S2 Standardizing the Normal Distribution

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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 :smile:

Reply 21

metaltron

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

\sigma

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. :smile:

Reply 22

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.

Reply 23

Asklepios

Original post by metaltron

Maybe I haven't explained myself properly. If you have:

X \sim N(6,25)

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

Z \sim N(0,1)

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

f(X) = \frac{1}{\sigma\sqrt{2\pi}}e^{-\frac{(X-\mu)^2}{2\sigma^2}}

You have done a substitution to get:

f(Z) = \frac{1}{\sigma\sqrt{2\pi}}e^{-\frac{Z^2}{2}}

This is correct, but as

Z \sim N(0,1)

, this means

\sigma^2=1

\sigma=1

.

So, substitute in sigma =1,

f(Z) = \frac{1}{1 \times\sqrt{2\pi}}e^{-\frac{Z^2}{2}}

Reply 24

davros

Original post by metaltron

Thank you! This is precisely what I wanted. :smile:

no problem...I normally keep well away from stats, unless it's the "pure maths" aspect of it :smile:

Reply 25

aznkid66

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.

Reply 26

metaltron

Original post by Asklepios

f(X) = \frac{1}{\sigma\sqrt{2\pi}}e^{-\frac{(X-\mu)^2}{2\sigma^2}}

You have done a substitution to get:

f(Z) = \frac{1}{\sigma\sqrt{2\pi}}e^{-\frac{Z^2}{2}}

This is correct, but as

Z \sim N(0,1)

, this means

\sigma^2=1

\sigma=1

.

So, substitute in sigma =1,

f(Z) = \frac{1}{1 \times\sqrt{2\pi}}e^{-\frac{Z^2}{2}}

Thanks, your explanation isn't quite right, there's an explanation above from davros which I am very grateful for

Reply 27

WhiteGroupMaths

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.

Reply 28

WhiteGroupMaths

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 :smile:

I guess the "looks can be deceiving" adage is applicable in mathematics. Peace.

Reply 29

Felix Felicis

Original post by metaltron

Hi,

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

X \sim N(\mu,\sigma^2)

The pdf f(X) is:

f(X) = \frac{1}{\sigma\sqrt{2\pi}}e^{-\frac{(X-\mu)^2}{2\sigma^2}}

and:

if \ Z = \frac{X-\mu}{\sigma}

f(Z) = \frac{1}{\sigma\sqrt{2\pi}}e^{-\frac{Z^2}{2}}

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 :smile:

I may be speaking complete rubbish here but
if

X \sim N(\mu , \sigma^{2})

Then

\mathbb{P}(a < X < b) = \displaystyle\int_{a}^{b} \dfrac{1}{\sigma \sqrt{2\pi}} \exp \left(- \dfrac{1}{2} \left(\dfrac{X - \mu}{\sigma}\right)^{2} \right) dX

Z = \dfrac{X - \mu}{\sigma} \Rightarrow dX = dZ \sigma

and

b \to \dfrac{b - \mu}{\sigma}

and

a \to \dfrac{a - \mu}{\sigma}

\therefore \displaystyle\int_{a}^{b} \dfrac{1}{\sigma \sqrt{2\pi}} \exp \left(- \dfrac{1}{2} \left(\dfrac{X - \mu}{\sigma}\right)^{2} \right) dX = \displaystyle\int_{\frac{a- \mu}{\sigma}}^{\frac{b - \mu}{\sigma}} \dfrac{1}{\sqrt{2 \pi}} \exp \left(- \dfrac{Z^{2}}{2} \right) dZ

(edited 11 years ago)

Reply 30

metaltron

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

Reply 31

davros

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

(x - \mu) / \sigma

is a straightforward one.

And they can't really discuss transformations of infinite integrals in S1 :smile:

Reply 32

metaltron

Original post by Felix Felicis

I may be speaking complete rubbish here but
if

X ~ N(\mu , \sigma^{2})

Then

P(a < X < b) = \displaystyle\int_{a}^{b} \dfrac{1}{\sigma \sqrt{2\pi}} \exp \left(- \dfrac{1}{2} \left(\dfrac{X - \mu}{\sigma}\right)^{2} \right)

Z = \dfrac{X - \mu}{\sigma} \Rightarrow dX = dZ \sigma

and

b \to \dfrac{b - \mu}{\sigma}

and

a \to \dfrac{a - \mu}{\sigma}

\displaystyle\int_{a}^{b} \dfrac{1}{\sigma \sqrt{2\pi}} \exp \left(- \dfrac{1}{2} \left(\dfrac{X - \mu}{\sigma}\right)^{2} \right) = \displaystyle\int_{\frac{a- \mu}{\sigma}}^{\frac{b - \mu}{\sigma}} \dfrac{1}{\sqrt{2 \pi}} \exp \left(- \dfrac{Z^{2}}{2} \right) dZ

Yes you're completely correct, unfortunately davros beat you to it! Thank you too

Reply 33

Felix Felicis

Original post by metaltron

Yes you're completely correct, unfortunately davros beat you to it! Thank you too

Curses...was the hairy latex :tongue:

Anyway, you're welcome :P

(edited 11 years ago)

Reply 34

spex

Formally

[br]F_Z(z) = P(Z \leq z) = P((X-\mu)/\sigma \leq z) = P(X \leq z \sigma + \mu) = F_X(z \sigma + \mu)[br]

[br]f_Z(z) = d/dz F_Z(z) = d/dz F_X(z \sigma + \mu) = \sigma f_X(z \sigma + \mu)[br][br]

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

Apparently, everyone in this thread is chatting ****. The answer your textbook gave you is incorrect

(edited 11 years ago)

Reply 35

davros

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 :tongue:

Anyway, you're welcome :P

Yeah, I'm impressed with the Latex - it would have taken me another hour to type that beast out :smile:

Reply 36

aznkid66

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.