6
$\begingroup$

Consider $N$ items with associated weights $w_i$. Each time, we sample one item from the remainder without replacement and the sampling probability is proportional to the weights. Continue sampling until all items are selected and we acquire a sequence. What's the distribution of this sequence? Does it belong to the exponential family?

$\endgroup$
1
$\begingroup$

Assume the item weights $W = \{w(1), \dots, w(N)\}$ are nonnegative and sum to one. Let $X=\{x_1, \dots, x_N\}$ denote a particular sequence, where $x_i$ is an integer from 1 to $N$ representing the index of the $i$th item drawn. Let $X_{i:j} = \{x_i, \dots, x_j\}$ denote a particular subsequnce. The probability of a sequence can be written as:

$$p(X \mid W) = p(x_1 \mid W) \prod_{i=2}^N p(x_i \mid X_{1:i-1}, W)$$

$p(x_1 \mid W)$ is the probability for the first item drawn. This is simply the item's weight:

$$p(x_1 \mid W) = w(x_1)$$

For subsequent draws ($i \ge 2$), $p(x_i \mid X_{1:i-1}, W)$ is the conditional probability of the $i$th item drawn, given all previously drawn items. It can be expressed as follows. The probability of drawing an item is zero if it has already been drawn. Otherwise, it's equal to the item's weight, divided by the sum of weights for all remaining items. This sum is equal to one minus the sum of weights for previously drawn items. Therefore:

$$p(x_i \mid X_{1:i-1}, W) = \frac{w(x_i)}{1 - \sum_{j=1}^{i-1} w(x_j)} \prod_{j=1}^{i-1} \mathcal{I}(x_i \ne x_j)$$

where $\mathcal{I}(\cdot)$ is the indicator function, which returns 1 if its argument is true, otherwise 0. Note that the product of indicator functions evaluates to one if and only if $x_i$ isn't a duplicate of previous items.

Final answer

Combine everything above to obtain the following. Sequences containing duplicate items have probability 0. Otherwise, if there are no duplicates:

$$p(X \mid W) = w(x_1) \prod_{i=2}^N \frac{w(x_i)}{1 - \sum_{j=1}^{i-1} w(x_j)}$$

| cite | improve this answer | |
$\endgroup$
0
$\begingroup$

I'm working on a similar problem. This is not a direct answer to your question but maybe it's close enough that it gives you some ideas about how to start.

My weights form an exponential (specifically, log-linear) family $w_i = \exp(\beta^\top a_i)/Z(\beta)$, where $Z(\beta) = 1 + \exp(\beta^\top a_i)$, $\beta$ is the vector of natural parameters, and $a_i$ is the $i$th column of the design matrix.

In my model, you first select $W = w$ uniformly at random from $\{0, 1, \ldots, N\}$, and then sample $w$ of the $N$ items without replacement according to the weights $w_i$ until you have collected $w$ of them. I believe the distribution of the set $x$ of items you collect this way is $\exp(\beta^\top \sum_{i \in x}a_i)/(N Z_w(\beta))$, where

$$Z_w(\beta) = \sum_{\substack{y \in \{0, 1\}^N \\ |y| = w}} \exp\left(\beta^\top \sum_{i \in y}a_i\right).$$

The marginal distribution is still $w_i$, but the samples are not independent, with that $1/N$ factor from the joint distribution turning into a

$$\frac{N - j - 1}{N - j}$$

factor in the probability of choosing the $(j+1)$th item when you condition on having already selected $j$ items. When you multiply out the successive conditional probabilities to get back to the joint probability, you end up with a factor of

$$ \frac{(N - 1)!}{N!} = \frac{1}{N} $$

as you would want.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.