Hey there! Sign in to join this conversationNew here? Join for free
    Offline

    2
    ReputationRep:
    (Original post by davros)
    Yeah, I'm impressed with the Latex - it would have taken me another hour to type that beast out
    Offline

    4
    ReputationRep:
    (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?
    wrong.
    Offline

    1
    ReputationRep:
    All normal distributions follow this function:

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

    This function has a mean/maximum at \mu and a shape such that the area from \mu-\sigma to \mu+\sigma is 68%, the area from \mu-2\sigma to \mu+2\sigma 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 do this concretely, let's start with a distribution X~N(m,s).

    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.

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

     f(Z) = \frac{1}{(1)\sqrt{2\pi}}e^{-\frac{(Z-(0))^2}{2(1)^2}}

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




    spex, why do you think that's wrong?
    • Thread Starter
    Offline

    10
    ReputationRep:
    (Original post by aznkid66)
    All normal distributions follow this function:

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

    This function has a mean/maximum at \mu and a shape such that the area from \mu-\sigma to \mu+\sigma is 68%, the area from \mu-2\sigma to \mu+2\sigma 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 do this concretely, let's start with a distribution X~N(m,s).

    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.

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

     f(Z) = \frac{1}{(1)\sqrt{2\pi}}e^{-\frac{(Z-(0))^2}{2(1)^2}}

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




    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
    Offline

    4
    ReputationRep:
    (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, f_Z(z) you do it to the distribution F_Z(z) In any case, it's extremely easy to check that what you quoted,
      f(Z)=f(\frac{x-\mu} {\sigma})
    is incorrect
    Offline

    1
    ReputationRep:
    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 F_X(x)=F_Z(z), which is an equality that doesn't even hold most of the time, but F_X(\sigma_x)=F_Z(\sigma_z).
    Offline

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

    Z = \dfrac{X - \mu}{\sigma} \Rightarrow E(Z) = E \left( \dfrac{1}{\sigma} X - \dfrac{\mu}{\sigma} \right)

     = \dfrac{1}{\sigma} E(X) - \dfrac{\mu}{\sigma} but E(X) = \mu \Rightarrow E(Z) = \mu_Z = 0

    Var(Z) = Var \left( \dfrac{1}{\sigma} X - \dfrac{\mu}{\sigma} \right) = \dfrac{1}{\sigma^{2}} Var(X) but Var(X) = \sigma^{2} \Rightarrow Var(Z) = 1
 
 
 
  • See more of what you like on The Student Room

    You can personalise what you see on TSR. Tell us a little about yourself to get started.

  • Poll
    Brussels sprouts
    Useful resources

    Make your revision easier

    Maths

    Maths Forum posting guidelines

    Not sure where to post? Read the updated guidelines here

    Equations

    How to use LaTex

    Writing equations the easy way

    Student revising

    Study habits of A* students

    Top tips from students who have already aced their exams

    Study Planner

    Create your own Study Planner

    Never miss a deadline again

    Polling station sign

    Thinking about a maths degree?

    Chat with other maths applicants

    Can you help? Study help unanswered threads

    Groups associated with this forum:

    View associated groups
  • See more of what you like on The Student Room

    You can personalise what you see on TSR. Tell us a little about yourself to get started.

  • The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

    Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

    Quick reply
    Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.