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Math Help - [SOLVED] Finding distribution

  1. #1
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    [SOLVED] Finding distribution

    Let Y_1, Y_2, ..., Y_5 be a random sample of size 5 from a normal distribution with mean 0 and variance 1, and let

    \overline{Y}=\frac{1}{5}\sum\limits_{i=1}^5 Y_i

    Let Y_6 be another independent observation from the same distribution

    a) What is the distribution of W=\sum\limits_{i=1}^5 Y_i^2? I think it's \chi_5
    b) What is the distribution of U= \sum\limits_{i=1}^5 (Y_i-\overline{Y})^2 ?

    c) What is the distribution of \sum\limits_{i=1}^5(Y_i-\overline{Y})^2 + Y_6^2?
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  2. #2
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    Quote Originally Posted by noob mathematician View Post
    Let Y_1, Y_2, ..., Y_5 be a random sample of size 5 from a normal distribution with mean 0 and variance 1, and let

    \overline{Y}=\frac{1}{5}\sum\limits_{i=1}^5 Y_i

    Let Y_6 be another independent observation from the same distribution

    a) What is the distribution of W=\sum\limits_{i=1}^5 Y_i^2? I think it's \chi_5
    b) What is the distribution of U= \sum\limits_{i=1}^5 (Y_i-\overline{Y})^2 ?

    c) What is the distribution of \sum\limits_{i=1}^5(Y_i-\overline{Y})^2 + Y_6^2?
    You will be flattered; there's so much information given you in the first two lines. It's a give away for you to earn easy scores.

    Hint: the shape of you distribution is a perfect symmety.
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  3. #3
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    Quote Originally Posted by novice View Post
    You will be flattered; there's so much information given you in the first two lines. It's a give away for you to earn easy scores.

    Hint: the shape of you distribution is a perfect symmety.
    Pardon me if i don't get your hint: so the answer is \chi_5, \chi_5, \chi_6?
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  4. #4
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    Quote Originally Posted by noob mathematician View Post
    Pardon me if i don't get your hint: so the answer is \chi_5, \chi_5, \chi_6?
    That is the correct distribution followed by W (I think novice misread the question ....)
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  5. #5
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    Quote Originally Posted by mr fantastic View Post
    That is the correct distribution followed by W (I think novice misread the question ....)
    Oh, the answers I gave were pure guessing, however i hope to know the explantion: I was stuck half way through my working as followed.

    b)Find the distribution of U=\sum\limits_{i=1}^5 (Y_i-\overline{Y})^2.

    I actually expanded the question:

    since \sum\limits_{i=1}^n (X_i-\overline{X})^2
    = \sum\limits_{i=1}^n (X_i^2-2X_i\overline{X}+\overline{X}^2)
    = \sum\limits_{i=1}^nX_i^2-2n\overline{X}^2+n\overline{X}^2
    = \sum\limits_{i=1}^nX_i^2-n\overline{X}^2

    Applying to my question will yield:

    \sum\limits_{i=1}^5Y_i^2-5\overline{Y}^2,
    then since it was defined initially that \overline{Y}=\frac{1}{5}\sum\limits_{i=1}^5 Y_i, then it should have a distribution of N(0,1) too.

    Subsequently, I will have: U={\chi_5}^2-5(Z)^2={\chi_5}^2-Z_1^2-Z_2^2-Z_3^2-Z_4^2-Z_5^2, where Z is the standard normal. Isn't the equation just cancelled out each other? How does it still remain as {\chi_5}^2?
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  6. #6
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    Quote Originally Posted by noob mathematician View Post
    Oh, the answers I gave were pure guessing, however i hope to know the explantion: I was stuck half way through my working as followed.

    b)Find the distribution of U=\sum\limits_{i=1}^5 (Y_i-\overline{Y})^2.

    I actually expanded the question:

    since \sum\limits_{i=1}^n (X_i-\overline{X})^2
    = \sum\limits_{i=1}^n (X_i^2-2X_i\overline{X}+\overline{X}^2)
    = \sum\limits_{i=1}^nX_i^2-2n\overline{X}^2+n\overline{X}^2
    = \sum\limits_{i=1}^nX_i^2-n\overline{X}^2

    Applying to my question will yield:

    \sum\limits_{i=1}^5Y_i^2-5\overline{Y}^2,
    then since it was defined initially that \overline{Y}=\frac{1}{5}\sum\limits_{i=1}^5 Y_i, then it should have a distribution of N(0,1) too.

