Expected Value E[Y] Where X Is Uniformly Sampled On (0,1) And Y Is Uniformly Sampled On (X², X)

This article delves into a fascinating probability problem involving continuous random variables. We are given that a random variable X is sampled uniformly from the interval (0,1). Subsequently, another random variable Y is sampled uniformly from the interval (X², X). Our primary objective is to determine the expected value of Y, denoted as E[Y], and express the result as an irreducible fraction p/q, where p and q are integers. This problem elegantly combines concepts of uniform distributions, conditional probability, and expectation, providing a rich exercise in probabilistic reasoning.

Understanding the Problem

Before diving into the solution, let's dissect the problem statement to ensure we grasp the underlying principles. The key elements are:

• Uniform Distribution: A uniform distribution implies that every value within a given interval has an equal probability of being selected. For X, this means any value between 0 and 1 is equally likely. Similarly, for Y, any value between X² and X is equally probable, given a specific value of X.
• Conditional Probability: The distribution of Y is conditional on the value of X. In other words, the range from which Y is sampled (X², X) depends directly on the value that X takes. This dependency is crucial for understanding the joint behavior of X and Y.
• Expectation: The expectation E[Y] represents the average value of Y over many repeated trials. It's a weighted average, where the weights are the probabilities of observing each possible value of Y. In this context, we need to consider the distribution of Y across all possible values of X.

Methodology

To solve this problem, we will leverage the concept of iterated expectations. The law of iterated expectations (also known as the tower rule) states that for random variables X and Y, the expected value of Y can be calculated as:

E[Y] = E[E[Y|X]]

Where E[Y|X] is the conditional expectation of Y given X. This rule is particularly powerful when dealing with conditional distributions, as it allows us to break down a complex expectation into simpler steps.

In our case, we will first find E[Y|X], which is the expected value of Y given a specific value of X. Since Y is uniformly distributed between X² and X, its conditional expectation is simply the midpoint of this interval. Then, we will compute the expectation of E[Y|X] over all possible values of X, which are uniformly distributed between 0 and 1. This will give us the final value of E[Y].

Step-by-Step Solution

1. Finding the Conditional Expectation E[Y|X]

Given that Y is uniformly distributed on the interval (X², X), the conditional expectation of Y given X is the average of the interval's endpoints. Therefore:

E[Y|X] = (X² + X) / 2

This represents the average value of Y for a specific value of X. For instance, if X = 0.5, then Y is uniformly distributed between 0.25 and 0.5, and its expected value is (0.25 + 0.5) / 2 = 0.375.

2. Computing the Expectation of E[Y|X]

Now that we have the conditional expectation, we need to find the expectation of this expression over all possible values of X. Since X is uniformly distributed on (0,1), we can compute this expectation using an integral:

E[E[Y|X]] = E[(X² + X) / 2] = ∫₀¹ [(x² + x) / 2] dx

This integral represents the average value of the function (x² + x) / 2 over the interval from 0 to 1. To evaluate this integral, we first find the antiderivative of the integrand:

∫ [(x² + x) / 2] dx = (1/2) ∫ (x² + x) dx = (1/2) [x³/3 + x²/2] + C

Where C is the constant of integration. Now, we evaluate the definite integral by plugging in the limits of integration:

∫₀¹ [(x² + x) / 2] dx = (1/2) [(1³/3 + 1²/2) - (0³/3 + 0²/2)]

= (1/2) [1/3 + 1/2] = (1/2) [2/6 + 3/6] = (1/2) [5/6] = 5/12

Therefore, the expected value of E[Y|X] is 5/12.

3. The Expected Value of Y

By the law of iterated expectations, E[Y] = E[E[Y|X]]. We have computed E[E[Y|X]] to be 5/12. Thus:

E[Y] = 5/12

This result signifies that, on average, the value of Y will be 5/12 when X is sampled uniformly from (0,1) and Y is subsequently sampled uniformly from (X², X).

Final Answer

The expected value E[Y] is 5/12. This fraction is already in its irreducible form, as 5 and 12 have no common factors other than 1. Therefore, p = 5 and q = 12.

In solving this problem, we utilized several fundamental concepts and theorems from probability theory:

1. Uniform Distribution: The uniform distribution plays a central role in this problem. Understanding that X and Y are sampled uniformly within their respective intervals is crucial for determining their probability densities and expected values.
2. Conditional Probability: The distribution of Y is conditional on the value of X, which means the probability of Y taking a particular value depends on the value of X. This conditional dependency is essential for setting up the problem correctly.
3. Expectation: The concept of expectation is fundamental to probability and statistics. The expected value of a random variable represents its average value over many repeated trials. In this case, we sought to find the expected value of Y.
4. Law of Iterated Expectations: This powerful theorem allows us to compute the expectation of a random variable by first conditioning on another random variable and then taking the expectation over the conditioning variable. The formula E[Y] = E[E[Y|X]] is the cornerstone of our solution.
5. Integration: To compute the expectation of E[Y|X] over the continuous range of X, we employed integration. This is a standard technique in probability theory for dealing with continuous random variables.

When tackling problems involving conditional expectations, it's easy to make mistakes if one isn't careful. Here are a few common pitfalls to avoid:

1. Incorrectly Determining the Conditional Expectation: The conditional expectation E[Y|X] must be computed correctly. In this case, since Y is uniformly distributed on (X², X), the conditional expectation is simply the midpoint of the interval. Mistakes can arise if one doesn't fully grasp the properties of uniform distributions.
2. Forgetting the Law of Iterated Expectations: This theorem is the key to solving this problem. Neglecting to use it or applying it incorrectly will lead to an incorrect solution.
3. Errors in Integration: When computing the expectation of E[Y|X] over the range of X, it's crucial to perform the integration accurately. Mistakes in finding the antiderivative or evaluating the definite integral can result in an incorrect answer.
4. Ignoring the Conditional Dependency: Failing to recognize that the distribution of Y is conditional on X is a significant error. This dependency must be considered when setting up the problem and applying the law of iterated expectations.

This problem serves as an excellent starting point for exploring more complex probability scenarios involving conditional distributions and expectations. Here are some avenues for further investigation:

1. Different Distributions: Consider what happens if X and Y follow different distributions. For example, what if X is exponentially distributed and Y is uniformly distributed on an interval that depends on X?
2. Multiple Variables: Extend the problem to involve more than two random variables. For instance, introduce a third random variable Z that is conditionally dependent on Y. How would you compute E[Z] in this case?
3. Applications in Statistics: Explore how these concepts apply to real-world statistical problems. Conditional expectations are used extensively in regression analysis, Bayesian inference, and other statistical techniques.
4. Simulation: Use computer simulations to verify the analytical solution. Generate many samples of X and Y according to the given distributions and compute the average value of Y. This can provide a practical check on your theoretical calculations.

In conclusion, we have successfully computed the expected value of Y, E[Y], given that X is sampled uniformly from (0,1) and Y is sampled uniformly from (X², X). By leveraging the law of iterated expectations and performing the necessary integration, we found that E[Y] = 5/12. This problem underscores the importance of understanding conditional distributions, expectations, and key theorems in probability theory. Furthermore, it provides a foundation for exploring more intricate probabilistic models and their applications in various fields.

This exercise demonstrates how mathematical principles can be applied to analyze probabilistic scenarios and derive meaningful results. By carefully dissecting the problem, applying the appropriate tools, and avoiding common pitfalls, we can navigate the complexities of probability theory and gain a deeper appreciation for its power and elegance.