Conditioning on independent random variables I am in a situation where I have to compute:
$$E(u(x_1)|\bar{X},S^2)$$
where $X_1$ is a normally distributed random variable and $u(.)$ some function. I know that by the student's theorem the sample mean and the sample variance are independent and moreover that $\frac{(n-1)S^2}{\sigma^2}\sim \chi^2 (n-1)$. 
Can I simplify the expectation with the information I possess? Is perhaps the bivariate normal distribution of use here?
Thanks.
EDIT: Yes, an iid sample on $X$ is assumed here, hence the subscript $1$ on $u(x_1)$. If the expectation cannot be simplified, what is the conditional distribution of $X_1$ given $\bar{X}$ and $S^2$?
 A: Given the information in the question, we have a sample of $n$ i.i.d normals (which also means that their joint distribution is also normal), each following $N(\mu, \sigma^2)$. Therefore $\bar X \sim N(\mu, \sigma^2/n)$.
One can easily calculate that $\operatorname{Cov}(X_1, \bar X) = \sigma^2/n$. So their correlation coefficient is $\rho = 1/\sqrt{n}$. Then, since their joint distribution is bi-variate normal, the conditional expected value of $X_1\mid \bar X$ is
$$E(X_1\mid \bar X) = \mu + \rho \frac{\sigma}{\sigma/\sqrt n}(\bar X - \mu)= \bar X$$.
...as should be expected
Now the reason I mentioned in the comments a 2nd-order Taylor expansion, is the completely arbitrary nature of $u(\cdot)$, which does not really permit us to do anything much. But with the Taylor expansion around the sample mean, we have (write $M$ for $\{\bar X, S^2\}$)
$$E(u(X_1)\mid M)\approx E\Big(u(\bar{X})+u^{\prime} (\bar{X}) (X_1-\bar{X})+\frac 12u^{\prime \prime} (\bar X)(X_1-\bar{X})^2\mid M\Big)$$
Since when we condition a function of a random variable on the random variable, we have $E(u(Y)\mid Y) = u(Y)$, $E(Y \mid Y) = Y$, $E(ZY\mid Y) = YE(Z\mid Y)$ in our case this implies
$$E(u(X_1)\mid M) \approx u(\bar{X})+u^{\prime} (\bar{X}) (E(X_1\mid M)-E(\bar{X}\mid M))+\frac 12u^{\prime \prime}(\bar X)E\Big[(X_1-\bar{X})^2\mid M\Big]$$
and using the previous results, the second term is zero, and so we obtain
$$E(u(X_1)\mid M) \approx u(\bar{X})+\frac 12u^{\prime \prime}(\bar X)E\Big[(X_1-\bar{X})^2\mid M\Big]$$
But
$$E\Big[(X_1-\bar{X})^2\mid M\Big] = \text{Var}(X_1\mid M) = S^2$$ so
$$E(u(X_1)\mid M) \approx u(\bar{X})+\frac 12u^{\prime \prime}(\bar X)S^2 \qquad [1]$$
Of course, one should deal also with the Remainder (the expected value of it). The "Peano form" of the Remainder is the most convenient here, meaning that we need to consider
$$E[R_2(X_1;\bar X) \mid M]=E\left[h_2(X_1)\cdot \left(X_1-\bar X\right)^2 \mid M\right] \qquad [2]$$
where $h_2()$ is some function with the property that $h_2(X_1) \rightarrow 0$ as $X_1 \rightarrow \bar X$.  
Then, take the full Taylor expansion of $h_2()$ around $\bar X$ using again the peano form to write this Remainder too:
$$h_2(X_1) = h_2(\bar X) + h_2'(\bar X)(X_1-\bar X) + g_1(X_1)\cdot (X_1-\bar X)$$
$h_2(\bar X) = 0$. Insert the rest into eq. $[2]$ to obtain:
$$E[R_2(X_1;\bar X) \mid M]=E\left[\Big(h_2'(\bar X)+ g_1(X_1)\Big)\cdot\left(X_1-\bar X\right)^3 \mid M\right]$$
$$=h_2'(\bar X)E\left[\left(X_1-\bar X\right)^3 \mid M\right] + E\left[g_1(X_1)\cdot\left(X_1-\bar X\right)^3 \mid M\right]$$
The conditional on $M$ distribution of $X_1$ will also be normal, and has mean $\bar X$ as we have seen. So the third central moment, being odd, will be zero. Then the 1st term in the above expression is zero and we are left with
$$E[R_2(X_1;\bar X) \mid M]= E\left[g_1(X_1)\cdot\left(X_1-\bar X\right)^3 \mid M\right] \qquad [3]$$
which I would say, will be a rather small amount, since $g_1$ belongs to the Remainder of the Remainder.
Note that if we can say that $u'''(X_1)$ is bounded above and below in a neighborhood of $\bar X$ and belongs to the interval, say $[a, A]$ then the expected value of the Remainder is sandwiched to zero, because we have
$$a\frac {(X_1-\bar X)^3}{3!} \le R_2(X_1;\bar X) \le A\frac {(X_1-\bar X)^3}{3!}$$
$$\Rightarrow E\left (a\frac {(X_1-\bar X)^3}{3!} \mid M\right) \le E[R_2(X_1;\bar X) \mid M] \le E\left (a\frac {(X_1-\bar X)^3}{3!} \mid M\right)$$
$$\Rightarrow 0 \le E[R_2(X_1;\bar X) \mid M] \le 0 \Rightarrow E[R_2(X_1;\bar X) \mid M]=0$$
If we cannot say anything about $u()$ then we are left with the approximation $[1]$ and the approximation error $[3]$.
