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    (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
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    (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.
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    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.
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    (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 so \sigma=1.

    So, substitute in sigma =1,

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

    if 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
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    (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
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    (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
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    (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)

    if 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
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    (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
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    Formally

    

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


    

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

    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
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    (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
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    (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.
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    (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.
    Talk about it:

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

    is so obviously the transformation:

     \sigma f(\sigma x+\mu)
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    No, it's the transformation f(Z)=f(\frac{x-\mu} {\sigma}).

    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?
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    (Original post by aznkid66)
    No, it's the transformation f(Z)=f(\frac{x-\mu} {\sigma}).

    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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