If Xi ~ N(0,1), derive the probability density function of Xi^2. Write down the probability density function of SigmaXi^2

>>Jacques

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- Mar 20th 2008, 04:52 AMJacquesRouxDistribution Theory
If Xi ~ N(0,1), derive the probability density function of Xi^2. Write down the probability density function of SigmaXi^2

>>Jacques - Mar 20th 2008, 05:02 AMmr fantastic
I don't have the time right now to post a solution. But if you read number 10 of this you just might figure it out yourself ....

If you're still stuck, please say where - if no-one else replies I'll post some help in the morning (*my*morning!) - Mar 20th 2008, 08:25 AMJacquesRoux
Thanks very much will let you know if I need more help!

- Mar 23rd 2008, 10:39 AMJacquesRouxDistribution Theory
C = Integral R exp(-z2 / 2)dz = (2 )1/2.

Hint: Express C^2 as a double integral over R^2 and then convert to polar coordinates:

Is this the correct method for the above problem. If it is the correct one how will I determine the intervals of the double integral. I am really stuck with this problem and really will appreciate guidedance with this problem. (Headbang) Please help me!! - Mar 23rd 2008, 02:40 PMmr fantastic
I'm sorry but

C = Integral R exp(-z2 / 2)dz = (2 )1/2.

Hint: Express C^2 as a double integral over R^2 and then convert to polar coordinates:

is unclear to me. What is R? Does exp(-z2 / 2) mean $\displaystyle e^{-z^2/2}\,$ ? What does (2 )1/2 mean? It would help if you could typeset these sorts of complex expressions using latex. Where has the hint come from? - Mar 23rd 2008, 03:10 PMJacquesRouxDistribution Theory
Sorry this was the question:

If Xi ~ N(0,1), derive the probability density function of Xi^2. Write down the probability density function of SigmaXi^2

and I have no clue how to even start the problem. Can you please give me some guidedance on how to even begin solving this problem.

Jacques - Mar 23rd 2008, 04:19 PMmr fantastic
Let $\displaystyle F(x) = \Pr(\chi^2 < x)$

$\displaystyle \Rightarrow F(x) = \Pr(-\sqrt{x} < \chi < \sqrt{x})$

$\displaystyle = \frac{1}{\sqrt{2 \pi}} \, \int_{-\sqrt{x}}^{\sqrt{x}} e^{-u^2/2} \, du$

$\displaystyle = \frac{2}{\sqrt{2 \pi}} \, \int_{0}^{\sqrt{x}} e^{-u^2/2} \, du$.

Therefore $\displaystyle f(x) = \frac{dF}{dx} = \frac{2}{\sqrt{2 \pi}} \, e^{-x/2} \, \left (\frac{1}{2 \sqrt{x}} \right) \, $ , $\displaystyle x > 0$

where the Fundamental Theorem of Calculus and the chain rule have been used to get the derivative

$\displaystyle = \frac{1}{\sqrt{2 \pi}} \, e^{-x/2} \, \frac{1}{\sqrt{x}} \, $ , $\displaystyle x > 0$

and f(x) = 0 for x < 0.

As expected, this is the pdf for the chi-square distribution of degree 1. Note: $\displaystyle \frac{1}{\sqrt{2 \pi}} = \frac{1}{\Gamma(1/2) 2^{1/2}}$.

You could also find the moment generating function of $\displaystyle \chi^2$ and recognise it as being the moment generating function of the chi-square distibution of degree 1. - Mar 24th 2008, 02:13 AMJacquesRouxDistribution Theory
Thanks very much!!

Jacques(Rock)