Question

Let us choose at random a point from the interval \((0,1)\) and let the random variable \(X_{1}\) be equal to the number which corresponds to that point. Then choose a point at random from the interval \(\left(0, x_{1}\right)\), where \(x_{1}\) is the experimental value of \(X_{1}\); and let the random variable \(X_{2}\) be equal to the number which corresponds to this point. (a) Make assumptions about the marginal pdf \(f_{1}\left(x_{1}\right)\) and the conditional pdf \(f_{2 \mid 1}\left(x_{2} \mid x_{1}\right)\) (b) Compute \(P\left(X_{1}+X_{2} \geq 1\right)\) (c) Find the conditional mean \(E\left(X_{1} \mid x_{2}\right)\).

Short answer

The marginal pdf \(f_{1}(x_{1}) = 1\), the conditional pdf \(f_{2 | 1}(x_{2} | x_{1}) = \frac{1}{x_{1}}\), the probability \(\P(X_{1}+X_{2} \geq 1) = 0.5\), and the conditional mean \(E\left(X_{1} | x_{2}\right) = \frac{1 - x_{2}}{2}\).

Step-by-step solution

Assumptions about pdfs

First, assume that the marginal pdf of \(X_{1}\), \(f_{1}(x_{1})\), is uniformly distributed over (0,1). Therefore, \(f_{1}(x_{1}) = 1\) for \(0 \leq x_{1} \leq 1\). Secondly, given \(X_{1} = x_{1}\), the conditional pdf of \(X_{2}\), \(f_{2 | 1}(x_{2} | x_{1})\), is uniformly distributed over (0, \(x_{1}\)). Therefore, \(f_{2 | 1}(x_{2} | x_{1}) = \frac{1}{x_{1}}\) for \(0 \leq x_{2} \leq x_{1}\).

Calculate the Probability

The probability that the sum of \(X_{1}\) and \(X_{2}\) is greater or equal to 1, is equivalent to the integral of their joint pdf over a specific region. By applying the marginal and conditional pdf previously deducted, the joint pdf is given by \(f_{1,2}(x_{1},x_{2}) = f_{1}(x_{1})f_{2 | 1}(x_{2} | x_{1}) = \frac{1}{x_{1}}\) for \(0 \leq x_{2} \leq x_{1} \leq 1\). From the restriction \(X_{1}+X_{2} \geq 1\), we deduce that the region we are interested in is upper bounded by the line \(x_{1} + x_{2} = 1\). Thus, \(\P(X_{1}+X_{2} \geq 1) = \int_{0}^{1/2}\int_{1 - x_{1}}^{x_{1}} \frac{1}{x_{1}} dx_{2} dx_{1}\). Integrating, we find that the desired probability is 1/2.

Find the Conditional Mean

The conditional expectation of \(X_{1}\) given \(X_{2}\) is obtained via the joint pdf re-formulated in terms of \(x_{2}\) and \(x_{1} + x_{2}\), with the limits of \(z(= x_{1} + x_{2})\) running from \(x_{2}\) to 1 and of \(x_{1}\) running from 0 to \(z\). It is given by \(E\left(X_{1} | x_{2}\right) = \int_{x_{2}}^{1}\int_{0}^{z}x_{1}f_{1,2}(x_{1},z - x_{1}) dx_{1} dz\). After calculating, we find that \(E\left(X_{1} | x_{2}\right) = \frac{1 - x_{2}}{2}\).

Key concepts

Probability Density Function (PDF)

Understanding the probability density function (PDF) is fundamental in statistics, especially when dealing with continuous random variables like in our exercise, where a point is chosen at random within an interval. A PDF is like a recipe that describes the likelihood of a random variable taking on values within a continuous range.

In our exercise, we assume the PDF of the random variable X₁, denoted as f₁(x₁), is uniformly distributed over the interval (0,1). This means that the probability is evenly spread throughout the interval, and any point within this range is equally likely to be chosen. Mathematically, for a uniform distribution in an interval (a, b), the PDF is given by f(x) = 1/(b-a). For our interval (0,1), the PDF simplifies to f₁(x₁) = 1, because b-a is equal to 1.

This concept is integral to computing probabilities and expectations involving continuous random variables. Once the PDF is defined, it can be used to calculate probabilities for different outcomes and can be integrated to find cumulative probabilities for ranges of values.

Conditional Probability

Conditional probability is a measure of the probability of an event occurring given that another event has already occurred. This concept extends into continuous probability with the conditional PDF, where we are interested in the likelihood of one variable given some condition on another variable.

In the exercise, the conditional PDF, f_2|1(x₂ | x₁), represents the probability density of X₂ under the condition that X₁ has already taken on a certain value, x₁. Since the second point is chosen from the interval (0, x₁), the conditional PDF is also uniform but with a changing interval that depends on the value of X₁. Therefore, for 0 ≤ x₂ ≤ x₁, the conditional PDF is f_2|1(x₂ | x₁) = 1/x₁.

Understanding conditional probability helps us tackle more complex problems that involve several stages or layers of randomness, like the selection process described in our exercise.

Expectation

Expectation or expected value is a powerful concept in probability that represents the average outcome of a random variable if the experiment were repeated numerous times. For discrete variables, it is calculated by summing the products of each outcome and its probability. For continuous variables, we integrate the product of a variable and its PDF over the possible values.

In our exercise, to find the conditional mean E(X₁ | x₂), we needed to understand the expectation of X₁ given that X₂ has already been selected. The conditional mean represents the expected value of X₁ under those conditions. After integrating the joint PDF over the appropriate range, as indicated in the solution, we obtain the conditional mean as (1 - x₂)/2.

Expectation is not just an average; it's a theoretical mean that serves as a predictive model. In a stochastic or random process, the expectation tells us the long-run average if we could observe the random variable under the same conditions infinitely many times. It's an essential tool for understanding the long-term behavior of random systems.