If Xi ~ N(0,1), derive the probability density function of Xi^2. Write down the probability density function of SigmaXi^2
>>Jacques
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!)
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. Please help me!!
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?
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
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.