Bayesian inference on a sum of iid real-valued random variables Let $X_1$, $X_2$, ..., $X_n$ be iid RV's with range $[0,1]$ but
unknown distribution.  (I'm OK with assuming that the distribution
is continuous, etc., if necessary.)
Define $S_n = X_1 + \cdots + X_n$.
I am given $S_k$, and ask: What can I infer, in a Bayesian manner, about
$S_n$?
That is, I am given the sum of a sample of size $k$ of the RV's, and I would
like to know what I can infer about the distribution of the sum of all the RV's,
using a Bayesian approach (and assuming reasonable priors about the distribution).
If the support were $\{0,1\}$ instead of $[0,1]$, then this problem is well-studied, and (with uniform priors) you get beta-binomial compound distributions for the inferred distribution on $S_n$.  But I'm not sure how to approach it with $[0,1]$ as the range...
Full disclosure: I already posted this on MathOverflow, but was told it would be
better posted here, so this is a re-post.
 A: Consider the following Bayesian nonparametric analysis.
Define $\mathscr{X}=[0,1]$ and let $\mathscr{B}$ be the Borel subsets of $\mathscr{X}$. Let $\alpha$ be a nonzero finite measure over $(\mathscr{X},\mathscr{B})$.
Let $Q$ be a Dirichlet process with parameter $\alpha$, and suppose that $X_1,\dots,X_n$ are conditionally i.i.d., given that $Q=q$, such that $\mu_{X_1}(B)=P\{X_1\in B\} = q(B)$, for every $B\in\mathscr{B}$.
From the properties of the Dirichlet process, we know that, given $X_1,\dots,X_k$, the predictive distribution of a future observation like $X_{k+1}$ is the measure $\beta$ over $(\mathscr{X},\mathscr{B})$ defined by
$$
  \beta(B) = \frac{1}{\alpha(\mathscr{X})+k} \left( \alpha(B) + \sum_{i=1}^k I_B(X_i)\right) \, .
$$
Now, define $\mathscr{F}_k$ as the sigma-field generated by $X_1,\dots,X_k$, and use measurability and the symmetry of the $X_i$'s to get
$$
  E\left[ S_n \mid \mathscr{F}_k \right] = S_k + E\left[ \sum_{i=k+1}^n X_i \,\Bigg\vert\, \mathscr{F}_k \right] = S_k + (n-k) E\left[ X_{k+1} \mid \mathscr{F}_k \right] \, ,
$$
almost surely.
To find an explicit answer, suppose that $\alpha(\cdot)/\alpha(\mathscr{X})$ is $U[0,1]$. Defining $c=\alpha(\mathscr{X})>0$, we have
$$
  E\left[ S_n \mid X_1=x_1,\dots,X_k=x_k \right] = s_k + \frac{n-k}{c+k}\left(\frac{c}{2}+s_k\right) \, ,
$$
almost surely $[\mu_{X_1,\dots,X_k}]$ (the joint distribution of $X_1,\dots,X_k$), where $s_k=x_1+\dots+x_k$. In the "noninformative" limit of $c\to 0$, the former expectation reduces to $n\cdot (s_k/k)$, which means that, in this case, your posterior guess for $S_n$ is just $n$ times the mean of the first $k$ observations, which looks like as intuitive as possible.
A: Forgive the lack of measure theory and abuses of notation in the below...
Since this is Bayesian inference, there must be some prior on the unknown in the problem, which in this case is the distribution of $X_1$, an infinite-dimensional parameter taking values in the set of distributions on $[0, 1]$ (call it $\pi$). The data distribution $S_k|\pi$ converges to a normal distribution, so if $k$ is large enough (Berry-Esseen theorem) we can just slap in that normal as an approximation. Furthermore, if the approximation is accurate the only aspect of the prior $p(\pi)$ that matters in practical terms is the induced prior on $(\text{E}_\pi(X_1),\text{Var}_\pi(X_1))=(\mu,\sigma^2)$.
Now we do standard Bayesian prediction and put in the approximate densities. ($S_n$ is subject to the same approximation as $S_k$.)
$p(S_n|S_k) = \int p(\pi|S_k)p(S_n|\pi,S_k)d\pi$
$p(S_n|S_k) = \int \frac{p(\pi)p(S_k|\pi)}{p(S_k)}p(S_n|\pi,S_k)d\pi$
$p(S_n|S_k) \approx \frac{\int p(\mu,\sigma^2)\text{N}(S_k|k\mu,k\sigma^2)\text{N}(S_n|(n-k)\mu + S_k, (n-k)\sigma^2) d(\mu,\sigma^2)}{\int p(\mu,\sigma^2)\text{N}(S_k|k\mu,k\sigma^2) d(\mu,\sigma^2)}$ 
For the limits of the integral, $\mu \in [0, 1]$, obviously; I think $\sigma^2 \in [0,\frac{1}{4}]$? 
Added later: no, $\sigma^2 \in [0,\mu(1-\mu)].$ This is nice -- the allowed values of 
$\sigma^2$ depend on $\mu$, so info in the data about $\mu$ is relevant to $\sigma^2$ too.
A: Let each $X_i$ belong to distribution family $F$ and have parameters $\theta$.
Given, $S_k$, we have a distribution on $\theta$:
\begin{align}
\Pr(\theta \mid S_k) &= \frac1Z \Pr(\theta)\Pr(S_k \mid \theta)
\end{align}
And, our distribution on $S_n$, $n \ge k$ is
\begin{align}
\Pr(S_n = i \mid S_k) &= \Pr(S_{n-k} = i - S_k | S_k) \\
&= \int \Pr(S_{n-k} = i - S_k | \theta)\Pr(\theta \mid S_k)d\theta \\
\end{align}
(and similarly for $n < k$)
Both of these equations have nice forms when $F$ is a distribution in the exponential family that is closed under summation of iid elements like the normal distribution, the gamma distribution, and the binomial distribution.  It also works for their special cases like the exponential distribution and the Bernoulli distribution.
It might be interesting to consider $F$ is the family of scaled (by $\frac1n$) binomial distributions with known "trials" $n$, and taking the limit as $n$ goes to infinity.
