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)
10th Edition
ISBN:9780134753119
Author:Sheldon Ross
Publisher:Sheldon Ross
Chapter1: Combinatorial Analysis
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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.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F76eec8ec-d87c-4fdd-96a3-96e8325f09a2%2Fd2f48467-11cb-4fdf-b3bb-3a327a94788b%2Fb49cgt_processed.png&w=3840&q=75)
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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