ketikadansecara independen


12

X dan adalah variabel acak yang didistribusikan secara independen di mana dan . Apa distribusi ?YXχ(n1)2YBeta(n21,n21)Z=(2Y1)X

Kepadatan sambungan diberikan oleh(X,Y)

fX,Y(x,y)=fX(x)fY(y)=ex2xn1212n12Γ(n12)yn22(1y)n22B(n21,n21)1{x>0,0<y<1}

Using the change of variables (X,Y)(Z,W) such that Z=(2Y1)X and W=X,

I get the joint density of (Z,W) as

fZ,W(z,w)=ew22wn3(14z24w2)n222n12Γ(n12)B(n21,n21)1{w>0,|z|<w}

Marginal pdf of Z is then fZ(z)=|z|fZ,W(z,w)dw, which does not lead me anywhere.

Again, while finding the distribution function of Z, an incomplete beta/gamma function shows up:

FZ(z)=Pr(Zz)

=Pr((2Y1)Xz)=(2y1)xzfX,Y(x,y)dxdy

What is an appropriate change of variables here? Is there another way to find the distribution of Z?

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.


4
Looks like a generalization of Box-Muller transformation
Francis

Jawaban:


10

Here's an algebraic proof. I'm going to instead let Xχn1 (not squared) so that we need to find Z:=(2Y1)X. These are all guaranteed to be valid densities so I'm not going to track normalization constants. We have

fX,Y(x,y)xn2ex2/2[y(1y)]n/221{0<x,0<y<1}.
Let Z=(2y1)X and W=X so the inverse transforms are x(z,w)=w and y(z,w)=z+w2w=z2w+12. This gives us |J|=12w. This leads us to
fZ,W(z,w)wn1ew2/2[z+w2w(1z+w2w)]n/221{0<w,1<zw<1}
wew2/2(w2z2)n/221{|z|<w}.
Thus
fZ(z)w>|z|wew2/2(w2z2)n/22dw.

For convenience let m=n/22. Multiply both sides by ez2/2 to get

ez2/2fZ(z)|z|we(w2z2)/2(w2z2)mdw.
Now let 2u=w2z2 so du=wdw. This gives us
ez2/2fZ(z)2m0umeudu=2mΓ(m+1).
Because this final integral doesn't depend on z, we have shown that ez2/2fZ(z)1, therefore
ZN(0,1).

1
+1. I am glad you restored this answer, because it covers all values of n, not just integral ones.
whuber

@whuber thanks, I somehow put z2w2 instead of w2z2 and it took me a while to figure out why I was getting weird behavior when n is odd
jld

9

2Y1 is distributed like one coordinate of a uniform distribution on the n1 sphere; X has the distribution of the sum of squares of n1 iid standard Normal variates; and these two quantities are independent. Geometrically (2Y1)X has the distribution of one coordinate: that is, it must have a standard Normal distribution.

(This argument applies to integral n=2,3,4,.)

If you need some numerical convincing (which is always wise, because it can uncover errors in reasoning and calculation), simulate:

Figure showing four histograms for n=2,3,4,5

The agreement between the simulated results and the claimed standard Normal distribution is excellent across this range of values of n.

Experiment further with the R code that produced these plots if you wish.

n.sim <- 1e5
n <- 2:5
X <- data.frame(Z = c(sapply(n, function(n){
  y <- rbeta(n.sim, n/2-1, n/2-1)  # Generate values of Y
  x <- rchisq(n.sim, n-1)          # Generate values of X
  (2*y - 1) * sqrt(x)              # Return the values of Z
})), n=factor(rep(n, each=n.sim)))

library(ggplot2)
#--Create points along the graph of a standard Normal density
i <- seq(min(z), max(z), length.out=501)
U <- data.frame(X=i, Z=dnorm(i))

#--Plot histograms on top of the density graphs
ggplot(X, aes(Z, ..density..)) + 
  geom_path(aes(X,Z), data=U, size=1) +
  geom_histogram(aes(fill=n), bins=50, alpha=0.5) + 
  facet_wrap(~ n) + 
  ggtitle("Histograms of Simulated Values of Z",
          paste0("Sample size ", n.sim))

1
Thank you, @Stubborn. It does matter that the parameters are consistent, for otherwise the conclusion is incorrect. I'll fix it.
whuber

3

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 have n>2 for the density of Y to be valid).

Let us consider the transformation (X,Y)(U,V) such that U=(2Y1)X and V=X.

This implies x=v and y=12(uv+1).

Now, x>0v>0 and 0<y<1v<u<v,

so that the bivariate support of (U,V) is simply S={(u,v):0<u2<v<,uR}.

Absolute value of the Jacobian of transformation is |J|=12v.

Joint density of (U,V) is thus

fU,V(u,v)=ev2vn121(uv+1)n22(12u2v)n22Γ(n2)(2v)2n12+n22Γ(n12)(Γ(n21))21S

=ev2vn42(v+u)n22(vu)n22Γ(n2)22n32+n22(v)n4Γ(n12)(Γ(n22))21S

Now, using Legendre's duplication formula,

Γ(n2)=2n3πΓ(n22)Γ(n22+12)=2n3πΓ(n22)Γ(n12) where n>2.

So for n>2,

fU,V(u,v)=2n3ev2(vu2)n22π23n72Γ(n21)1S

Marginal pdf of U is then given by

fU(u)=12n12πΓ(n21)u2ev2(vu2)n22dv

=eu222n12πΓ(n21)0et2t(n211)dt

=12n12π(12)n21eu22

=12πeu2/2,uR

2

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 of Z is a standard normal distribution.

(* Transformation *)
f = {(2 y - 1) Sqrt[x], Sqrt[x]};
sol = Solve[{z == (2 y - 1) Sqrt[x], w == Sqrt[x]}, {x, y}][[1]]
(*{x -> w^2,y -> (w+z)/(2 w)} *)
(* Jacobian *)
J = D[f, {{x, y}}]

(* Joint pdf of Z and W *)
{jointpdf, conditions} = FullSimplify[PDF[BetaDistribution[n/2 - 1, n/2 - 1], y] 
  PDF[ChiSquareDistribution[n - 1], x] Abs[Det[J]] /. sol,
  Assumptions -> {w >= 0, 0 <= y <= 1}][[1, 1]]

(* Integrate over W *)
Integrate[jointpdf Boole[conditions], {w, 0, \[Infinity]}, Assumptions -> {n > 2, z == 0}]
(* 1/Sqrt[2 \[Pi]] *)

Integrate[jointpdf Boole[conditions], {w, 0, \[Infinity]}, Assumptions -> {n > 2, z > 0}]
(* Exp[-(z^2/2)]/Sqrt[2 \[Pi]] *)

Integrate[jointpdf Boole[conditions], {w, 0, \[Infinity]}, Assumptions -> {n > 2, z < 0}]
(* Exp[-(z^2/2)]/Sqrt[2 \[Pi]] *)

1

Not an answer per se, but it may be worthwhile to point out the connection to Box-Muller transformation.

Consider the Box-Muller transformation Z=2lnUsin(2πV), where U,VU(0,1). We can show that lnUExp(1), i.e. 2lnUχ22. 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 (2Y1)X when n=3.

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