Suggestions for the optimal estimator in "one-shot" prediction problems? Assume you have a prediction distribution for a quantity. What point on this distribution should you use if the process you are predicting will end after the next observation and you want to be within $\epsilon$ units of that value? 
I have seen the mean touted as an optimal predictor, but that only makes sense if you want to minimize your expected mean square error (and assuming stationarity of the process being predicted!). I've considered two other obvious options: the median and the mode. In principle, it seems that the mode is ideal since the mode  will maximize the probability that the observation will be within the desired range if the underlying distribution is unimodal (i.e., P($X\;\in\;(a-\epsilon,\;a+\epsilon))$ is maximized if a=mode). 
Specifically, I am trying to find the optimal predictor for the following problem:
Let $Y =\{X_{1},X_{2}...X_{N}\}$ be a set of $N$ random quantities with assocated set of distributions $F=\{F_{1},F_{2}...F_{N}\}$ and acceptable absolute errors $\epsilon = \{\epsilon_{1},\epsilon_{2}...\epsilon_{N}\}$. You are to develop a single-point prediction set $\hat X =\{\hat X_{1},\hat X_{2}...\hat X_{N}\}$ which contains one point from the possible range of each random quantity such that you maximize  $P(\sum\limits_{i=1}^{N}\textbf{1}_{|X_{i}-\hat X_{i}|\leq \epsilon_{i}}= N)$.
Anyone have experience with these types of predictive situations and can share insights or good heuristics? Thanks :)
 A: This question brings on the surface a fact about prediction that tends to be forgotten: that a prediction is "optimal" only relative to the specific objective function being optimized -and this objective function must represent the real world situation in which the "users" of the prediction are in, and how the prediction error will affect them.
The ubiquitous "mean squared error" criterion is not so much an objective function that was conceived to represent the case of "repeated predictions", but mainly it reflects a situation where the costs of prediction error are quadratic. In obvious notation, minimizing $E(x-\hat x)^2$ is the same as minimizing $E\left[A(x-\hat x)^2\right], \; A>0$, wich means that we implicitly assume that our error cost function is $A(x-\hat x)^2$. If for example, the error cost function is linear, zero at zero error, but not symmetric around zero, then the optimal predictor is not the expected value -the median is. If we move away from quadratic functions, then in general we need the error cost function to be symmetric around zero and the distribution to be symmetric around its mean, so that the expected value remains the optimal predictor (essentially because it then equals the median).
That said, you need to think on what error cost function represents better your situation. Does under-prediction has "equal costs" to you compared to over-prediction by the same amount? (if "prediction accuracy" is the only thing that you are after, then you could argue that your error cost function is symmetric around zero, for example, since the direction of the error is not important to you, only its magnitude).  
After that, and since the error cost function will contain the unknown quantity of the actual future value of the variable you want to predict, you need to consider how you will nevertheless end up with a predictor that can be calculated/estimated (considering the expected value is one way to do that).  
I am willing to expand on this question if you provide feedback on these matters. 
EXPANSION (after OP's own answer and comments).  
Since we assume independent r.v's, maximizing the sum of probabilities is indeed equivalent to maximize each probability separately. So 
$$\max_{\hat X_i} P(\textbf{1}_{|X_{i}-\hat X_{i}|\leq \epsilon_{i}}) = \max_{\hat X_i}\left\{F_i(\hat x_i+\epsilon_i) - F_i(\hat x_i-\epsilon_i)\right\} =\max_{\hat X_i}\int_{\hat x_i-\epsilon_i}^{\hat x_i+\epsilon_i}f(x_i)dx_i $$ and the first order condition will give us
$$\hat x_i^* : f(\hat x_i^*+\epsilon_i) - f(\hat x_i^*-\epsilon_i) =0 $$
What about the 2nd-order conditions? They are
$$f'(\hat x_i^*+\epsilon_i) - f'(\hat x_i^*-\epsilon_i) <0 $$
In the nice case of a unimodal and symmetrical density, the above conditions lead to $\hat x_i^*$ being the mode.  
Assume now that the distribution is unimodal but not symmetric (as are many well-known distributions of non-negative r.v's, like the Gamma family). Then can you find a point in its support that will satisfy the first order condition?
Next, assume that the distribution is symmetric but not unimodal, but has one central global mode, and (being symmetric), it also has two other local modes to the left and to the right of the global modal, placed symmetrically, but also that the pre-determined error includes the two local modes in the permissible set of values. Then?
A: You should use the median as the value for $c$ if you want to minimize $E[|X-c|].$
You can derive this by starting with the definition of $E[|X-c|]$and minimizing it with respect to $c.$ 
A: The answer is to optimize the probability within each interval.
See: Guessing Game Stochastic Optimization
