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# S2 Standardizing the Normal Distribution Watch

1. (Original post by WhiteGroupMaths)
Like how you phrased things. Legendary. Nearly drove me nuts trying to crack it previously. Peace.
Well it's "legendary" on this forum because every so often people keep asking how to evaluate the indefinite integral, thinking they can work it out
2. (Original post by davros)
Sorry if I'm telling you something you already know, but it's not just a case of relabelling the variable in the pdf. You start with the condition that the pdf must integrate to 1 over the entire range, so you're actually changing a variable inside the integral. That means when you convert from dX to dZ you get an extra factor which cancels the one you have.

Basically, you always need the appropriate normalisation constant for your pdf.
Thank you! This is precisely what I wanted.
3. I still wonder why the textbook is saying that its a simple C1 transformation, is it? Because what I can make out it is C4 integration by substitution.
4. (Original post by metaltron)
Maybe I haven't explained myself properly. If you have:

you can use the standardisation formula to get another variable Z which has normal distribution:

However I'm struggling to see how the standardisation formula is derived.

You have done a substitution to get:

This is correct, but as , this means so .

So, substitute in sigma =1,

5. (Original post by metaltron)
Thank you! This is precisely what I wanted.
no problem...I normally keep well away from stats, unless it's the "pure maths" aspect of it
6. (Original post by metaltron)
I still wonder why the textbook is saying that its a simple C1 transformation, is it? Because what I can make out it is C4 integration by substitution.
You could interpret it as a horizontal shift by -m and a horizontal stretch by 1/s.

You have done a substitution to get:

This is correct, but as , this means so .

So, substitute in sigma =1,

Thanks, your explanation isn't quite right, there's an explanation above from davros which I am very grateful for
8. Did you attempt a blunt substitution to discover the pdf of Z based on the pdf of X?

Because it seems you have done that in the very first post of this thread.

Peace.
9. (Original post by davros)
Well it's "legendary" on this forum because every so often people keep asking how to evaluate the indefinite integral, thinking they can work it out
I guess the "looks can be deceiving" adage is applicable in mathematics. Peace.
10. (Original post by metaltron)
Hi,

I'm a bit confused about how the normal standardisation formula is derived. We know if:

The pdf f(X) is:

and:

My question is why is there still a sigma on the bottom of the fraction? Have I done something wrong with the algebra? My S2 book insists that it is a simple C1 transformation which makes me feel a bit stupid not understanding where this standardisation formula comes from! Can somebody please explain it for me? Thanks
I may be speaking complete rubbish here but
if

Then

if and and

11. (Original post by aznkid66)
You could interpret it as a horizontal shift by -m and a horizontal stretch by 1/s.
Yeah, it isn't obvious from the formula though. Thanks, for your replies
12. (Original post by metaltron)
I still wonder why the textbook is saying that its a simple C1 transformation, is it? Because what I can make out it is C4 integration by substitution.
the only thing I can think of is that they mean from the point of view of getting a Z value you can look up in tables, the calculation of is a straightforward one.

And they can't really discuss transformations of infinite integrals in S1
13. (Original post by Felix Felicis)
I may be speaking complete rubbish here but
if

Then

if and and

Yes you're completely correct, unfortunately davros beat you to it! Thank you too
14. (Original post by metaltron)
Yes you're completely correct, unfortunately davros beat you to it! Thank you too
Curses...was the hairy latex Anyway, you're welcome :P
15. Formally

If you sub everything in then you get the standard normal pdf, no sigmas in sight

16. (Original post by metaltron)
Yes you're completely correct, unfortunately davros beat you to it! Thank you too

(Original post by Felix Felicis)
Curses...was the hairy latex Anyway, you're welcome :P
Yeah, I'm impressed with the Latex - it would have taken me another hour to type that beast out
17. (Original post by metaltron)
Yeah, it isn't obvious from the formula though. Thanks, for your replies
Hm? It's obvious that because the shape of the normal curve with respect to the stdev (and thus the integral with bounds in terms of the stdev) won't change from these transformations, you can apply these obvious transformations to get a curve with the correct probabilities, a mean of 0, and a standard deviation of 1. Simple C1 stuff.
18. (Original post by aznkid66)
Hm? It's obvious that because the shape of the normal curve with respect to the stdev (and thus the integral with bounds in terms of the stdev) won't change from these transformations, you can apply these obvious transformations to get a curve with the correct probabilities, a mean of 0, and a standard deviation of 1. Simple C1 stuff.

is so obviously the transformation:

19. No, it's the transformation .

We've already show how it's a simple substitution after knowing the rule. Why are you trying to get the rule from the function?
20. (Original post by aznkid66)
No, it's the transformation .

We've already show how it's a simple substitution after knowing the rule. Why are you trying to get the rule from the function?
Explain it then, using C1 transformations, nothing on integration.

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