# How does Kernel density estimation lead to normalization? Does it at all?

I am looking at a report of a company. They are saying they are normalizing data. They have time-series data (i.e. data for certain parameters for sequence of time). They first generate new variables as $100\cdot(x/lag(x)-1)$ Then they do Kernel density estimation. I thought normalizing is subtracting mean and dividing by sd, wasn't it?

1) So, is $100\cdot(x/lag(x)-1)$ a normalization for time-series? EDIT: as I understand this is just calculating the ratio of one year by another year. Is it considered normalization?

2) What is the kernel used for in here then? Just smoothing?

• If you write $lag$, do you mean logarithm $\log$? Commented Oct 23, 2015 at 13:40
• "Normalization" is a very overloaded term -- much like "linear."
– Sycorax
Commented Oct 23, 2015 at 13:56
• Subtracting the mean and dividing by the SD is usually called standardization (although some people overload that term, too).
– whuber
Commented Oct 23, 2015 at 14:33
• @user777, could you clarify what you mean by "overloaded"? Do you mean that there are different meanings to it? Commented Nov 15, 2015 at 12:56
• @user3349993 I mean that there are many meanings of "normalization." It sounds like the company's definition and yours are not the same -- and since companies' procedures tend to be etched in stone, changing their mind will be like shouting into a hurricane. 0-mean, unit variance normalization is one method; but some people "normalize" by putting everything in the $[-1,1]$ or $[0,1]$ intervals. Or they are referring to normalized vectors... Or any of a number of other examples.
– Sycorax
Commented Nov 15, 2015 at 18:26

I assume that $lag$ is the lagged variable. Somehow $lag(x_t)=x_{t-1}$.
Therefore it is not normalization that is performed. They just turn a (price/value?) series into a series of (returns?) ratios : $\tilde{x}_t=\frac{x_t-x_{t-1}}{x_{t-1}}$. The 100 multiplier is just a way to express these ratios in percentage I guess.