91Ó°ÊÓ

A Conditional Probability Density Function (p.d.f.) describes the probability of a random variable given that another variable is known. In the exercise, we look at \(f(x_1 | x_2) = c_1 \frac{x_1}{x_2^2}\) for the range \(0 < x_1 < x_2\). The conditional p.d.f. assumes you already know the value of \(x_2\), and it provides the likelihood of each \(x_1\) within the defined range. To ensure it is a valid probability density function, it must integrate to 1 over \(x_1\). By solving the integral \(\int_0^{x_2} c_1 \frac{x_1}{x_2^2} \, dx_1 = 1\), we determine that the constant \(c_1 = 1\). This indicates that, given \(x_2\), the p.d.f. equally weighs possibilities for \(x_1\) from 0 to \(x_2\). Conditional p.d.fs are valuable for narrowing the focus from a wider probability distribution to a specific case.

A Marginal Probability Density Function (p.d.f.) represents the distribution of a single random variable by integrating over the possible values of another variable. In this problem, we have \(f_2(x_2) = c_2 x_2^4\) for \(0 < x_2 < 1\). It gives the probability of various values of \(x_2\) while ignoring the influence of \(x_1\).To make sure \(f_2(x_2)\) is a legitimate p.d.f., the function should integrate to 1 over its entire range \(0 < x_2 < 1\). Calculating \(\int_0^1 c_2 x_2^4 \, dx_2 = \frac{c_2}{5}\), and equating it to 1, we find \(c_2 = 5\). This constant ensures that the p.d.f. accounts for all possible scenarios involving \(x_2\). Marginal p.d.fs are essential in simplifying complex multidimensional problems by focusing on individual random variables.

Once we have the joint and conditional p.d.fs, we can calculate probabilities of specific events. The joint p.d.f. would be the product of the conditional and marginal p.d.fs. Thus, \(f(x_1, x_2) = 5x_1x_2^2\).For probability questions such as \(\operatorname{Pr}\left(\frac{1}{4} < X_1 < \frac{1}{2} \mid X_2 = \frac{5}{8}\right)\), we integrate the conditional p.d.f. over the range \(\left(\frac{1}{4}, \frac{1}{2}\right)\) using the given \(x_2\) value. Here, the result is 0.96, indicating a high likelihood for this range under the condition.To find \(\operatorname{Pr}\left(\frac{1}{4} < X_1 < \frac{1}{2}\right)\), the joint p.d.f. is integrated over \(X_1: \frac{1}{4}\) to \(\frac{1}{2}\) and \(X_2: x_1\) to 1. The final answer of 0.3205 suggests that while reasonably likely, it's not as probable as within the conditioned range. Using integrals with p.d.fs enables precise probability determination for continuous random variables.