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Math Help - Random Variable

  1. #1
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    Random Variable

    1. Suppose that X ~ N (mu, sig^2). Show

    ((x-mu)/sig)^2 ~ Gamma(1/2, 1/2.) = Ki-square subscripte 1



    2. Suppose that X1,X2.... Xm are iid N(mu, sig^2). Show that

    summation i = 1 to n ( (Xi - mu)/sig)^2 ~ Ki-square subscript n
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    Quote Originally Posted by weakmath View Post
    1. Suppose that X ~ N (mu, sig^2). Show

    ((x-mu)/sig)^2 ~ Gamma(1/2, 1/2.) = Ki-square subscripte 1



    2. Suppose that X1,X2.... Xm are iid N(mu, sig^2). Show that

    summation i = 1 to n ( (Xi - mu)/sig)^2 ~ Ki-square subscript n
    This thread contains the ideas you need to use: http://www.mathhelpforum.com/math-he...tml#post119542
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    does it make a difference that your paremeters are unknown while solving and also, the result of subscript 1 and n?
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    Quote Originally Posted by weakmath View Post
    does it make a difference that your paremeters are unknown while solving and also, the result of subscript 1 and n?
    Your parameters are not unknown. The mean is \mu and the standard deviation is \sigma. All you have to do is generalise the solution given in the thread I referenced.

    The subscripts 1, 2, .... n are just part of the symbol used for the random variable.
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    Quote Originally Posted by weakmath View Post
    [snip]
    2. Suppose that X1,X2.... Xm are iid N(mu, sig^2). Show that

    summation i = 1 to n ( (Xi - mu)/sig)^2 ~ Ki-square subscript n
    I'd suggest using a moment-generating function approach for Q2.

    I don't have time now but if you're stuck, post your work and I'll give some details later.

    (You could also use this approach for Q1)
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    Quote Originally Posted by weakmath View Post
    does it make a difference that your paremeters are unknown while solving and also, the result of subscript 1 and n?
    In fact, at second glance Q1 is even easier than I thought. The thread I referenced actually gives a solution to Q1 since

    \frac{X - \mu}{\sigma} = Z ~ N(0, \, 1).


    Q2:

    The moment generating function of Z^2 is:

    M_{Z^2}(t) = E\left[e^{tZ^2}\right] = \frac{1}{\sqrt{2 \pi} } \int_{-\infty}^{+\infty}e^{tz^2} e^{-z^2/2} \, dz = \frac{1}{\sqrt{2 \pi} } \int_{-\infty}^{+\infty} e^{-z^2(1 - 2t)/2} \, dz = \frac{1}{(1 - 2t)^{1/2}}

    (which by the way is recognised as the moment generating function for the chi-squared distribution with 1 degree of freedom).

    Therefore the moment generating function of  Z_{1}^2 + Z_{2}^2 + \, .... \, + Z_{n}^2 = M_{Z_{1}^2}(t) \cdot M_{Z_{2}^2}(t) \cdot \, .... \cdot M_{Z_{n}^2}(t) = \frac{1}{(1 - 2t)^{n/2}}

    (since Z_{1}^2, \, Z_{2}^2, \, .... \, Z_{n}^2 are i.i.d. random variables).

    This is recognised as the moment generating function for the chi-squared distribution with n degrees of freedom. And since when a moment generating function exists there is a unique distribution corresponding to that moment generating function, ....
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