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A normal distribution is one of the most important concepts in statistics and is often called a Gaussian distribution. It is a continuous probability distribution characterized by a symmetric, bell-shaped curve.

• The distribution is defined by two parameters: the mean \(\mu\), which determines its center, and the variance \(\sigma^2\), which determines its spread.
• The probability density function (pdf) of a normal distribution is given by: \[ f_X(x) = \frac{1}{\sqrt{2\pi}\sigma} e^{-\frac{(x-\mu)^2}{2\sigma^2}}. \]
• This distribution is commonly used in statistics because of the central limit theorem, which states that the sum of a large number of independent, identically distributed variables is approximately normally distributed, regardless of the original distribution.

In the exercise, \(X\) is assumed to be normally distributed with mean \(\mu\) and variance \(\sigma^2\). This assumption plays a crucial role while applying integration techniques to solve expectation problems.

Integration by parts is a fundamental calculus technique used to integrate the product of two functions. Like the product rule for differentiation, it provides a way to break down a complex integration problem into simpler parts. The formula for integration by parts is expressed as:

• \[ \int u \, dv = uv - \int v \, du. \]

Here, \(u\) and \(dv\) are parts of the original integrand. The choice of \(u\) and \(dv\) is strategic, often simplifying \(\int v \, du\) relative to the original integral.
In the provided solution, integration by parts is employed to handle the expectation \(\mathbb{E}[(X-\mu)g(X)]\). By selecting \(u = g(x)\) and \(dv = (x-\mu)f_X(x)\,dx\), the problem becomes manageable, allowing the transformation of a complex integral into one involving the derivative of \(g(x)\). The trick is to eliminate the difficult term by transferring the differentiation from \(x - \mu\) to \(g(x)\).

Expectation is a fundamental concept in probability that provides the average or mean value of a random variable over repeated trials. When we talk about expectations involving functions of random variables, integration comes into play. For a continuous random variable \(X\) with pdf \(f_X(x)\), the expectation \(\mathbb{E}[h(X)]\) of a function \(h(x)\) is given by:

• \[ \mathbb{E}[h(X)] = \int_{-\infty}^{\infty} h(x) f_X(x) \, dx. \]

In the problem at hand, the expectation \(\mathbb{E}[(X-\mu)g(X)]\) is transformed into an integral, leveraging both the pdf of a normal distribution and integration by parts. The aim is to show its equivalence to \(\sigma^2 \mathbb{E}[g'(X)]\), effectively transitioning from an expectation involving \(X-\mu\) to one concerning the derivative \(g'(x)\). This not only demonstrates the power of integration in simplifying expectations but also highlights their interdependencies, using properties of the normal distribution.