Re: Knives (was Re: Lee Murray in Critical Condition After Stabbing



"GreenDistantStar" <Greendistantstar@xxxxxxxxxxx> wrote in news:nLJ%
e.4783$U51.3508@xxxxxxxxxxxxxxxxxxxxxxxxxx:

>> You are not competent to discus mathematics either publicly or offline
-
>> and I'm not going to give you the chance to escape. Either stay
public
>> and
>> take your ass-whupping like a man or slink off. But you'll get no
email
>> from me.
>
> Suit yourself.
>
> Explain zero correlation / determinacy from this ...
>
> y = x^2 for -1 < x < 1
>
> GDS


You really are a dunce, aren't you? It has already been explained once,
but I'll go over a few features for you.

First of all note that for my two variables, x and y, I gave an equation
relating them - a continuous equation, I emphasize! No paired sets of
discrete data. Because they aren't necessary! (Although that's the way
some undergrads and their equally ignorant teachers always think when
thinking of correlation - because they are concerned with mere
arithmetical manipulation rather than truly understanding the underlying
mathematical concepts.)

No, correlation is DEFINED as:

E((x-u)*(y-v))/(stddev(x)*stddev(y))

where u,v are the means of x,y repectively (I won't try to use Greek
fonts). E is the "expectation operator."

No need for discrete values. In fact, manipulating paired sets of
sampled discrete values usually gives only an *estimate* of correlation
between the underlying variables, an estimate subject to error. But in
my illustrative case I can give the correlation *exactly* as zero!

Notice there were no least-squares best-line fits in my exposition.
Aside from being unnecessary they could be actually misleading to the
mathematically weak (like you!). Best-fit linear equations assume a
dependence of one variable on another, not the symmetrical and coequal
relationship of correlation! (and there are other differences as well
such as that changes in u,v would matter to the calculation of a least-
squares best-fit line but not to correlation, etc.)

Regards,

PS I could explain the expectation operator by saying that the
"expected value" of a variable x (if it exists) is the Lebesgue integral
of x.dP over O in the probability space (O,P) but that is so far beyond
your meagre comprehension that I will leave you to absorb the elementary
facts I have already presented.

PPS Note that the expected value (and therefore correlation which
depends upon it) is defined in terms of *all* possible outcomes in the
probability space and not a *sample* drawn from it. Not that you will
understand the significance of that statement, mind you.

.



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