Normalizing to a benchmark

One often needs to compare data across samples, say, when one or more samples constitute a benchmark. For example, consider comparing interview scores across panels when one of the panels is considered to be the benchmark.

The objective is to transform the data of a sample ${Y}$ so as to match its mean $(μ_{Y})$ and standard deviation $(σ_{Y})$ to that of the benchmark sample. That is, we begin with:

\begin{aligned} E [Y] & = μ_{Y} \\ E [(Y - μ)^{2}] & = σ_{Y}^{2} \end{aligned}

and we need to transform data ${Y}$ as ${Z = c Y + d}$ such that $μ_{Z} = μ_{B}$ and $σ_{Z}^{2} = σ_{B}^{2} .$ where $μ_{B}$ and $σ_{B}$ respectively represent the mean and standard deviation of the benchmark sample.

The method of moments approach in statistics consists of finding parameters $c$ and $d$ such that:

\begin{aligned} E [c Y + d] & = μ_{B} \\ \Rightarrow c^{2} μ_{Y}^{2} + d^{2} + 2 c d μ_{Y} & = μ_{B}^{2} \\ E [((c Y + d) - μ_{B})^{2}] & = σ_{B}^{2} \end{aligned}

Simplifying the third equation, and after a bit a algebra, we get:

\begin{aligned} c & = \frac{σ_{B}}{σ_{Y}} \\ d & = μ_{B} - c E [Y] = μ_{B} - c μ_{Y} \\ = μ_{B} - \frac{σ_{B}}{σ_{Y}} μ_{Y} \end{aligned}

Plugging the finding back into $Z = c Y + d$ tells us that transforming $Y$ as

Z = \frac{σ_{B}}{σ_{Y}} (Y - μ_{Y}) + μ_{B}

ensures that $Z$ has the same mean and standard deviation as the benchmark data. Needless to say that this will work best for situations where the underlying data can be safely assumed to be coming from the Normal distribution or any of its kin (why?).

Looking at the transformation, upon a moment’s reflection it almost appears obvious, doesn’t it? To the point that this whole post seems moot. So, why bother? Well, while this is a simple enough example, the method of moments is important enough in economics and finance for its generalized version to be granted the Nobel prize a couple of years ago.