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# S2 Standardizing the Normal Distribution Watch

1. (Original post by davros)
Yeah, I'm impressed with the Latex - it would have taken me another hour to type that beast out
2. (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?
wrong.
3. All normal distributions follow this function:

This function has a mean/maximum at and a shape such that the area from to is 68%, the area from to is 95%, etc. and all that jazz with symmetry and what-not. In other words, the function has a shape such that f(x), where x=g(s), has the same value for a given s no matter what g(s) is.

We want the substitution for the standard normal curve, which is a normal curve where m=0 and s=1.

This is all given.

So the question is, which transformations can we use? If we do a horizontal translation, then the shape of the curve doesn't change - it's still a normal curve, and so we can apply any horizontal translation we want. If we do a horizontal stretch, then the f(x) at a given s is still the same, and so we can apply any horizontal stretch we want. Also note that horizontal stretches centered around the mean don't change the mean, which is nice because then we can first set m=0 and then use a stretch to set m=0 and s=1.

To set the mean to 0, we would first shift all the points m units to the left. This gives us a substitution y=x-m and a distribution Y~N(0,s)

Right now, one standard deviation from the mean is equal to 's' y-units away from the mean. So to set the standard deviation to 1, we would bring all the points 's' times closer to the mean. This gives us a substitution z=(x-m)/s and a distribution Z~N(0,1).

Oh hey look, it's the standard normal curve!

And what's the equation that follows from that substitution/transformation? Why, it's just the normal curve where the mean is 0 and the standard deviation is 1.

spex, why do you think that's wrong?
4. (Original post by aznkid66)
All normal distributions follow this function:

This function has a mean/maximum at and a shape such that the area from to is 68%, the area from to is 95%, etc. and all that jazz with symmetry and what-not. In other words, the function has a shape such that f(x), where x=g(s), has the same value for a given s no matter what g(s) is.

We want the substitution for the standard normal curve, which is a normal curve where m=0 and s=1.

This is all given.

So the question is, which transformations can we use? If we do a horizontal translation, then the shape of the curve doesn't change - it's still a normal curve, and so we can apply any horizontal translation we want. If we do a horizontal stretch, then the f(x) at a given s is still the same, and so we can apply any horizontal stretch we want. Also note that horizontal stretches centered around the mean don't change the mean, which is nice because then we can first set m=0 and then use a stretch to set m=0 and s=1.

To set the mean to 0, we would first shift all the points m units to the left. This gives us a substitution y=x-m and a distribution Y~N(0,s)

Right now, one standard deviation from the mean is equal to 's' y-units away from the mean. So to set the standard deviation to 1, we would bring all the points 's' times closer to the mean. This gives us a substitution z=(x-m)/s and a distribution Z~N(0,1).

Oh hey look, it's the standard normal curve!

And what's the equation that follows from that substitution/transformation? Why, it's just the normal curve where the mean is 0 and the standard deviation is 1.

spex, why do you think that's wrong?
That's quite nice, thanks I think this is what the book was trying to get to, but I think I will remember the integration method better. I think I might have been a bit sceptical so sorry
5. (Original post by aznkid66)

spex, why do you think that's wrong?
When you do a location scale transformation you don't do it to the density, you do it to the distribution In any case, it's extremely easy to check that what you quoted,
is incorrect
6. Sure, but I don't see what's wrong with the equation standing by itself. Z=(x-m)/s, and so f(Z)=f((x-m)/s).

In addition, for normal distributions, we don't really care about cumulative distribution , which is an equality that doesn't even hold most of the time, but .
7. (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

but

but

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