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.