6
$\begingroup$

Would you also be able to provide an example? I have very little mathematical/statistical knowledge and have never understood normalization.

$\endgroup$

2 Answers 2

6
$\begingroup$

Unity just means 1, so they have presumably normalized their values so that they all sum to 1 instead of whatever their "natural" total is. I could imagine a few specialized normalization schemes, but this is typically done by dividing, and that's what I would assume in the absence of a more detailed description. If they had normalized so that the values summed to 100 instead, they'd be expressing it as a percentage.

Suppose there is a substance made of three chemicals: 5L of Chemical A, 2L of Chemical B, and 3L of Chemical C. You could do a similar normalization and say that each litre of substance contains 0.5L of A, 0.2L of B, and 0.3L of C (each value has been divided by 10, the total, so all the values together sum to one). If you normalized to 100 instead of unity, then you could also say that the substance is 50% A, 20% B, and 30% C.

One of the most common uses of this technique is to turn event counts into probabilities. By definition, probabilities have in [0,1] (i.e., greater than or equal to zero and less than or equal to one). Suppose you have an urn with 10 balls in it, seven of which are red and three of which are blue. You could normalize these counts so that they sum to unity and restate this as the probability that a randomly chosen ball is red, $P(ball=\textrm{red}) = \frac{7}{7+3} = \frac{7}{10} = 0.7$ and $P(ball=\textrm{blue}) = \frac{3}{3+7} = \frac{3}{10}=0.3$.

$\endgroup$
0
$\begingroup$

A natural application is conditional probabilities. If I roll a die, the unconditional probability of each outcome is ${1 \over 6}.$ But suppose I roll it and tell you that the outcome is at least 4. You can find the new conditional probabilities for rolls of 4, 5, or 6 by dividing ${1 \over 6}$ by ${1 \over 2}$ for each of the outcomes of 4, 5, or 6. This process of division ensures the conditional probabilities sum to unity as they must.

$\endgroup$

Your Answer

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

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