Functions of the standard normal
The random variable 𝑋 has a standard normal distribution. Find the PDF of the random variable 𝑌, where:
𝑌=5𝑋−7.
ANSWER: 𝑓𝑌(𝑦)=1/5𝑓_𝑋((𝑦+7)//5)
𝑌=𝑋^2−2𝑋. For 𝑦≥−1,
ANSWER: (𝑓_𝑋(1+√(𝑦+1))+𝑓_𝑋(1−√(𝑦+1)))/(2√(𝑦+1))
or
ANSWER: (𝑓_𝑋(1+sqrt(𝑦+1))+𝑓_𝑋(1−sqrt(𝑦+1)))/(2*sqrt(𝑦+1))
To find the PDF of the random variable 𝑌, we need to express 𝑌 in terms of the standard normal random variable 𝑋 and apply the appropriate transformation.
1. 𝑌 = 5𝑋 - 7:
- We start with the equation 𝑌 = 5𝑋 - 7.
- To find the PDF of 𝑌, we need to transform it to an expression of 𝑋.
- Rearranging the equation, we get 𝑋 = (𝑌 + 7) / 5.
- Now, we substitute this value of 𝑋 into the PDF of the standard normal distribution.
- The PDF of the standard normal distribution, denoted by 𝑓_𝑋(𝑥), is conventionally represented as 𝑓(𝑥) = (1/√(2π)) * e^(-𝑥^2/2).
- Substituting 𝑋 = (𝑌 + 7) / 5 into 𝑓(𝑥), we get 𝑓(𝑌) = (1/5𝑓_𝑋((𝑌 + 7) / 5)).
- Simplifying, 𝑓(𝑌) = (1/5𝑓_𝑋((𝑌 + 7) // 5)), where "//" denotes integer division.
2. 𝑌 = 𝑋^2 - 2𝑋, for 𝑦 ≥ -1:
- We have the equation 𝑌 = 𝑋^2 - 2𝑋.
- Expressing 𝑋 in terms of 𝑌, we need to complete the square.
- Rearrange the equation to 𝑌 = 𝑋^2 - 2𝑋 + 1 - 1.
- This simplifies to 𝑌 = (𝑋 - 1)^2 - 1.
- The left-hand side of the equation represents 𝑌, while the right-hand side represents (𝑋 - 1)^2 - 1.
- Solving for 𝑋, we get 𝑋 = 1 ± √(𝑌 + 1).
- For 𝑦 ≥ -1, we choose the positive value of √(𝑌 + 1) to ensure 𝑋 ≥ -1.
- Substituting 𝑋 = 1 + √(𝑌 + 1) into the PDF of the standard normal distribution, we get 𝑓(𝑌) = 𝑓_𝑋(1 + √(𝑌 + 1)).
- Similarly, substituting 𝑋 = 1 - √(𝑌 + 1), we get 𝑓(𝑌) = 𝑓_𝑋(1 - √(𝑌 + 1)).
- Taking the average of these two probabilities (since there are two possible values of 𝑋), we obtain 𝑓(𝑌) = (𝑓_𝑋(1 + √(𝑌 + 1)) + 𝑓_𝑋(1 - √(𝑌 + 1))) / 2.
- Simplifying, 𝑓(𝑌) = (𝑓_𝑋(1 + √(𝑌 + 1)) + 𝑓_𝑋(1 - √(𝑌 + 1))) / (2√(𝑌 + 1)).
Note: The PDF 𝑓_𝑋(𝑥) is the probability density function of the standard normal distribution, which is given by the equation 𝑓_𝑋(𝑥) = (1/√(2π)) * e^(-𝑥^2/2).