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The Normal Distribution, often called the Gaussian distribution, is a fundamental concept in statistics. It describes a continuous probability distribution characterized by its bell-shaped curve, which is symmetric about its mean.

• The mean, often noted as \(\mu\), represents the average or center of the distribution.
• The variance, noted as \(\sigma^2\), indicates the spread or dispersion of the data within the distribution.

For a standard normal distribution, the mean is 0, and the variance is 1.

In the context of the problem, we have a normal distribution with mean 0 and variance \(\sigma^2\). This means our bell curve is centered at zero, and the spread of data is determined by \(\sigma\). The probability density function (pdf), given for this normal distribution, is:\[f_X(x) = \frac{1}{\sqrt{2\pi \sigma^2}} e^{-\frac{x^2}{2\sigma^2}}\]This formula quantifies the likelihood of a random variable falling within a particular range. The integral of this function over all possible values equals 1, which signifies that the area under the curve, representing the total probability, is complete.

An Absolute Value Transformation involves modifying a random variable by taking its absolute value. In mathematical terms, if you have a variable \(X\), the transformation \(Y = |X|\) converts any negative value of \(X\) into positive while keeping zero and positive values unchanged.

• This transformation is particularly interesting with symmetric distributions, like the normal distribution, because it results in a 'folded' version over the vertical axis at \(x = 0\).
• The absolute value concept in this context generates a probability density function for \(Y\), which describes how the absolute values are distributed.

The resulting pdf for \(Y\) is calculated by considering both the positive and negative halves of the normal distribution. For this reason, the pdf of the original distribution (on which the transformation is applied) is multiplied by 2, reflecting contributions from both sides of the axis:\[f_Y(y) = \frac{2}{\sqrt{2\pi \sigma^2}} e^{-\frac{y^2}{2\sigma^2}}, \quad y > 0\]At \(y = 0\), however, the pdf is zero because continuous distributions do not assign probability to single points.

Continuous Distributions are used to model random variables that can take any value within a specified range. Unlike discrete distributions, which are concerned with distinct values, continuous distributions function over intervals.

• The probability that a continuous random variable is exactly equal to a single value is zero.
• Instead, we look at the probability of the variable falling within a particular range, which is determined by the area under the curve of its probability density function (pdf).

The normal distribution is a prime example of a continuous distribution. In our exercise, converting \(X\) to \(Y = |X|\) results in a continuous distribution for \(Y\) as well. This is evident from the derived pdf for \(Y\), which is expressed as:\[f_Y(y) = \begin{cases} \frac{2}{\sqrt{2\pi \sigma^2}} e^{-\frac{y^2}{2\sigma^2}}, & y > 0 \0, & y = 0\end{cases}\]This function describes how the probability is distributed across possible values of \(Y\), emphasizing that the continuous probability spreads over an interval, excluding individual points.