How to spread probability in a "gradient", i.e. so that it decreases with distance from a point on a dimension?

Question

I have data tasks which involve a little bit of math - which is really not my strong suit - hope you can help.

Background
Respondent has been given a question that they answer with options on a discrete 0:10 scale. One of the answers is "the true" response, so if each of the (11) options is assigned a (subjective) probability the sum of these probabilities must be 100%. So...

Options on the scale: 0, 1 , 2 , 3, 4, 5, 6, 7, 8, 9, 10.

$p(o_0)+p(o_1)+...p(o_i)...+p(o_9)+p(o_{10})$ = 1.0 = 100%

What I plan to do is the following:

• First. If a respondent answers some option $o_i$. I want to assign, to the respondent's option, a subjective probability, let's call it $p(o_k)$.

• Then. Spread the remaining $1-p(o_k)$ probability over the other options; with the probability of each option declining as the distance from $o_k$ increase. All options must be assigned a non-zero probability.

Problem
What I can't solve is what function would give me the probability of an option on the scale, given:

1. The probability of the respondent's option.
2. The distance, on the scale, of an option from the respondent's option.
3. That the sum of the probabilities of all options, on the scale, is 1.

Example
The respondent has answered option 4. I assign to option 4 a 25% chance of being true. Then there's a 75% probability of one of the other options is the true option. But what probability to assign each individual option if it is to decline with the distance (absolute difference) from 4?

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I hope this was sufficiently clear. Have a great day.

Answer 1

If you definitely want to keep the first probability $p(o_k)$ unchanged, then you can just pick an arbitrary relative assignment to the others and scale it.

For example, suppose you fix $p(o_k)$. Then for $j\neq k$, put $n_j=1/|j-k|$ (which decreases as $j$ gets further from $k$, but you can pick anything that does this; the scale will be adjusted to compensate below).

The total for these other picks is $S=\sum\limits_{j\neq k}n_j$. Scale the sum to the desired remainder by multiplying by $(1-p(o_k))/S$. Then the desired probabilities for the other items are $$p(o_j) = (1-p(o_k))\frac{n_j}{S}$$ for each $j\neq k$.

Note that the total probability over all items (assuming $N$ items altogether) is $$p(o_1)+p(o_2)+\cdots+p(o_N)$$ $$=p(o_k)+\sum_{j\neq k}p(o_j)$$ $$=p(o_k)+\sum_{j\neq k}\left((1-p(o_k))\frac{n_j}{S}\right)$$ $$=p(o_k) + \left(\frac{1-p(o_k)}{S}\right)\sum_{j\neq k}n_j$$ $$=p(o_k) + \left(\frac{1-p(o_k)}{S}\right)S$$ $$=p(o_k) + (1-p(o_k))$$ $$=1$$

Answer 2

The problem is ill-posed, namely the solution may not exist or is not unique in the general case. Let $k$ be the chosen index henceforward, as an examle if you assign $p(k)=0$ the problem has no solution. Let me go into more details, if $p(k) =1$, the problem has a unique solution, which would correspond to a delta peak in the continuous case. Now I will derive the condition for this problem to have a solution and the structure of the solutions. Let us consider a more general problem with $o_a,...,o_b; a\leq b$, that is, there are $b-a+1$ answers (plug in $a=0, b=10$ and you get the original problem). Whenever a user picks an index $k \in \{a,...,b\}$, there are $l = k-a$ indices to its left and $r = b-k$ indices to its right. The following must hold $(1-p(k))/(l+r) \leq p(k)$, or equivalently $1/(1+l+r) \leq p(k)$, otherwise it is not possible that $p(k)\geq p(i), k<i\leq b$ and $p(k) \geq p(j), a\leq j<k$ (your requirement that the probability is non-increasing on both sides). Now that we have found the condition for this problem to have a solution, we should consider the possible solutions: $$p = \{ p_i, i \in \{a,...,b\} | p_k = p(k); \phi, \psi : \{a,...,b\} \rightarrow [0,1]; \phi(x), \psi(x) \leq p_k; \phi(x)\leq \phi(y), x<y; \psi(x) \leq \psi(y), x>y; \sum_{i=a}^{k-1}{\phi(i)} + \sum_{j=k+1}^{b}{\psi(j)} = 1-p_k; p_i = \phi(i), a\leq i < k; p_i = \psi(i), k<i\leq b\}$$ There are infinitely many such functions $\phi, \psi$ in the general case. Note thar $\phi$ is monotonically non-decreasing and $\psi$ is monotonically non-increasing, with the additional requirement $\phi(x) \leq p_k, \psi(x) \leq p_k$, and the requirement on the sums. Note that it is enough that $\phi$ is defined on $\{a,...,k-1\}$, and $\psi$ on $\{k+1,...,b\}$.