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About LinkBacks1. 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#post119542Originally Posted by weakmath
Your parameters are not unknown. The mean isand the standard deviation is
. 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.
In fact, at second glance Q1 is even easier than I thought. The thread I referenced actually gives a solution to Q1 since
~
.
Q2:
The moment generating function ofis:
![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](http://latex.codecogs.com/png.latex?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)
(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
(sinceare 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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