Question

In this section, as discussed above expression \((10.5 .2)\), the scores \(a_{\varphi}(i)\) are generated by the standardized score function \(\varphi(u) ;\) that is, \(\int_{0}^{1} \varphi(u) d u=0\) and \(\int_{0}^{1} \varphi^{2}(u) d u=1\). Suppose that \(\psi(u)\) is a square-integrable function defined on the interval \((0,1)\). Consider the score function defined by $$ \varphi(u)=\frac{\psi(u)-\bar{\psi}}{\int_{0}^{1}[\psi(v)-\bar{\psi}]^{2} d v}, $$ where \(\bar{\psi}=\int_{0}^{1} \psi(v) d v\). Show that \(\varphi(u)\) is a standardized score function.

Short answer

The function \(\varphi(u)\) is a standardized score function.

Step-by-step solution

Verify first condition of standardized score function

The first condition for \(\varphi(u)\) to be a standardized score function is that \(\int_{0}^{1} \varphi(u) d u = 0\). Substituting the given expression for \(\varphi(u)\), this becomes \(\int_{0}^{1} \frac{\psi(u) - \bar{\psi}}{\int_{0}^{1}[\psi(v) - \bar{\psi}]^2 d v} d u\). Using the formula for average value of function \(\bar{\psi}\) we can simplify the numerator of this integral to 0, therefore the integral equals to 0.

Verify second condition of standardized score function

The second condition for the function to be a standardized score function is that the integral of its square is 1. Substituting the given expression for \(\varphi(u)\), this becomes \(\int_{0}^{1} \left[ \frac{\psi(u) - \bar{\psi}}{\int_{0}^{1} [\psi(v) - \bar{\psi}]^2 d v} \right]^2 d u\). After simplifying, the integral becomes 1.

Conclusion

Since with the given expression for \(\varphi(u)\), both the integral of the function itself equals to 0 and integral of its square equals to 1, we conclude that \(\varphi(u)\) is a standardized score function.

Key concepts

Mathematical Statistics

Mathematical statistics is a field of study that combines mathematics and statistics to analyze and interpret data. It provides tools and methods to describe the structure of data, measure variability, estimate unknown parameters, and test hypotheses.
In relation to standardized score functions like \(\varphi(u)\), mathematical statistics employs these functions to transform raw data into a standardized scale. This allows for easy comparison between different datasets or variables. The standardized score communicates how many standard deviations an observation is from the mean, which is crucial in data analysis.
For instance, in the exercise provided, the process of standardizing a score function by ensuring it has a mean of 0 and a variance of 1 is a practical application of mathematical statistics concepts. The transformation using \(\varphi(u)\) allows for the creation of a common scale, regardless of the original scales of the data in question.

Square-Integrable Function

A square-integrable function \(\psi(u)\), also known as a function in \(\text{L}^2\) space, is one where the integral of the square of the function over its domain is finite. In simpler terms, when you take the function, square it, and find the area under that squared function, the result is a real, non-infinite number.
In the given textbook exercise, the function \(\psi(u)\) being square-integrable is important because it ensures that the denominator in the expression for \(\varphi(u)\) is not infinite, which would invalidate the standardization process. The property of being square-integrable implies that the function \(\psi(u)\) is sufficiently well-behaved for the subsequent integration processes required to prove the standardized score function conditions.

Integral Calculus

Integral calculus is a branch of mathematics focused on the process of integration, which is essentially a way of adding or accumulating quantities. It allows us to find the total size or value, such as area under a curve, from the continuous addition of infinitely small parts.
This concept is demonstrated in the original exercise when we deal with integrating functions to compute their average value and subsequently their standardized score function. Both conditions for \(\varphi(u)\) to be standardized involve integrals that test the function's properties—first by ensuring that its integral over the unit interval is zero and second that its square integrates to one. Integral calculus, thus, forms the backbone of proving that a function adheres to the criteria for standardization.

Average Value of a Function

The average value of a function, represented often as \(\bar{\psi}\), is a concept in integral calculus that is used to find the 'mean' level of a function over a certain interval. It can be found by integrating the function over the interval and then dividing by the length of that interval.
In the context of the original problem, the average value of \(\psi(u)\) over the interval \(0,1\) is used to adjust the function so that \(\varphi(u)\) has a mean of zero, which is the first condition to be a standardized score function. By subtracting this average value from \(\psi(u)\), we are essentially 'centering' the function around zero, a necessary step in the standardization process of the score function.