | X_{(n)} = x)\tag{1}$$
but I get stuck on the density.
I guess we should have something like
$$E(h(X_1,\ldots,X_n) | X_{(n)} = x) = \int_0^M\ldots\int_0^M h(x_1,\ldots,x_n) \psi(x_1,\ldots,x_n, x) / (nF^{n-1}(x)f(x))dx_1\ldots dx_n,$$
where $nF^{n-1}(\cdot)f(\cdot)$ is the density of $X_{(n)}$ and the joint density $\psi$ should be zero outside the rectangle $[0,x]^n.$
I tried to get $\psi$ by differentiating the joint cdf with respect to $x_1,\ldots,x_n,x$ but this does not appear to work: Note that if $x_1,\ldots,x_n\le x$ we get that
$$P(X_1\le x_1,\ldots X_n\le x_n, X_{(n)} \le x) = P(X_1\le x_1,\ldots X_n\le x_n) = F(x_1)\ldots F(x_n)$$
and if $x_j > x$ for some $j$ we get 
$$P(X_1\le x_1,\ldots X_n\le x_n, X_{(n)} \le x) = 0.$$
But then differentiating with respect to $x$ yields zero in any case, and $\psi=0$ everywhere cannot be correct.
Note that if $X_1$ were discrete we could simply write
$$E(h(X_1,\ldots,X_n) | X_{(n)} = x) \\
= \sum_{x_1,\ldots,x_n:x_{(n)} = x} h(x_1,\ldots,x_n) Pr(X_1=x_1,\ldots,X_n=x_n, X_{(n)}=x)/Pr(X_{(n)}=x)\\
= \sum_{x_1,\ldots,x_n:x_{(n)} = x} h(x_1,\ldots,x_n) Pr(X_1=x_1,\ldots,X_n=x_n)/Pr(X_{(n)}=x)\\
= \sum_{x_1,\ldots,x_n:x_{(n)} = x} h(x_1,\ldots,x_n) Pr(X_1=x_1)\ldots Pr(X_n=x_n)/Pr(X_1=x)^n$$
but I'm not sure how to generalize that intuition to the continuous case because the set $\{x_1,\ldots,x_n:x_{(n)} = x\}$ has probability zero.
In addition to the properties mentioned above I know that $h$ is symmetric in all components, that is $h(x_1,\ldots,x_n) = h(x_{\pi(1)},\ldots,x_{\pi(n)})$ for all permutations $\pi$ of $n$ letters. But I'm not sure if this is helpful for this calculation.
 A: We have :
\begin{align}
\mathbb E&\left[h(X_1,\ldots,X_n)\mathbf{1}_{X_{(n)} \le x}\right] \\
&= \mathbb E\left[h(X_1,\ldots,X_n)\mathbf{1}_{X_{(n)} \le x}\sum_{k=1}^n \mathbf{1}_{X_k = X_{(n)}}\right]\\
&= \sum_{k=1}^n \mathbb E \left[h(X_1,\ldots,X_n)\mathbf{1}_{X_{(n)} \le x}\mathbf 1_{X_k = X_{(n)}}\right]\\
&= \sum_{k=1}^n \mathbb E\left[h(X_1,\ldots,X_n)\mathbf{1}_{X_k \le x}\prod_{j\neq k}\mathbf{1}_{X_j \le X_k}\right]\\
&= \sum_{k=1}^n \int_{\left[0,M\right]^n} h(x_1,\ldots, x_n)\mathbf{1}_{x_k \le x}\prod_{j\neq k}\mathbf{1}_{x_j \le x_k} \prod_{j=1}^n f(x_j) \prod_{j=1}^n \mathrm dx_j\\
&= \sum_{k=1}^n \int_{0}^x \left\{\int_{[0, x_k]^{n-1}} h(x_1,\ldots,x_{k-1}, x_k, x_{k+1}, \ldots, x_n) \prod_{j\neq k} f(x_j) \prod_{j\neq k} \mathrm dx_j\right\} f\left(x_k\right)\mathrm dx_k\\
&= \sum_{k=1}^n \int_{0}^x \left\{\int_{[0, u]^{n-1}} h(u,x_{1}, \ldots, x_{n-1}) \prod_{j=1}^{n-1} f(x_j) \prod_{j=1}^{n-1}\mathrm dx_j\right\}f(u)\mathrm du\\
&= n\int_{0}^x \left\{\int_{[0, u]^{n-1}} h(u,x_{1}, \ldots, x_{n-1}) \prod_{j=1}^{n-1} f(x_j) \prod_{j=1}^{n-1}\mathrm dx_j\right\} f(u)\mathrm du
\end{align}
Let $$g(x) = \mathbb E\left[h(X_1,\ldots,X_n)\Big|X_{(n)} = x\right]$$ and since :
\begin{align}
\mathbb E\left[h(X_1,\ldots,X_n)\mathbf{1}_{X_{(n)}\le x}\right] &= \mathbb E\left[\mathbb E \left[h(X_1,\ldots,X_n)\mathbf{1}_{X_{(n)}\le x}\Big | \sigma \left(X_{(n)}\right)\right]\right]\\
&= \mathbb E\left[\mathbb E \left[h(X_1,\ldots,X_n)\Big | \sigma \left(X_{(n)}\right)\right] \mathbf{1}_{X_{(n)}\le x}\right]\\
&= \mathbb E\left[g\left(X_{(n)}\right) \mathbf{1}_{X_{(n)}\le x}\right]\\
&= \int_{0}^x g(u) f_{X_{(n)}}(u)\mathrm d u
\end{align}
this proves that :
 $$\frac{\mathrm d}{\mathrm dx} \mathbb E\left[h(X_1,\ldots,X_n)\mathbf{1}_{X_{(n)}\le x}\right] = g(x) f_{X_{(n)}}(x) = ng(x) f(x) \left\{F(x)\right\}^{n-1}$$
By differentiating :
\begin{align}
\mathbb E\left[h(X_1,\ldots,X_n) \Big | X_{(n)} = x\right] &= \frac{n f(x)\left\{\displaystyle\int_{[0, x]^{n-1}} h(x,x_{1}, \ldots, x_{n-1}) \prod_{j=1}^{n-1} f(x_j) \prod_{j=1}^{n-1}\mathrm dx_j\right\}}{n f(x) \left\{F(x)\right\}^{n-1}}\\
&= \frac1{\left\{F(x)\right\}^{n-1}}\left\{\displaystyle\int_{[0, x]^{n-1}} h(x,x_{1}, \ldots, x_{n-1}) \prod_{j=1}^{n-1} f(x_j) \prod_{j=1}^{n-1}\mathrm dx_j\right\}
\end{align}
