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?

  • $\begingroup$ If you write $lag$, do you mean logarithm $\log$? $\endgroup$ Oct 23, 2015 at 13:40
  • 3
    $\begingroup$ "Normalization" is a very overloaded term -- much like "linear." $\endgroup$
    – Sycorax
    Oct 23, 2015 at 13:56
  • 2
    $\begingroup$ Subtracting the mean and dividing by the SD is usually called standardization (although some people overload that term, too). $\endgroup$
    – whuber
    Oct 23, 2015 at 14:33
  • $\begingroup$ @user777, could you clarify what you mean by "overloaded"? Do you mean that there are different meanings to it? $\endgroup$ Nov 15, 2015 at 12:56
  • $\begingroup$ @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. $\endgroup$
    – Sycorax
    Nov 15, 2015 at 18:26

1 Answer 1


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.

Then there is nothing wrong with fitting any distribution on the returns. Kernel estimation is indeed smoother (than producing the empirical distribution).


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.