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Thread: Moment Generating Function of a Chi-Squared distribution

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    Moment Generating Function of a Chi-Squared distribution




    Hey guys, need help with this question.

    Random Variables V and U are independent with N~(0,4).
    Let W = U^2 + V^2 . Calculate P(W>24).

    Is the Degree of Freedom here 2 ?

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    Re: Moment Generating Function of a Chi-Squared distribution

    Why does your title mention the moment generating function?
    I don't have emotions and sometimes that makes me very sad.

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    Re: Moment Generating Function of a Chi-Squared distribution

    It happens to be at the end of my lecture notes. More specifically x^2 distribution

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    Re: Moment Generating Function of a Chi-Squared distribution

    First, I would ask you to explain how to arrived at 2 for the degrees of freedom.

    What needs to be done is to find the distribution of W, which you can ALSO do with moment generating functions, or by using relations (which can also be derived from moment generating functions). On this matter you must first do something to see what the distributions of (or rather functions of) U^2 and V^2 are.
    “When things get too complicated, it sometimes makes sense to stop and wonder: Have I asked the right question?” ― Enrico Bombieri

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    Re: Moment Generating Function of a Chi-Squared distribution

    Since they are independent, they follow a x^2 (n) distribution. How do i find out what is n ?

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    Re: Moment Generating Function of a Chi-Squared distribution

    Quote Originally Posted by saternal View Post
    Since they are independent, they follow a x^2 (n) distribution. How do i find out what is n ?
    The distribution will be Chi-square with 2 degrees of freedom. It will also have a mean of 8 and variance of 64. You should be able to know what to do from this point.

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    Re: Moment Generating Function of a Chi-Squared distribution

    U and V have to be normalized first
    So we take, U / 2 and V /2, from here you square both (which gives what distribution for them respectively, with how many respective degrees of freedom) and then you can add them together to give you W /4.

    The probability in the question must be slightly rewritten in order to use above results.
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    Re: Moment Generating Function of a Chi-Squared distribution

    I still don't know what to do. Moreover, i don't know how you found the Mean to be 8 with Variance 64 =/

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    Re: Moment Generating Function of a Chi-Squared distribution

    Quote Originally Posted by chippy View Post
    U and V have to be normalized first
    So we take, U / 2 and V /2, from here you square both (which gives what distribution for them respectively, with how many respective degrees of freedom) and then you can add them together to give you W /4.

    The probability in the question must be slightly rewritten in order to use above results.
    Both have distributions of N[0,4] so can i say P(W>6) ?

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    Re: Moment Generating Function of a Chi-Squared distribution

    Since you want to calculate \Pr\{W > 24\}, you need to know the distribution of W = U^2 + V^2.

    If you can only use a chi-square table, or you need to express the above probability in terms of a chi-square random variable, then you probably need to use the following facts to exploit the relationships:

    1. If U \sim \mathcal{N}(0, \sigma^2), then \frac {U} {\sigma} \sim \mathcal{N}(0, 1).

    2. If Z \sim \mathcal{N}(0, 1), then Z^2 \sim \chi^2(1)

    3. If U \sim \chi^2(m), V \sim \chi^2(n) and U, V are independent, then U + V \sim \chi^2(m + n)

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    Re: Moment Generating Function of a Chi-Squared distribution

    If I read your statement as a direct implication
    Both have distributions of N[0,4] so can i say P(W>6)
    , then I'd say it's incorrect.

    The square of a standard normal random variable is a chi-squared RV with 1 degree of freedom - thats why we first normalize U & V.

    Now, you can add the two of them : (\frac{U}{2})^2\  (\frac{V}{2})^2 to obtain yet another chi-squared random variable with 2 (1 + 1) degrees of freedom - this results you can easily prove with MGF.

    This leads to \frac{W}{4} = (\frac{U}{2})^2 + (\frac{V}{2})^2

    From which P[W > 24] = P[W/4 > 6], where W/4 \sim  \chi^2 (2)
    “When things get too complicated, it sometimes makes sense to stop and wonder: Have I asked the right question?” ― Enrico Bombieri

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    Re: Moment Generating Function of a Chi-Squared distribution

    Oh, sorry about the double post above...I see BGM answered, as I was still typing.
    “When things get too complicated, it sometimes makes sense to stop and wonder: Have I asked the right question?” ― Enrico Bombieri

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    Re: Moment Generating Function of a Chi-Squared distribution

    It's obvious that the distribution is exponentional with parameter theta=8. Use this distribution to compute the probability W>24 = 0.950213.

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    Re: Moment Generating Function of a Chi-Squared distribution

    @Dragan : Hi Dragan, I would just like to know where the "obvious" part comes into play? A more direct route is always welcome.
    “When things get too complicated, it sometimes makes sense to stop and wonder: Have I asked the right question?” ― Enrico Bombieri

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    Re: Moment Generating Function of a Chi-Squared distribution


    Quote Originally Posted by chippy View Post
    If I read your statement as a direct implication , then I'd say it's incorrect.

    The square of a standard normal random variable is a chi-squared RV with 1 degree of freedom - thats why we first normalize U & V.

    Now, you can add the two of them : (\frac{U}{2})^2\  (\frac{V}{2})^2 to obtain yet another chi-squared random variable with 2 (1 + 1) degrees of freedom - this results you can easily prove with MGF.

    This leads to \frac{W}{4} = (\frac{U}{2})^2 + (\frac{V}{2})^2

    From which P[W > 24] = P[W/4 > 6], where W/4 \sim  \chi^2 (2)
    Hey Chippy, are you saying without normalizing them, the given distribution N(0,1) has only 1 degree of freedom?
    Then, is it a must to normalize U and V always?

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