Estimate the second moment of a latent variable using a conditionally unbiased proxy The Setup: Let $X_t$ denote an unobservable stochastic sequence where the first two unconditional moments are finite constants; ie $\mathbb{E} X_t = \mu < \infty$ and $\mathbb{E} X_t^2 = \gamma < \infty$. Assume that $X_t$ obeys some minimum set of regularity conditions such that a weak law of large numbers applies to its first two sample moments; ie conditions that limit the dependence in the sequence and bound the higher order moments. 
Let $Y_t$ denote a conditionally unbiased proxy for $X_t$. That is, $\mathbb{E} [Y_t | X_t] = X_t$, and hence $\mathbb{E} Y_t = \mathbb{E} X_t = \mu$. Assume that $Y_t$ also obeys a minimum set of regularity conditions such that a weak law of large numbers applies to its first two sample moments.
Let $S_T = T^{-1} \sum_{t=1}^T Y_t$. Given the assumptions on $X_t$ and $Y_t$, it follows that $\mathbb{E} S_T = \mu$ and $S_T \overset{\mathbb{P}}{\rightarrow} \mu$, hence $S_T$ is a consistent, unbiased estimator for $\mu$
My Question: Given the setup, how can I estimate $\mathbb{E} X_t^2 = \gamma$? 
Things I Know: From Slutsky's theorem, $(S_T)^2 \overset{\mathbb{P}}{\rightarrow} (\mathbb{E} X_t)^2 \neq \mathbb{E} X_t^2 = \gamma$, via Jensen's inequality, so in general $(S_T)^2$ will be a poor estimator for $\gamma$. Also, $T^{-1} \sum_{t=1}^T Y_t^2$ may not be a good estimator since the assumption set does not guarantee that $\mathbb{E} X_t^2 = \mathbb{E} Y_t^2$.
Final Note: I would be happy with any answer that improves upon the properties of $(S_T)^2$. So an approximation type estimator obtained via some sort of series expansion would be great. That's what I've mainly been looking into thus far, but with little luck. Even suggestions regarding expansions I might look into would be most welcome.
Thanks in advance to all responders.
 A: I'm not sure you can do much more with what you've got to work with.
I mean, you can lower bound $\gamma \geq \mu^2$, as
$$
\mathbb{Var}(X_t) = \mathbb{E}(X_t^2)-\mathbb{E}(X_t)^2 = \gamma-\mu^2 \geq 0
$$
then estimate $\mu$ from the arithmetic mean of your set of samples $S_T$ as usual. 
If we have the situation where $Y_t$ is equal to $X_t$ plus some uncorrelated noise, then you can upper bound it too as
$$
\mathbb{E}(Y_t^2) \geq \mathbb{E}(X_t^2).
$$
To see why, look at Innuo's splendid comment on your original post, and note that the variance of $Z_t$ must be positive. If the covariance between it and $X_t$ is zero, then the inequality becomes pretty clear. If the correlation between $X_t$ and $Z_t$ is positive, then it still holds, albeit more weakly.
So we have $\mathbb{E}(Y_t^2) \geq \gamma \geq \mu^2$ for uncorrelated or positively correlated additive noise. I think that's all that can be said. If those bounds are sufficiently tight then maybe that will do? If not, then either $X_t$ or $Z_t$ has a big variance, and beyond taking more direct measurements somehow I'm not sure if you can tell which.
