expectation of sample variance $s_{t+1}$ given sample data $y_t$? Assume that data $Y_t$ is a collection of i.i.d. variables $y_1,y_2,...,y_t$ that are sampled from a normal distribution with mean $\mu$ and variance $\sigma^2$. 
Assume now that we have all the data up to time $t$, and that we want to predict the sample variance that we get when we collect one more datapoint from time $t+1$:
$$E(s^2_{t+1}|Y_t)$$
My question is: are the mean and sample variance $\bar y_t, s^2_t$ sufficient statistics for calculating this expectation? i.e.
$$E(s^2_{t+1}|Y_t)\overset ? =E(s^2_{t+1}|\bar y_t, s^2_t)$$
And secondly, how do we calculate this? I know that the calculation has to invoke the assumption that the data is normally distributed, but I am not sure how to get to the desired result. 
 A: Yes they’re sufficient, in both senses of that word. This was mentioned in the comments. Here are a couple comments on some of things being mentioned in this thread.
Comment 1: you probably wouldn’t want a Kalman filter.
Following up on @BenOgorek's comment, you can use a Kalman filter if you write your model as 
$$ 
x_t = \Phi x_{t-1} + \Upsilon u_t + w_t
$$
and
$$
y_t = x_t + v_t = x_t
$$
where $\Phi = 0$ and $\text{Var}(v_t) = 0$, $u_t = 1$ and $\Upsilon = \mu$. Also, $\text{Var}(w_t) = Q$, and you're assuming Normal errors. Sorry for the extra notation, I am just following "Time Series And Its Applications with R Examples".
The Kalman recursions give you recursive formulas for $E[X_t \mid y_{1:t}]$, $\text{Var}[X_t \mid y_{1:t}]$, $E[X_{t+1} \mid y_{1:t}]$, $\text{Var}[X_{t+1} \mid y_{1:t}]$. Specifically they are


*

*$E[X_{t} \mid y_{1:t-1}] = \Upsilon = \mu$

*$\text{Var}[X_{t} \mid y_{1:t-1}] = Q$

*$E[X_t \mid y_{1:t} ] = \mu + (y_t - \mu) = y_t$

*$\text{Var}[x_t \mid y_{1:t}] = 0$.


This is the "recursive" formula that doesn’t have much to do with what you’re looking for. It assumes you know the parameters, and gives you the common sense answer for the next data point, but it doesn’t say much about the average one step ahead sample variance.
Comment 2: the posterior predictive distribution
Another thing, @Zhanxiong's answer looks good (+1), but it might not work in a Bayesian setting, where you're using the posterior predictive distribution instead of assuming the parameters are known. In a Bayesian setting, you would have (by the law of total expectation)
$$
E[S_{t+1}^2 \mid y_{1:t}] = E[E(S_{t+1}^2 \mid \mu, \sigma^2) \mid y_{1:t}] = E[\sigma^2 \mid y_{1:t}],
$$
which is the posterior mean of $\sigma^2$. Whether or not this has a nice recursive formula depends on your choice of priors for $\mu$ and $\sigma^2$.
A: A direct calculation is possible (we can come up with a solution without bothering any linear algebra). Intuitively, it would be helpful to separate the new observation $y_{t + 1}$ from the history records $y_1, \ldots, y_t$, i.e., try to decompose $s_{t + 1}^2$ as follows: 
\begin{align}
& s_{t + 1}^2 = \frac{1}{t}\sum_{i = 1}^{t + 1}(y_i - \bar{y}_{t + 1})^2 = \frac{1}{t}\left[\sum_{i = 1}^{t + 1}y_i^2 - (t + 1)\bar{y}_{t + 1}^2\right] \\
= & \frac{1}{t}\sum_{i = 1}^t y_i^2 + \frac{1}{t}y_{t + 1}^2 - \frac{t}{t + 1}\bar{y}_t^2 - \frac{2\bar{y}_ty_{t + 1}}{t + 1} - \frac{1}{t(t + 1)}y_{t + 1}^2 \\
= & \frac{1}{t}\sum_{i = 1}^t y_i^2 + \frac{1}{t + 1}y_{t + 1}^2 - \frac{t}{t + 1}\bar{y}_t^2 - \frac{2\bar{y}_ty_{t + 1}}{t + 1} 
\end{align}
Using $(1)$, let's evaluate $I_1 := E[s_{t + 1}^2 | y_1, \ldots, y_t]$ and $I_2 := E[s_{t + 1}^2 |\bar{y}_t, s_t^2]$ respectively, and check if they agree with each other.
By $(1)$, the linearity of conditional expectation, and the condition that $y_1, \ldots, y_{t + 1}$ are i.i.d., it follows that
\begin{align}
& I_1 = \frac{1}{t}E\left[\sum_{i = 1}^t y_i^2 |Y_t\right] + \frac{1}{t + 1}E[y_{t + 1}^2 | Y_t] - \frac{t}{t + 1}E[\bar{y}_t^2|Y_t] - \frac{2}{t + 1}E[\bar{y}_ty_{t + 1}|Y_t] \\
= & \frac{1}{t}\sum_{i = 1}^t y_i^2 + \frac{1}{t + 1}E[y_{t + 1}^2] - \frac{t}{t + 1}\bar{y}_t^2 - \frac{2}{t + 1} \bar{y}_tE[y_{t + 1}] \tag{2} \\
= & \frac{1}{t}\sum_{i = 1}^t y_i^2 + \frac{1}{t + 1}(\mu^2 + \sigma^2) - \frac{t}{t + 1}\bar{y}_t^2 - \frac{2}{t + 1}\mu\bar{y}_t.
\end{align}
On the other hand, in view of
$$\sum_{i = 1}^t y_i^2 = (t - 1)s_t^2 + t\bar{y}_t^2,$$
similar calculation to $I_1$ leads to 
\begin{align}
& I_2 = \frac{1}{t}E[(t - 1)s_t^2 + t\bar{y}_t^2|\bar{y}_t, s_t^2] + \frac{1}{t + 1}E[y_{t + 1}^2 | \bar{y}_t, s_t^2] - \frac{t}{t + 1}E[\bar{y}_t^2|\bar{y}_t, s_t^2] - \frac{2}{t + 1}E[\bar{y}_ty_{t + 1}|\bar{y}_t, s_t^2] \\
= & \frac{1}{t}(t - 1)s_t^2 + \bar{y}_t^2 + \frac{1}{t + 1}E[y_{t + 1}^2] - \frac{t}{t + 1}\bar{y}_t^2 - \frac{2}{t + 1} \bar{y}_tE[y_{t + 1}] \tag{3} \\
= & \frac{1}{t}\sum_{i = 1}^t y_i^2 + \frac{1}{t + 1}(\mu^2 + \sigma^2) - \frac{t}{t + 1}\bar{y}_t^2 - \frac{2}{t + 1}\mu\bar{y}_t.
\end{align}
Therefore, $I_1 = I_2$. The above calculation also shows that the equality holds even we cancelled the normality assumption (of course, in that case, the final result would not be expressed in terms of $\mu$ and $\sigma^2$). 
In $(2)$ and $(3)$, apart from the linearity, the following properties of conditional expectation are applied:


*

*$E[X | Y] = E[X]$ if $X$ and $Y$ are independent.

*$E[f(X) | X] = f(X)$ and $E[Yf(X)|X] = f(X)E[Y | X]$ for any measurable function $f: \mathbb{R} \to \mathbb{R}$. 

