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I have a question in which it asks to verify whether if the Uniform distribution (${\rm Uniform}(a,b)$) is normalized.

  1. For one, what does it mean for any distribution to be normalized?
  2. And two, how do we go about verifying whether a distribution is normalized or not?

I understand by computing $$ \frac{X-\text{mean}}{\text{sd}} $$ we get normalized data, but here it's asking to verify whether a distribution is normalized or not.

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    $\begingroup$ What it means for a distribution to be normalized is not so simple (and it's usually not the distribution itself being normalized, but the random variable). For example, in the case of the uniform, some people may mean "linearly rescaled so as to get a standard uniform" (i.e. to get $a=0$ and $b=1$) ... while another person might mean "linearly rescaled so as to get mean 0 and sd 1". For the uniform, I'd normally assume the first, but as you see from an answer below, other people may take it to mean something else. The best option is to ask the person using the term to be less ambiguous. $\endgroup$
    – Glen_b
    Sep 20, 2013 at 0:20
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    $\begingroup$ The more conventional terms are standardized (to achieve a mean of zero and SD of one) and normalized (to bring the range to the interval $[0,1]$ or to rescale a vector norm to $1$). Thus the re-expression $X\to (X-\text{mean})/SD$ is a standardization whereas multiplying a density $f$ by a constant $C$ to make $\int_{-\infty}^\infty Cf(x)dx=1$ is a normalization, because $\int f(x)dx$ is the $L^1$ norm of $f$. $\endgroup$
    – whuber ♦
    Sep 20, 2013 at 1:39
  • $\begingroup$ Also asked on math.SE. $\endgroup$
    – Dilip Sarwate
    Sep 20, 2013 at 3:16
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    $\begingroup$ Please don't cross-post, @Ada. That is against SE policy. If you post a Q on 1 site & then think you should have posted it on another, flag your Q & ask the moderators to migrate it for you. $\endgroup$
    – gung - Reinstate Monica
    Sep 20, 2013 at 13:32

3 Answers 3

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Unfortunately, terms are used differently in different fields, by different people within the same field, etc., so I'm not sure how well this can be answered for you here. You should make sure you know the definition that your instructor / the textbook is using for "normalized". However, here are some common definitions:

Centered: $$ X-{\rm mean} $$ Standardized: $$ \frac{X-\text{mean}}{\text{sd}} $$ Normalized: $$ \frac{X-\min(X)}{\max(X)-\min(X)} $$ Normalizing in this sense rescales your data to the unit interval. Standardizing turns your data into $z$-scores, as @Jeff notes. And centering just makes the mean of your data equal to $0$.

It is worth recognizing here that all three of these are linear transformations; as such, they do not change the shape of your distribution. That is, sometimes people call the $z$-score transformation "normalizing" and believe, because of $z$-scores' association with the normal distribution, that this has made their data normally distributed. This is not so (as @Jeff also notes, and as you could tell by plotting your data before and after). Should you be interested, you could change the shape of your data using the Box-Cox family of transformations, for example.

With respect to how you could verify these transformations, it depends on what exactly is meant by that. If they mean simply to check that the code ran properly, you could check means, SDs, minimums, and maximums.

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    $\begingroup$ I have seen normalized used to suggest standardized or to suggest fitted onto a standard normal distribution i.e. $\Phi^{-1}(F(X))$, so of the three normalized is most likely to be misunderstood. Ada's comment of the application of a normalizing constant to a likelihood function is yet another possible interpretation. $\endgroup$
    – Henry
    Sep 23, 2013 at 20:16
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By using the formula you provided on each score in your sample, you are converting them all to z-scores.

To verify that you computed all the z-scores correctly, find the new mean and standard deviation of your sample. If the mean is $0$ and the standard deviation is $1$, you've done everything correctly.

The purpose of doing this is to put everything in units relative to the standard deviation of your sample. This may be useful for a variety of purposes, such as comparing two different data sets that were scored using different units (centimeters and inches, perhaps).

It is important not to get this confused with asking whether a distribution is normal, i.e. whether it approximates a Gaussian distribution.

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  • $\begingroup$ so to check whether or not the Uniform distribution was normalized would it be equivalent to say E(X) = 0 and Var(X) = 1 where X~Uniform(a,b)? $\endgroup$
    – user25658
    Sep 20, 2013 at 0:14
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    $\begingroup$ the data do not even have to be from a uniform distribution, they can be from any distribution. also, this is only true using the formula you provided; data can be normalized in ways other than using z-scores. for instance, IQ scores are said to be normalized with a score of 100 and standard deviation of 15. $\endgroup$
    – Jeff
    Sep 20, 2013 at 0:20
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After consulting the TA, what the question was asking was whether if

$$ \int_{-\infty}^{\infty}f(x)dx=1 $$

where $f(x)$ in this case is the density of the uniform(a,b).

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    $\begingroup$ The terminology to use here is that the probability density function of the distribution is normalized. Because this reflects the axiomatic fact that the total probability must equal $1$, asking whether any distribution itself is normalized (in this sense) always has the same trivial answer: of course. $\endgroup$
    – whuber ♦
    Sep 23, 2013 at 20:07
  • $\begingroup$ This is what we are asked to verify. f(x) doesn't really have to be a pdf, and it can be any non-negative function. For any non-negative function where the above doesn't satisfy, we can always multiply by a normalizing constant $\endgroup$
    – Ada
    Sep 23, 2013 at 20:16
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    $\begingroup$ Not always. For instance, let $f(x)=e^{-x}$, a non-negative function defined on all the real numbers: there is no normalizing constant. But when you are told, as in your question statement, that "so-and-so is the PDF for such-and-such a distribution" then there is nothing whatsoever to verify: by definition it integrates to unity. $\endgroup$
    – whuber ♦
    Sep 23, 2013 at 20:22
  • $\begingroup$ It's true not any non-negative function where we can make it satisfy the above condition even if we multiply by a normalizing constant. $\endgroup$
    – Ada
    Sep 23, 2013 at 20:29

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