2. AllElectronics caries 1000 products, P1, … P1000. Consider customers Ada, Bob, and Cathy such that Ada and Bob purchase three products in common, P1, P2, and P3. For the other 997 products, Ada and Bob independently purchase seven of them randomly. Cathy purchases 10 products, randomly selected from the 1000 products. In Euclidean distance, what is the probability that dist(Ada, Bob) > dist(Ada, Cathy)? What if Jaccard similarity (Chapter 2) is used? What can you learn from this example? (Problem 11.2, Page 539- 509)
solution?
That's a lot.
To solve this problem, we need to calculate the probabilities based on the given information.
1. Euclidean Distance:
To find the distance between Ada and Bob, we need to consider the products they have in common. In this case, they share three products: P1, P2, and P3. The remaining 997 products are purchased independently.
- Ada and Bob's distance calculation:
They have 3 products in common, so their distance for those products is 0. For the remaining 997 products, each product can either be purchased by Ada or by Bob. Since they are chosen independently, the probability of a specific product purchased by Ada is 1/2 and by Bob is also 1/2. Therefore, the probability that Ada and Bob have the same product for each of the 997 items is (1/2)^997.
- Ada and Cathy's distance calculation:
They don't share any common products. For the remaining 1000 products, each product can either be purchased by Ada or by Cathy. Since they are chosen independently, the probability of a specific product purchased by Ada is 1/2 and by Cathy is 1/10. Therefore, the probability that Ada and Cathy have the same product for each of the 1000 items is (1/2)^1000 * (1/10)^1000.
Now, we can calculate the probability that dist(Ada, Bob) > dist(Ada, Cathy):
This is the probability that the distance between Ada and Bob (considering common products) is greater than the distance between Ada and Cathy. Since the distances are calculated independently for each product, we can find the probability as the product of individual probabilities.
P(dist(Ada, Bob) > dist(Ada, Cathy)) = P(Ada and Bob have the common products) * P(Ada and Bob have different products for 997 items) * P(Ada and Cathy have different products for 1000 items)
P(dist(Ada, Bob) > dist(Ada, Cathy)) = 1 * (1/2)^997 * (1 - (1/2)^1000 * (1/10)^1000)
2. Jaccard Similarity:
Jaccard similarity measures the intersection divided by the union of two sets. In this case, the sets represent the products purchased by Ada, Bob, and Cathy.
For Jaccard similarity, we need to calculate the intersection and union of Ada & Bob's purchases and Ada & Cathy's purchases.
- Ada and Bob's Jaccard similarity:
They have 3 common products (intersection) and a total of 997 + 7 = 1004 products (union).
- Ada and Cathy's Jaccard similarity:
They have 0 common products (intersection) and a total of 1000 + 10 = 1010 products (union).
The Jaccard similarity is calculated as follows:
Jaccard similarity = intersection / union
P(dist(Ada, Bob) > dist(Ada, Cathy)) = P(Ada and Bob's Jaccard similarity > Ada and Cathy's Jaccard similarity)
Now, we need to find the probability that Ada and Bob's Jaccard similarity is greater than Ada and Cathy's Jaccard similarity based on the provided information.
What can we learn from this example?
This example showcases the calculation of probabilities based on Euclidean distance and Jaccard similarity. It demonstrates how probabilities can be computed when considering common and different items within sets. Additionally, it highlights the importance of considering the choice of distance metric and how it can affect the probabilities and outcomes.
To calculate the probability that dist(Ada, Bob) > dist(Ada, Cathy) in Euclidean distance, we need to consider the distance between Ada and Bob and the distance between Ada and Cathy.
First, let's calculate the distance between Ada and Bob. Since they have three products in common, the Euclidean distance between them for those three products is 0. For the other 997 products, Ada and Bob independently purchase seven of them randomly. This means that for each of the remaining 997 products, the probability that Ada and Bob both purchase it is 7/1000. So the average distance between Ada and Bob for the remaining products is the square root of (7/1000) raised to the power of 997. Let's call this distance AB.
Next, let's calculate the distance between Ada and Cathy. Cathy purchases 10 products randomly selected from the 1000 products, so the probability that a specific product is purchased by both Ada and Cathy is 10/1000. Therefore, the average distance between Ada and Cathy for these 10 common products is the square root of (10/1000) raised to the power of 10. Let's call this distance AC.
Now, to calculate the probability that dist(Ada, Bob) > dist(Ada, Cathy), we compare the values of AB and AC. If AB is greater than AC, then dist(Ada, Bob) is greater than dist(Ada, Cathy). The probability of this happening can be calculated by finding the probability that AB is greater than AC.
The Jaccard similarity measures the intersection of two sets divided by the union of the two sets. In this case, the intersection of Ada and Bob's purchases is three products, and the intersection of Ada and Cathy's purchases is also three products. The union of Ada and Bob's purchases is 1000 products for Bob and 1000 products for Ada, while the union of Ada and Cathy's purchases is 1000 products for Cathy and 1000 products for Ada. Therefore, the Jaccard similarity in this case is 3/1000 for both cases.
From this example, we can learn that the Euclidean distance considers the actual distance between two points in a multi-dimensional space, while the Jaccard similarity measures the similarity between two sets by looking at their intersections and unions. Depending on the context and the type of data, either measure can be appropriate.