Question

In the list problem, when the \(P_{i}\) are known, show that the best ordering (best in the sense of minimizing the expected position of the element requested) is to place the elements in decreasing order of their probabilities. That is, if \(P_{1}>P_{2}>\cdots>P_{n}\) show that \(1,2, \ldots, n\) is the best ordering.

Short answer

In summary, we proved that the best ordering to minimize the expected position of the element requested is to place the elements in decreasing order of their probabilities. This becomes evident when comparing the expected positions of two different orderings and applying the given conditions (\(P_{1}>P_{2}>\cdots>P_{n}\)). We demonstrated this by comparing the expected positions of two elements with different probabilities in both orderings and showed that placing elements in decreasing order of their probabilities leads to a smaller expected position.

Step-by-step solution

Define the expected position and understand how order affects this value

Let's define the expected position as the sum of the products of each element's probability and its position in the list, which can be expressed as: \[E = \sum_{i=1}^{n} P_{i}x_{i}\] where \(P_{i}\) is the probability of each element and \(x_{i}\) is its position in the list. Intuitively, the smaller an element's position in the list, the smaller the contribution of its probability to the expected position. Thus, we should order the elements in such a way that those with higher probabilities have smaller positions in the list.

Consider two elements with different probabilities

Let's take two elements with different probabilities, say \(P_{a}\) and \(P_{b}\), where \(P_{a} > P_{b}\). We want to prove that placing element \(a\) before element \(b\) in the list leads to a smaller expected position. Let's assume that the positions of elements \(a\) and \(b\) in the sequence are \(x_{a}\) and \(x_{b}\) respectively. Note that \(x_{a} < x_{b}\) since we are placing element \(a\) before element \(b\).

Compare the expected positions of both orderings

Now let's compare the expected position when elements \(a\) and \(b\) are placed in the proposed order in the list (with element \(a\) before element \(b\)) and when they are placed in the opposite order (with element \(b\) before element \(a\)). 1. In the first ordering, the combined contribution of elements \(a\) and \(b\) to the expected position: \[E_{ab} = P_{a}x_{a} + P_{b}x_{b}\] 2. In the second ordering, the combined contribution of elements \(a\) and \(b\) to the expected position: \[E_{ba} = P_{b}(x_{a}) + P_{a}(x_{b})\]

Prove that the first ordering leads to a smaller expected position

We want to prove that \(E_{ab} < E_{ba}\) since we are looking for the minimum expected position. \[P_{a}x_{a} + P_{b}x_{b} < P_{b}x_{a} + P_{a}x_{b}\] We can simplify the inequality by canceling out equal terms on both sides: \[P_{a}x_{a} - P_{b}x_{a} < P_{b}x_{b} - P_{b}x_{a}\] Notice that we can factor both left and right sides of the inequality: \[(P_{a} - P_{b})x_{a} < (P_{b} - P_{a})x_{b}\] Remember that we assumed \(P_{a} > P_{b}\) and \(x_{a} < x_{b}\). These conditions indicate that the left side of the inequality represents a positive value and the right side represents a negative value. Since a positive value is always greater than a negative value, we have proven that placing elements in decreasing order of their probabilities leads to a smaller expected position.

Conclusion

In summary, we proved that the best ordering to minimize the expected position of the element requested is to place the elements in decreasing order of their probabilities. This becomes evident when comparing the expected positions of two different orderings and applying the given conditions.

Key concepts

Expected Position in Probability Models

The concept of the expected position becomes crucial when dealing with probability models that concern the ordering of items based on their chances of occurrence. Imagine you're responsible for organizing books in a library so that the most sought-after books are easiest to find. Similarly, in a probabilistic model, the expected position represents the average position of an element in an ordered list given its likelihood of being requested.

The formula to calculate the expected position, denoted as E, is given by:
\[E = \sum_{i=1}^{n} P_{i}x_{i}\]
where \(P_{i}\) is the probability of each element being requested, and \(x_{i}\) its position in the list. To minimize E, it's essential to order items so that those with higher probabilities (\(P_{i}\)) have lower positional values (\(x_{i}\)), minimizing the overall expected position value and thus ensuring a more efficient system, be it a library or any data structure wanting to optimize item retrieval times.

Probability Ordering

Understanding probability ordering is crucial in decision-making and optimization problems. Consider the scenario where you need to prioritize tasks with varying levels of importance—a common real-life situation. In mathematical terms, this is similar to arranging a set of probabilities in a specific order to optimize a particular outcome, such as minimizing search times in a database or reducing the cost in a scheduling problem.

Probability ordering aligns individual items according to their probabilities of occurrence in a descending sequence. This arrangement ensures that events or items with a higher chance of occurring are placed before those with a lesser chance. From a practical standpoint, this organization enables faster and more efficient access to high-priority items or resolving the most probable events first, analogous to a checkout line where customers with fewer items are served first to speed up the overall service.

List Problem Probability

The list problem probability is a classic example of using probability to solve an optimization issue—which item should come first to ensure the most efficient outcome? For instance, in the context of computer science, how should we order a list of processes given the probability of each process needing to be executed?

Given a series of probabilities \(P_{1}, P_{2}, ..., P_{n}\), the challenge lies in determining the order of processes from highest to lowest probability to reduce the expected position. The key takeaway is that the element with the highest probability should be placed first, and then followed consecutively by elements of descending probabilities. This is borne out by comparing the expected positions, as shown in the exercise solution. The mathematical proof reinforces that following this probability ordering not only makes theoretical sense but also provides a tangible method for minimizing operational delays, thus optimizing efficiency in practical applications.