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Let X be a random variable with CDF Fx(z). Find the CDF of aX+b first for a > 0, then for a < 0.

Let X be a random variable with CDF Fx(z). Find the CDF of aX+b first for a > 0, then for a < 0.

A First Course in Probability (10th Edition)
A First Course in Probability (10th Edition)
10th Edition
ISBN:9780134753119
Author:Sheldon Ross
Publisher:Sheldon Ross
Chapter1: Combinatorial Analysis
Section: Chapter Questions
Problem 1.1P: a. How many different 7-place license plates are possible if the first 2 places are for letters and...
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**Problem Statement:** Let \( X \) be a random variable with cumulative distribution function (CDF) \( F_X(x) \). Find the CDF of \( aX + b \) first for \( a > 0 \), then for \( a < 0 \). **Explanation:** This problem involves transforming a random variable \( X \) and finding the cumulative distribution function of the linear transformation \( aX + b \). The process involves: 1. **For \( a > 0 \):** - The CDF of \( aX + b \) is derived by considering the probability: \[ P(aX + b \leq y) = P\left(X \leq \frac{y-b}{a}\right) \] - Therefore, the CDF of \( aX + b \) is: \[ F_{aX+b}(y) = F_X\left(\frac{y-b}{a}\right) \] 2. **For \( a < 0 \):** - The inequality changes direction, so: \[ P(aX + b \leq y) = P\left(X \geq \frac{y-b}{a}\right) \] - Since \( P(X \geq z) = 1 - F_X(z) \), the CDF becomes: \[ F_{aX+b}(y) = 1 - F_X\left(\frac{y-b}{a}\right) \] This analysis helps in understanding how linear transformations affect the distribution of random variables and how to compute their cumulative distribution functions accordingly.
Transcribed Image Text:**Problem Statement:** Let \( X \) be a random variable with cumulative distribution function (CDF) \( F_X(x) \). Find the CDF of \( aX + b \) first for \( a > 0 \), then for \( a < 0 \). **Explanation:** This problem involves transforming a random variable \( X \) and finding the cumulative distribution function of the linear transformation \( aX + b \). The process involves: 1. **For \( a > 0 \):** - The CDF of \( aX + b \) is derived by considering the probability: \[ P(aX + b \leq y) = P\left(X \leq \frac{y-b}{a}\right) \] - Therefore, the CDF of \( aX + b \) is: \[ F_{aX+b}(y) = F_X\left(\frac{y-b}{a}\right) \] 2. **For \( a < 0 \):** - The inequality changes direction, so: \[ P(aX + b \leq y) = P\left(X \geq \frac{y-b}{a}\right) \] - Since \( P(X \geq z) = 1 - F_X(z) \), the CDF becomes: \[ F_{aX+b}(y) = 1 - F_X\left(\frac{y-b}{a}\right) \] This analysis helps in understanding how linear transformations affect the distribution of random variables and how to compute their cumulative distribution functions accordingly.
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