+ Reply to Thread
Results 1 to 5 of 5

Thread: Chi distribution link with standard normal distribution?

  1. #1
    Points: 1,205, Level: 19
    Level completed: 5%, Points required for next Level: 95

    Posts
    7
    Thanks
    2
    Thanked 0 Times in 0 Posts

    Smile Chi distribution link with standard normal distribution?




    Hello,

    The chi distribution is the sum of n random standard normal variables. It's used for example to know if a distribution fits the uniform distribution by taking the sum of the squared difference between the observed frequency of a value and the expected frequency of a value divided by the expected frequency of a value. By doing this I assume that this is because the difference between the observed frequency and the expected frequency must be standard normally distributed. My question is: why is that difference standard normally distributed? And if it's not standard normally distributed why is the sum of the squared differences chi-distributed?

    Thank you very much for your attention to this matter.

  2. #2
    Omega Contributor
    Points: 38,432, Level: 100
    Level completed: 0%, Points required for next Level: 0
    hlsmith's Avatar
    Location
    Not Ames, IA
    Posts
    7,006
    Thanks
    398
    Thanked 1,186 Times in 1,147 Posts

    Re: Chi distribution link with standard normal distribution?

    I am totally guessing and I am silly nave when it comes to these things, but is there a difference squared going on in formulae, since the chi-square is equal to standard normal squared.
    Stop cowardice, ban guns!

  3. #3
    Super Moderator
    Points: 13,151, Level: 74
    Level completed: 76%, Points required for next Level: 99
    Dragan's Avatar
    Location
    Illinois, US
    Posts
    2,014
    Thanks
    0
    Thanked 223 Times in 192 Posts

    Re: Chi distribution link with standard normal distribution?

    This is a confusing topic to most students and others.

    First, assume X~N(Mu, Sigma^2), then Z = (X - Mu)/Sigma is Standard Normal.

    Second, the Chi-Square Distribution can be empirically derived by summing independent Z's e.g. Z^2 is Chi square on one df; Z1^2 + Z2^2 is Chi square on two df, and so on.

    Third, the discrete Binomial Distribution with N independent ob's has a mean of N*p, where p is the probability of success, with variance of N*p*q where q is the probability of failure, and Stand. Dev. of Sqrt[N*p*q].

    Now, through the Central Limit Theorem (CLT) the Binomial Distribution will approximate the Normal distribution when N is large enough. A typical rule is N*p>=5 and N*q>=5.

    Next, "connect the dots." That is,

    Chi-Square (1df) = Z^2 = (X - Mu)^2 / (Sigma^2).

    Subsequently, impose the CLT using the binomial distribution parameters:

    Chi-Square (1df) = Z^2 = (X - N*p)^2 / (N*p*q). (Note that X is the number of successes.)

    And then through some not so obvious algebra we have:

    Chi-Square (1df) = (X - N*p)^2 / (N*p) + ([N-X] - N*q)^2 / (N*q).

    This is usually written as:

    Chi-Square (1df) = (O_1 - E_1)^2 / (E_1) + (O_2 - E_2)^2 / (E_2).

    Note that it is the Expected frequencies that must be greater than (or equal to) 5 and not the actual observations - to invoke the CLT.

  4. The Following User Says Thank You to Dragan For This Useful Post:

    rogojel (04-30-2017)

  5. #4
    Points: 1,974, Level: 26
    Level completed: 74%, Points required for next Level: 26

    Location
    New Zealand
    Posts
    227
    Thanks
    3
    Thanked 48 Times in 47 Posts

    Re: Chi distribution link with standard normal distribution?

    Quote Originally Posted by Eometh View Post
    Hello,

    The chi distribution is the sum of n random standard normal variables.
    The short answer is that don't just add the numbers. You square the numbers before you add them. If you take say 5 normal numbers at random, square each of them and add these five squares, the total you get is distributed chi squared with 5 df.
    The errors are normal, so the sum of the errors squared is chi squared.

  6. #5
    Points: 1,205, Level: 19
    Level completed: 5%, Points required for next Level: 95

    Posts
    7
    Thanks
    2
    Thanked 0 Times in 0 Posts

    Thumbs up Re: Chi distribution link with standard normal distribution?


    Quote Originally Posted by Dragan View Post
    Third, the discrete Binomial Distribution with N independent ob's has a mean of N*p, where p is the probability of success, with variance of N*p*q where q is the probability of failure, and Stand. Dev. of Sqrt[N*p*q].

    Chi-Square (1df) = (X - N*p)^2 / (N*p) + ([N-X] - N*q)^2 / (N*q).

    This is usually written as:

    Chi-Square (1df) = (O_1 - E_1)^2 / (E_1) + (O_2 - E_2)^2 / (E_2).
    Thank you very much for the detailed answer. I just have 2 more questions based on the explanation: Why does suddenly the binomial distribution appear? It counts the number of successes but what is a success in the case that we want to test if a distribution has a uniform distribution?

    My second question is why is X equal to the observed frequency? Is it due to the number of successes of the binomial distribution? Is it also valid if we want to test if random variable X has a uniform distribution?

+ Reply to Thread

           




Posting Permissions

  • You may not post new threads
  • You may not post replies
  • You may not post attachments
  • You may not edit your posts






Advertise on Talk Stats