Uncertainty modeling for samples I have instances of data that are composed of 6 attributes. Each attribute takes values between 1-10. Any instance has a utility calculated by some utility function:
Example: {1,1,2,5,5,2} , utility = 0.89
In some readings for these attribute values, undefined values may appear in the instance, I mark them as question mark '?'
Example: {1,1,?,5,?,2}
These question marks represents uncertainty in the instance.
How can I model this uncertainty in a statistical or mathematical way ? 
Thanks
 A: A general, rigorous way to handle arbitrary missing values for an arbitrary function is with interval analysis.  This technique extends arithmetic operations to sets of numbers based on the principle that the interval-valued version $\hat{f}$ of any function $f:\mathbb{R}^n\to \mathbb{R}$ is defined as
$$y \in \hat{f}\left([x_1^-, x_1^+], [x_2^-, x_2^+], \ldots, [x_n^-, x_n^+]\right)$$
if and only if there exist numbers $x_i, z_i \in [x_i^-, x_i^+]$ for which 
$$f(x_1, \ldots, x_n) \le y \le f(z_1, \ldots, z_n).$$
Interpreting "?" as the interval $[1,10]$ of all possible values that could have appeared, apply the interval-valued extension of the utility function to propagate the uncertainty in the arguments of $f$ into uncertainty about its value.
For example, suppose the utility function is given by 
$$f(x_1,\ldots,x_6) = x_1+\cdots+x_6 - x_1 (x_2 + \cdots + x_6) / 10 - x_3 x_5 / 5.$$
Then for instance 
$$f(1,1,2,5,5,2) = 1+1+2+5+5+2 - 1*(1+2+5+5+2)/10 - 2*5/5 = 12.5.$$ 
Using interval arithmetic (for integral arguments) we find that 
$$\hat{f}(1,1,?,5,?,2) = \hat{f}(1,1,[1,10],5,[1,10],2) = [6.2, 16.1].$$
The proof of this lies in demonstrating that $f(1,1,x,5,y,2)$ is maximized for $(x,y)=(10,1)$ and minimized for $(x,y) = (10,10)$ and the values it attains range from $16.1$ down to $6.2$.
The interpretation is that given the known information, we can say for certain that the utility could not be less than $6.2$ and not more than $16.1$.
This approach can occasionally be useful. Utilities are meant to be compared. If, subject to missing information you compute one utility to be $[6.2, 16.1]$ and another to be $[0.1, 3.9]$, then you can conclude the first is greater than the second, because every number in the first interval exceeds every number in the second interval.  If the second is instead $[0.1, 8.5]$ you cannot (from this information alone) conclude that either is larger than the other.
There are subtleties. It is quite possible in the previous instance that the difference in the two utilities could be the interval $[0.1, 14.9]$.  At first blush this does not appear to be consistent with subtracting $[0.1, 8.5]$ from $[6.2, 16.1]$--and it's not.  But the structure of the utility function could be such that the uncertainty in the difference is less than the difference in the uncertainties.  A simple, intuitive example concerns a monotonic utility.  Suppose, say, $\hat{f}(1,1,?,5,?,2) = [6.2, 16.1]$ and $\hat{f}(1,1,?,7,?,2) = [7.2, 17.1]$. Although the two values overlap substantially, nevertheless it is certain that the second utility is greater than the first if the missing values are the same. In effect, the point is that when those missing values are the same, you should compare the utilities by means of the interval-valued extension of the function $f(1,1,x,7,y,2) - f(1,1,x,5,y,2)$--and, for a monotonic utility $f$, this is guaranteed to be a positive interval.
For more about interval arithmetic, as well as some practical applications, please follow the link at the beginning.

It is important to know that often one can do much better than interval analysis by assuming probability distributions for the unknown values and applying Bayesian analysis to propagate those into distributions for the utilities.  This requires having some basis for such assumptions.  If absolutely nothing is known about the missing values--and perhaps if the mechanism of missingness changes from one time to another, possibly defeating attempts at statistical analysis--then the Bayesian approach may be invalid and interval analysis becomes attractive.
