Expectation of squared sample t statistics Given i.i.d data points $X_{1},...,X_{n}$ from unknown smooth distribution $f(x)$ with $EX=\theta$ with parameter dimension 1. Let $\bar{X}=\sum_{i=1}^{n}X_{i}/n$, and $\sigma_{n}^2=\sum_{i=1}^{n}(X_{i}-\bar{X})^2/n$. Based on classic asymptotic theory, we know that $$ \sqrt{n}(\bar{X}-\theta)/\sigma_{n}\rightarrow N(0,1) \space\ \text{in distribution},$$ and $$n(\bar{X}-\theta)^2/\sigma_{n}^2\rightarrow \chi_{1}^{2} \space\ \text{in distribution}.$$
My question is to calculate the following integral such as $$E\left(n(\bar{X}-\theta)^2/\sigma_{n}^2\right)=\int f(x)\frac{n(\bar{X}-\theta)^2}{\sigma_{n}^2}dx=?$$
Asymptotically I expect this integral to be 1 since the test statistic converges in distribution to $\chi_{1}^{2}$. 
 A: I don't know how to answer this totally generally but here's something.
We know that for two random variables $U$ and $V$ $Cov(U, V) = E(UV) - E(U)E(V)$ so
$$
E\left(n(\bar{X}-\theta)^2/\sigma_{n}^2\right) = nCov\left((\bar{X}-\theta)^2, \frac{1}{\sigma_{n}^2}\right) + nE\left((\bar{X}-\theta)^2\right)E\left(\frac{1}{\sigma_{n}^2}\right).
$$
We'll always have that 
$$
nE\left((\bar{X}-\theta)^2\right)E\left(\frac{1}{\sigma_{n}^2}\right) = n Var(\bar X) E\left(\frac{1}{\sigma_{n}^2}\right).
$$
The $X_i$ are iid so $Var(\bar X) = \frac{Var(X_1)}{n}$ so we have that
$$
E\left(n(\bar{X}-\theta)^2/\sigma_{n}^2\right) = nCov\left((\bar{X}-\theta)^2, \frac{1}{\sigma_{n}^2}\right) + \sigma^2 E\left(\frac{1}{\sigma_{n}^2}\right).
$$
where $\sigma^2 = Var(X_1)$.
Now suppose that $f$ is a distribution such that $\bar X$ is complete and sufficient for $\theta$ and $\sigma^2_n$ is ancillary to $\theta$. The normal distribution is an example of this.
Then by Basu's theorem we have that $\bar X \perp \sigma^2_n$ therefore the covariance term is 0 and we are left with 
$$
E\left(n(\bar{X}-\theta)^2/\sigma_{n}^2\right) = \sigma^2 E\left(\frac{1}{\sigma_{n}^2}\right).
$$
I'm not sure how to evaluate this final expectation without knowing more. 
