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A Probability Density Function (PDF) represents a continuous probability distribution. It helps in understanding how probabilities are distributed over the range of possible values.
The PDF of a random variable provides insight into the likelihood of different outcomes. For example, in the context of the truncated normal distribution, we have a PDF of the form:

• Let’s denote it as \(g(y) = \frac{\phi(y)}{\Phi(b) - \Phi(a)} \), where \( a < y < b \).
• Outside the interval \((a, b)\), the value is zero, meaning the probability of \(Y\) being outside this interval is nil.
• The functions \(\phi(x)\) and \(\Phi(x)\) represent the standard normal distribution's PDF and cumulative distribution function (CDF), respectively.

In the exercise, this truncated nature of the PDF focuses only on values of \(Y\) between limits \(a\) and \(b\), redistributing the probabilities defined by the standard normal distribution only within this interval. Understanding this redistribution is key to mastering concepts related to truncated distributions.

The Expected Value, or mean of a random variable, gives us the average value we expect to observe in a probability distribution. In this exercise concerning a truncated normal distribution,
the expected value is calculated by integrating the product of the variable \(y\) and its PDF \(g(y)\) over the interval \([a, b]\). The equation for the expected value \(E(Y)\) is:

• \(E(Y) = \int_{a}^{b} y \cdot \frac{\phi(y)}{\Phi(b) - \Phi(a)} \, dy\)
• Since \(g(y)\) only exists between \(a\) and \(b\), our limits of integration are also \(a\) to \(b\).

This approach takes into account only the non-zero probability areas. By performing this integral, we can isolate the expected value of \(Y\) using the given PDF.

Integration by Parts is a technique that simplifies the process of integrating products of functions by transforming one kind of integral into another, usually more manageable, integral. It is based on the product rule for differentiation.To implement Integration by Parts in our example, consider the formula:

• \( \int u \, dv = u \cdot v - \int v \, du \)

In the truncated distribution problem, it is applied as follows:

• Select \(u = y\) and let \(dv = \frac{\phi(y)}{\Phi(b)-\Phi(a)} \, dy\).
• This choice gives \(du = dy\) and the integral of \(dv\), which is \(v = -\frac{\phi(y)}{\Phi(b)-\Phi(a)}\).

Substituting these back into the integration by parts formula, the calculation simplifies with additional algebraic manipulation, integrating the constant term and recognizing the integration of the full PDF spans probability one.
This culminates in deriving the expected value \( E(Y) = \frac{\phi(a) - \phi(b)}{\Phi(b) - \Phi(a)} \), thus seamlessly applying this rule makes finding the expected value more intuitive and accessible.