Math Problem Statement

Let X be a continous random variable with pdf fX(x) and cdf FX(x). Let A be a subset of the real line. Let I_A(x) be the indicator function for A. Find an expression for the cdf of Y. first find the probablity mass function for Y.

Let X be a continuous random variable with pdf fX(x) and cdf FX(x). Let A be a subset of the real line. Let I_A(x) be the indicator function for A. Find an expression for the cdf of Y. First, find the probability mass function for Y.

Solution

You provided an image representing an indicator function IA(x)I_A(x) for a set ARA \subset \mathbb{R}, and you described a continuous random variable XX with a probability density function (PDF) fX(x)f_X(x) and cumulative distribution function (CDF) FX(x)F_X(x). Let Y=IA(X)Y = I_A(X) be the transformation you're interested in.

1. Understanding the Indicator Function:

The indicator function IA(x)I_A(x) is defined as:

1 & \text{if } x \in A, \\ 0 & \text{if } x \notin A. \end{cases}$$ Thus, $$Y = I_A(X)$$ will take the value 1 if $$X \in A$$, and 0 otherwise. ### 2. Probability Mass Function (PMF) of $$Y$$: Since $$Y$$ is derived from a continuous random variable $$X$$, but takes discrete values (either 0 or 1), we can define the probability mass function (PMF) for $$Y$$. $$P(Y = 1) = P(X \in A) = \int_A f_X(x) \, dx$$ This represents the probability that $$X$$ falls within the set $$A$$, which makes $$Y = 1$$. $$P(Y = 0) = P(X \notin A) = \int_{A^c} f_X(x) \, dx$$ This represents the probability that $$X$$ does not fall within the set $$A$$, meaning $$Y = 0$$. Since the total probability must sum to 1, you could also express it as: $$P(Y = 0) = 1 - P(X \in A)$$ ### 3. Cumulative Distribution Function (CDF) of $$Y$$: To find the cumulative distribution function $$F_Y(y)$$ of $$Y$$, we need to compute: $$F_Y(y) = P(Y \leq y)$$ For $$Y$$, which only takes values 0 and 1, the CDF will be: - $$F_Y(y) = 0$$ for $$y < 0$$ (since $$Y \geq 0$$), - $$F_Y(y) = P(Y = 0) = 1 - P(X \in A)$$ for $$0 \leq y < 1$$, - $$F_Y(y) = P(Y \leq 1) = 1$$ for $$y \geq 1$$ (since $$P(Y = 1) + P(Y = 0) = 1$$). Thus, the CDF of $$Y$$ is a step function: $$F_Y(y) = \begin{cases} 0 & \text{if } y < 0, \\ 1 - \int_A f_X(x) \, dx & \text{if } 0 \leq y < 1, \\ 1 & \text{if } y \geq 1. \end{cases}$$ ### Summary: 1. **PMF of $$Y$$:** - $$P(Y = 1) = \int_A f_X(x) \, dx$$, - $$P(Y = 0) = 1 - \int_A f_X(x) \, dx$$. 2. **CDF of $$Y$$:** $$F_Y(y) = \begin{cases} 0 & \text{if } y < 0, \\ 1 - \int_A f_X(x) \, dx & \text{if } 0 \leq y < 1, \\ 1 & \text{if } y \geq 1. \end{cases}$$ --- Would you like more details on how to apply this to specific cases or to discuss properties of the distribution? Here are some questions for further exploration: 1. What are some examples of sets $$A$$ on the real line that could change the form of the PMF and CDF? 2. How would this work if $$A$$ is an interval, say $$A = [a, b]$$? 3. How does this method generalize to other types of random variables besides continuous ones? 4. How would you compute the expected value $$E[Y]$$ based on the PDF of $$X$$? 5. Can this indicator function approach help with modeling other types of random processes? **Tip**: When working with indicator functions, always carefully check the definition of the set $$A$$ to ensure proper integration boundaries when calculating probabilities.

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Math Problem Analysis

Mathematical Concepts

Probability Theory
Continuous Random Variables
Indicator Functions
Probability Mass Function (PMF)
Cumulative Distribution Function (CDF)

Formulas

Indicator Function: I_A(x) = {1 if x ∈ A, 0 if x ∉ A}
PMF of Y: P(Y = 1) = ∫_A f_X(x) dx, P(Y = 0) = 1 - ∫_A f_X(x) dx
CDF of Y: F_Y(y) = {0 if y < 0, 1 - ∫_A f_X(x) dx if 0 ≤ y < 1, 1 if y ≥ 1}

Theorems

Properties of Probability Distributions
Fundamental Theorem of Calculus

Suitable Grade Level

Undergraduate Level