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A Probability Density Function (PDF) plays a crucial role in statistics, particularly in continuous probability distributions. It describes the likelihood of a random variable taking on a particular value. However, unlike probabilities in a discrete setup, for continuous variables, the exact probability of the variable taking a specific value is zero. Instead, the PDF is used to calculate the probability over an interval.

Key features of a PDF include:

• It must be non-negative for all possible values of the random variable.
• The integral of the PDF over its entire space must equal 1, reflecting a probability distribution that encompasses all possible outcomes.

In the given exercise, we check these properties for the function \(f(x) = c 2^{-x^2}\) to determine if it is a valid normal PDF. By transforming the function using exponential rules, the function is analyzed to satisfy these critical conditions. The importance lies in ensuring that \(f(x)\) represents a true probability distribution over the continuous range of \(x\).

The concept of non-negativity is vital for functions representing probability density functions. For any function to be considered a PDF, it must be non-negative for all inputs within its domain. Simply put, the probability cannot be negative since probabilities range between 0 and 1.

For the function \(f(x) = c 2^{-x^2}\), it is evident that:

• The term \(2^{-x^2}\) is always positive because raising 2 to any real power, even negative, results in a positive number.
• By choosing a positive constant \(c\), the entire function \(f(x)\) becomes non-negative, satisfying the first essential condition of a PDF.

This ensures that the function complies with the practical interpretation of probabilities as non-negative metrics.

Integration is a fundamental process in calculus used to find areas under curves, which directly applies to finding total probabilities under a PDF curve. When dealing with exponential functions, especially those of the form \(e^{-x^2}\), special techniques and known results become handy.

In the solved exercise, the integral of the function \(f(x) = c 2^{-x^2}\) is transformed using the hint \(2 = e^{\log 2}\). This allows the function to be rewritten as \(c e^{-x^2 \log 2}\). Then the integral \(\int_{-\infty}^{+\infty} e^{-x^2}\) is recognized as\(\sqrt{\pi}\), an essential result for Gaussian functions.

This integral result, once calculated, demonstrates that:

• By setting this integral equal to 1 and solving for \(c\), we find that \(c = \frac{1}{\sqrt{2 \pi}}\) provides the normalization constant ensuring \(f(x)\) integrates to 1 over its domain.
• This transformation and integration process shows not only how \(f(x)\) becomes a valid probability distribution but also highlights special techniques used in handling exponential integrals in mathematical statistics.

Hence, through integration and transformation, the property of the PDF totaling to unity is verified, fulfilling the second fundamental condition.