My Math Forum Transformation of a Random Variable (chi-square)

 November 8th, 2014, 06:29 AM #1 Newbie   Joined: Nov 2014 From: Brussels Posts: 2 Thanks: 0 Transformation of a Random Variable (chi-square) We have a random variable x with p.d.f. $\sqrt{\dfrac{\theta}{\pi x}}\exp(-x\theta)$, x>0 and θ a positive parameter. We are required to show that 2θx has a χ2 distribution with 1 degree of freedom and deduce that, if $x_1,\dots,x_n$ are independent r.v. with this p.d.f., then $2\theta\sum_{i=1}^n x_i$ has a χ2 distribution with n degrees of freedom. Using transformation y=2θx I found the pdf of $$y=\frac{1}{\sqrt{2\pi}}y^{-1/2}e^{-y/2}.$$ How do I find the distribution of $2\theta\sum_{i=1}^n x_i$? Do I need to find the likelihood function (which contains $\sum_{i=1}^n x_i$) first? How do I recognise the degrees of freedom of this distribution (Is it n because it involves $x_1,\dots,x_n$, i.e. n random variables?
November 8th, 2014, 08:07 AM   #2
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From: Erewhon

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Quote:
 Originally Posted by SGill We have a random variable x with p.d.f. $\sqrt{\dfrac{\theta}{\pi x}}\exp(-x\theta)$, x>0 and θ a positive parameter. We are required to show that 2θx has a χ2 distribution with 1 degree of freedom and deduce that, if $x_1,\dots,x_n$ are independent r.v. with this p.d.f., then $2\theta\sum_{i=1}^n x_i$ has a χ2 distribution with n degrees of freedom. Using transformation y=2θx I found the pdf of $$y=\frac{1}{\sqrt{2\pi}}y^{-1/2}e^{-y/2}.$$ How do I find the distribution of $2\theta\sum_{i=1}^n x_i$? Do I need to find the likelihood function (which contains $\sum_{i=1}^n x_i$) first? How do I recognise the degrees of freedom of this distribution (Is it n because it involves $x_1,\dots,x_n$, i.e. n random variables?
You use the result that if $v\sim N(0,1)$ then $v^2=y\sim \chi_1^2$ and that the sum of the squares of $N$ independent standard normal RVs is $\sim \chi_N^2$.

Otherwise, use the moment generating function of the characteristic function.

CB

Last edited by CaptainBlack; November 8th, 2014 at 08:18 AM.

 November 8th, 2014, 10:38 AM #3 Newbie   Joined: Nov 2014 From: Brussels Posts: 2 Thanks: 0 Thank you very much for your response. Could you, please, elaborate, on how using the moment generating function would help us in this respect ( i.e. finding the 2θΣx(i=1 to n) distribution?
November 8th, 2014, 12:12 PM   #4
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From: Erewhon

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Quote:
 Originally Posted by SGill Thank you very much for your response. Could you, please, elaborate, on how using the moment generating function would help us in this respect ( i.e. finding the 2θΣx(i=1 to n) distribution?
The moment generating function of the sum of independent RV is the product of their individual moment generating functions. So if the N power of the MGF of a $\chi^2$ distribution with 1 DF is the MGF of a $\chi^2$ distribution with N DF you are done.

CB

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