    Subsequently, I will have: U={\chi_5}^2-5(Z)^2={\chi_5}^2-Z_1^2-Z_2^2-Z_3^2-Z_4^2-Z_5^2, where Z is the standard normal. Isn't the equation just cancelled out each other? How does it still remain as {\chi_5}^2?
    Please accept my sincerely apology. Indeed I misread the question as Mr. Fantastic said.
    Now it is clear that the question is more interesting than I originally thought. Admittedly, I my self have something to gain by solving it, but I doubt that I could beat you to it.
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  7. #7
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    Quote Originally Posted by noob mathematician View Post
    Oh, the answers I gave were pure guessing, however i hope to know the explantion: I was stuck half way through my working as followed.

    b)Find the distribution of U=\sum\limits_{i=1}^5 (Y_i-\overline{Y})^2.

    I actually expanded the question:

    since \sum\limits_{i=1}^n (X_i-\overline{X})^2
    = \sum\limits_{i=1}^n (X_i^2-2X_i\overline{X}+\overline{X}^2)
    = \sum\limits_{i=1}^nX_i^2-2n\overline{X}^2+n\overline{X}^2
    = \sum\limits_{i=1}^nX_i^2-n\overline{X}^2

    Applying to my question will yield:

    \sum\limits_{i=1}^5Y_i^2-5\overline{Y}^2,
    then since it was defined initially that \overline{Y}=\frac{1}{5}\sum\limits_{i=1}^5 Y_i, then it should have a distribution of N(0,1) too.

    Subsequently, I will have: U={\chi_5}^2-5(Z)^2={\chi_5}^2-Z_1^2-Z_2^2-Z_3^2-Z_4^2-Z_5^2, where Z is the standard normal. Isn't the equation just cancelled out each other? How does it still remain as {\chi_5}^2?
    So any view on my above comment?
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  8. #8
    MHF Contributor matheagle's Avatar
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    a) is \chi^2_5

    next use {(n-1)S^2\over\sigma^2}\sim\chi^2_{n-1}

    with \sigma^2=1

    we have in (b) {(n-1)S^2\over\sigma^2}\sim\chi^2_4

    hence (c) is \chi^2_5

    since \chi^2_4 + \chi^2_1=\chi^2_5

    using independence, via the MGF technique
    Last edited by matheagle; October 6th 2009 at 09:26 PM.
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  9. #9
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    Quote Originally Posted by matheagle View Post
    a) is \chi^2_5

    next use {(n-1)S^2\over\sigma^2}\sim\chi^2_{n-1}

    with \sigma^2=1

    we have in (b) {(n-1)S^2\over\sigma^2}\sim\chi^2_4

    hence (c) is \chi^2_5

    since \chi^2_4 + \chi^2_1=\chi^2_5

    via independence, via the MGF technique
    Thanks, it's a cool solution.

    So now I'm interested to know the next distribution:

    2(5\overline{Y}^2+Y_6^2)/\chi_4^2 It seems like it's a F-distribution, but I can't seem to extract the equation to form pattern like F_{m,n}\rightarrow (U/m)/(V/n)

    Btw, just curious that for the nominator: (5\overline{Y}^2+Y_6^2), does it convert to \chi_6^2? Then does that mean that 2(5\overline{Y}^2+Y_6^2) will be \chi_{12}^2??
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  10. #10
    MHF Contributor matheagle's Avatar
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    no
    and you must have that chi-square independent of y bar and y6.

    It's an F2,4


    Y_i\sim N(0,1)

    so \bar Y\sim N(0,1/5)

    or \sqrt{5}\bar Y\sim N(0,1)

    Thus 5\bar Y^2\sim \chi^2_1

    Since Y_6^2\sim \chi^2_1 we have

    5\bar Y^2 +Y_6^2\sim \chi^2_2

    HENCE {{5\bar Y^2 +Y_6^2\over 2}\over {\chi^2_4\over 4}}\sim F_{2,4}


    The 2 in the numerator is just algebra, 4/2.
    Last edited by matheagle; October 6th 2009 at 05:46 PM. Reason: typo
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  11. #11
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    Quote Originally Posted by matheagle View Post
    no
    and you must have that chi-square independent of y bar and y6.

    It's an F2,4


    Y_i\sim N(0,1)

    so \bar Y\sim N(0,1/5)

    or \sqrt{5}\bar Y\sim N(0,1)

    Thus 5\bar Y^2\sim \chi^2_1
    Hi thanks again, I'm sure you are right.

    However, I was kinda confused with the above transformation, such as how it jumps from \bar Y\sim N(0,1/5) to \sqrt{5}\bar Y\sim N(0,1)
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  12. #12
    MHF Contributor matheagle's Avatar
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    you obtain a N(0,1) by subtracting off the mean and dividing by the st deviation

    Z={W-\mu\over \sigma}={\bar Y-0\over 1/{\sqrt 5}}={\sqrt 5}\bar Y\sim N(0,1)
    Last edited by matheagle; October 6th 2009 at 08:41 PM.
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