Understanding Summation: S_X and S^2_X

[Ted123]

Does

\displaystyle S^2_X = \sum^n_{i=1} (X_i - \overline{X})^2

mean

\displaystyle S_X = \sum^n_{i=1} (X_i - \overline{X}) (X is a random variable here)

and is this true of summations in general?

i.e. does \sqrt{\sum X_i^2} = \sum X_i ?

I thought it'd be \sqrt{\left( \sum X_i\right) ^2} = \sum X_i ?

 

[micromass]

NO, this is NOT true at all. It is NOT true that (a+b)^2=a^2+b^2, which is basically what you're claiming...

 

[LCKurtz]

Try a simple case with numbers. Does sqrt(a²+b²) = a + b? Put some numbers in and see what you think.

 

[Ted123]

So this is just a special case where S_X is defined to be the sample standard deviation and S_X^2 is the sample variance?

 

[Mark44]

\displaystyle S_X = \sqrt{\sum^n_{i=1} (X_i - \overline{X})^2}\neq \sum^n_{i=1} (X_i - \overline{X})

 

[micromass]

And I also have the feeling that you have to divide by n-1...

So

S_X^2=\frac{1}{n-1}\sum_{i=1}^n{(X_i-\overline{X})^2}

 

[HallsofIvy]

You can do this:
\sum (X_i- \overline{X})^2= \sum (X_i^2- 2\overline{X}X_i+ \overline{X}^2)
\sum X_i^2- 2\overline{X}\sum X_i+ \sum \overline X

Now, here,
\overline{X}= \frac{1}{n}\sum X_i
and, of course, since \overline{X} is a constant,
\sum \overline{X}= \overline{X}\sum 1= n\overline{X}

so
\sum (X_i- \overline{X})^2= \sum X_i^2- 2n \overline{X}^2+ n\overline{X}^2
= \sum X_i^2- n\overline{X}^2