dan adalah variabel acak yang didistribusikan secara independen di mana dan . Apa distribusi ?
Kepadatan sambungan diberikan oleh
Using the change of variables such that and ,
I get the joint density of as
Marginal pdf of is then , which does not lead me anywhere.
Again, while finding the distribution function of , an incomplete beta/gamma function shows up:
What is an appropriate change of variables here? Is there another way to find the distribution of ?
I tried using different relations between Chi-Squared, Beta, 'F' and 't' distributions but nothing seems to work. Perhaps I am missing something obvious.
As mentioned by @Francis, this transformation is a generalization of the Box-Müller transform.
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StubbornAtom
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Jawaban:
Here's an algebraic proof. I'm going to instead letX∼χn−1 (not squared) so that we need to find Z:=(2Y−1)X . These are all guaranteed to be valid densities so I'm not going to track normalization constants. We have
For convenience letm=n/2−2 . Multiply both sides by ez2/2 to get
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(This argument applies to integraln=2,3,4,… .)
If you need some numerical convincing (which is always wise, because it can uncover errors in reasoning and calculation), simulate:
The agreement between the simulated results and the claimed standard Normal distribution is excellent across this range of values ofn .
Experiment further with the
R
code that produced these plots if you wish.sumber
As user @Chaconne has already done, I was able to provide an algebraic proof with this particular transformation. I have not skipped any details.
(We already haven>2 for the density of Y to be valid).
Let us consider the transformation(X,Y)↦(U,V) such that U=(2Y−1)X−−√ and V=X .
This impliesx=v and y=12(uv√+1) .
Now,x>0⟹v>0 and 0<y<1⟹−v√<u<v√ ,
so that the bivariate support of(U,V) is simply S={(u,v):0<u2<v<∞,u∈R} .
Absolute value of the Jacobian of transformation is|J|=12v√ .
Joint density of(U,V) is thus
Now, using Legendre's duplication formula,
So forn>2 ,
Marginal pdf ofU is then given by
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This is more of a black box answer (i.e., the algebraic details are missing) using Mathematica. In short as @whuber states the answer is that the distribution ofZ is a standard normal distribution.
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Not an answer per se, but it may be worthwhile to point out the connection to Box-Muller transformation.
Consider the Box-Muller transformationZ=−2lnU−−−−−−√sin(2πV) , where U,V∼U(0,1) . We can show that −lnU∼Exp(1) , i.e. −2lnU∼χ22 . On the other hand, we can show that sin(2πV) has the location-scale arcsine distribution, which agrees with the distribution of 2B(1/2,1/2)−1 . This means Box-Muller transformation is a special case of (2Y−1)X−−√ when n=3 .
Related:
How to use Box-Muller transform to generate n-dimensional normal random variables
How to generate uniformly distributed points on the surface of the 3-d unit sphere?
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