https://file.aiquickdraw.com/mathbot/file/172654679995012bupbld.jpg

We are given that \( X \sim N(0, 1) \), meaning that \( X \) is a standard normal random variable, and \( Y = |X| \). We are asked to find the probability density function (p.d.f.), mean, and variance of \( Y \).

### 1. Finding the p.d.f. of \( Y \):
Since \( Y = |X| \), this is a transformation of the random variable \( X \). The distribution of \( Y \) is related to the absolute value of a standard normal distribution.

#### Step 1: CDF of \( Y \)
The cumulative distribution function (CDF) of \( Y \) is:
\[
F_Y(y) = P(Y \leq y) = P(|X| \leq y) = P(-y \leq X \leq y)
\]
For a standard normal variable \( X \), this is:
\[
F_Y(y) = P(-y \leq X \leq y) = \Phi(y) - \Phi(-y)
\]
where \( \Phi(y) \) is the CDF of the standard normal distribution. Since \( \Phi(-y) = 1 - \Phi(y) \), we have:
\[
F_Y(y) = 2\Phi(y) - 1
\]
for \( y \geq 0 \) (since \( Y \) cannot be negative).

#### Step 2: p.d.f. of \( Y \)
The p.d.f. \( f_Y(y) \) is obtained by differentiating the CDF:
\[
f_Y(y) = \frac{d}{dy} \left( 2\Phi(y) - 1 \right) = 2 \phi(y)
\]
where \( \phi(y) \) is the p.d.f. of the standard normal distribution:
\[
\phi(y) = \frac{1}{\sqrt{2\pi}} e^{-y^2 / 2}
\]
Thus, the p.d.f. of \( Y \) is:
\[
f_Y(y) = \frac{2}{\sqrt{2\pi}} e^{-y^2 / 2}, \quad y \geq 0
\]

### 2. Finding the mean of \( Y \):
The mean of \( Y \) is given by:
\[
\mathbb{E}[Y] = \int_0^\infty y f_Y(y) \, dy
\]
Substituting the p.d.f. \( f_Y(y) \):
\[
\mathbb{E}[Y] = \int_0^\infty y \cdot \frac{2}{\sqrt{2\pi}} e^{-y^2 / 2} \, dy
\]
This is a standard integral for the mean of the absolute value of a normal variable, and it is known that:
\[
\mathbb{E}[Y] = \sqrt{\frac{2}{\pi}}
\]

### 3. Finding the variance of \( Y \):
The variance of \( Y \) is:
\[
\text{Var}(Y) = \mathbb{E}[Y^2] - (\mathbb{E}[Y])^2
\]
We first compute \( \mathbb{E}[Y^2] \), which is:
\[
\mathbb{E}[Y^2] = \int_0^\infty y^2 f_Y(y) \, dy
\]
Substituting \( f_Y(y) \):
\[
\mathbb{E}[Y^2] = \int_0^\infty y^2 \cdot \frac{2}{\sqrt{2\pi}} e^{-y^2 / 2} \, dy
\]
This integral is known to evaluate to 1 (since \( Y^2 \) is simply \( X^2 \), and \( X^2 \) has mean 1 for a standard normal variable).

Thus:
\[
\text{Var}(Y) = 1 - \left( \sqrt{\frac{2}{\pi}} \right)^2 = 1 - \frac{2}{\pi}
\]
Therefore:
\[
\text{Var}(Y) = 1 - \frac{2}{\pi}
\]

### Final Results:
- The p.d.f. of \( Y \) is \( f_Y(y) = \frac{2}{\sqrt{2\pi}} e^{-y^2 / 2}, \quad y \geq 0 \)
- The mean of \( Y \) is \( \mathbb{E}[Y] = \sqrt{\frac{2}{\pi}} \)
- The variance of \( Y \) is \( \text{Var}(Y) = 1 - \frac{2}{\pi} \)

Let me know if you want further details or clarifications.

Here are 5 relative questions to think about:
1. How is the absolute value transformation of a random variable related to its original distribution?
2. What are the properties of the standard normal distribution?
3. How can the moment-generating function be used to compute moments like the mean and variance?
4. How would the results change if \( X \) followed a normal distribution with a nonzero mean?
5. Can you derive the p.d.f. of \( Z = |X| \) where \( X \sim N(\mu, \sigma^2) \) for arbitrary \( \mu \) and \( \sigma \)?

**Tip**: Knowing how to manipulate the CDF and p.d.f. of transformed variables can be extremely useful when dealing with non-standard distributions.

Let X ~ N(0, 1) and Y = |X|. Find the p.d.f. f_Y(y), the mean, and the variance of Y.

This problem focuses on finding the p.d.f., mean, and variance of Y = |X| where X follows a standard normal distribution (N(0,1)). Learn the detailed process involving transformations of variables and properties of the normal distribution.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation