12
$\begingroup$

$X$ and $Y$ are independently distributed random variables where $X\sim\chi^2_{(n-1)}$ and $Y\sim\text{Beta}\left(\frac{n}{2}-1,\frac{n}{2}-1\right)$. What is the distribution of $Z=(2Y-1)\sqrt X$ ?

Joint density of $(X,Y)$ is given by

$$f_{X,Y}(x,y)=f_X(x)f_Y(y)=\frac{e^{-\frac{x}{2}}x^{\frac{n-1}{2}-1}}{2^{\frac{n-1}{2}}\Gamma\left(\frac{n-1}{2}\right)}\cdot\frac{y^{\frac{n}{2}-2}(1-y)^{\frac{n}{2}-2}}{B\left(\frac{n}{2}-1,\frac{n}{2}-1\right)}\mathbf1_{\{x>0\,,\,0<y<1\}}$$

Using the change of variables $(X,Y)\mapsto(Z,W)$ such that $Z=(2Y-1)\sqrt X$ and $W=\sqrt X$,

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

$$f_{Z,W}(z,w)=\frac{e^{-\frac{w^2}{2}}w^{n-3}\left(\frac{1}{4}-\frac{z^2}{4w^2}\right)^{\frac{n}{2}-2}}{2^{\frac{n-1}{2}}\Gamma\left(\frac{n-1}{2}\right)B\left(\frac{n}{2}-1,\frac{n}{2}-1\right)}\mathbf1_{\{w>0\,,\,|z|<w\}}$$

Marginal pdf of $Z$ is then $f_Z(z)=\displaystyle\int_{|z|}^\infty f_{Z,W}(z,w)\,\mathrm{d}w$, which does not lead me anywhere.

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

$F_Z(z)=\Pr(Z\le z)$

$\quad\qquad=\Pr((2Y-1)\sqrt X\le z)=\displaystyle\iint_{(2y-1)\sqrt{x}\le z}f_{X,Y}(x,y)\,\mathrm{d}x\,\mathrm{d}y$

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.

$\endgroup$
  • 4
    $\begingroup$ Looks like a generalization of Box-Muller transformation $\endgroup$ – Francis Feb 8 '18 at 13:24
10
$\begingroup$

Here's an algebraic proof. I'm going to instead let $X \sim \chi_{n-1}$ (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 $$ f_{X,Y}(x,y) \propto x^{n-2}e^{-x^2/2}\left[y(1-y)\right]^{n/2-2} \mathbf 1_{\{0 < x, \, 0 < y < 1\}}. $$ Let $Z = (2y-1)X$ and $W=X$ so the inverse transforms are $x(z,w) = w$ and $y(z,w) = \frac{z+w}{2w} = \frac{z}{2w} + \frac 12$. This gives us $|J| = \frac 1{2w}$. This leads us to $$ f_{Z,W}(z,w) \propto w^{n-1}e^{-w^2/2} \left[\frac{z+w}{2w} \left( 1 - \frac{z+w}{2w}\right)\right]^{n/2-2} \mathbf 1_{\{0 < w, \, -1 < \frac zw < 1\}} $$ $$ \propto w e^{-w^2/2} \left(w^2-z^2\right)^{n/2-2} \mathbf 1_{\{|z| < w\}}. $$ Thus $$ f_Z(z) \propto \int_{w > |z|} w e^{-w^2/2} \left(w^2-z^2\right)^{n/2-2} \,\text dw. $$

For convenience let $m = n/2-2$. Multiply both sides by $e^{z^2/2}$ to get $$ e^{z^2/2} f_Z(z) \propto \int_{|z|}^\infty w e^{-(w^2-z^2)/2}(w^2-z^2)^m \,\text dw. $$ Now let $2u = w^2 - z^2$ so $\text du = w\,\text dw$. This gives us $$ e^{z^2/2} f_Z(z) \propto 2^m \int_0^\infty u^m e^{-u} \,\text du = 2^m \Gamma(m+1). $$ Because this final integral doesn't depend on $z$, we have shown that $e^{z^2/2} f_Z(z) \propto 1$, therefore $$ Z \sim \mathcal N(0, 1). $$

$\endgroup$
  • 1
    $\begingroup$ +1. I am glad you restored this answer, because it covers all values of $n$, not just integral ones. $\endgroup$ – whuber Feb 10 '18 at 13:34
  • $\begingroup$ @whuber thanks, I somehow put $z^2-w^2$ instead of $w^2-z^2$ and it took me a while to figure out why I was getting weird behavior when $n$ is odd $\endgroup$ – jld Feb 11 '18 at 3:28
9
$\begingroup$

$2Y-1$ is distributed like one coordinate of a uniform distribution on the $n-1$ sphere; $X$ has the distribution of the sum of squares of $n-1$ iid standard Normal variates; and these two quantities are independent. Geometrically $(2Y-1)\sqrt{X}$ has the distribution of one coordinate: that is, it must have a standard Normal distribution.

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

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))
$\endgroup$
  • 1
    $\begingroup$ Thank you, @Stubborn. It does matter that the parameters are consistent, for otherwise the conclusion is incorrect. I'll fix it. $\endgroup$ – whuber Feb 10 '18 at 13:32
3
$\begingroup$

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 $(X,Y)\mapsto (U,V)$ such that $U=(2Y-1)\sqrt{X}$ and $V=X$.

This implies $x=v$ and $y=\frac{1}{2}\left(\frac{u}{\sqrt{v}}+1\right)$.

Now, $x>0\implies v>0$ and $0<y<1\implies-\sqrt{v}<u<\sqrt{v}$,

so that the bivariate support of $(U,V)$ is simply $S=\{(u,v):0<u^2<v<\infty,\,u\in\mathbb{R}\}$.

Absolute value of the Jacobian of transformation is $|J|=\frac{1}{2\sqrt{v}}$.

Joint density of $(U,V)$ is thus

$$f_{U,V}(u,v)=\frac{e^{-\frac{v}{2}}v^{\frac{n-1}{2}-1}\left(\frac{u}{\sqrt{v}}+1\right)^{\frac{n}{2}-2}\left(\frac{1}{2}-\frac{u}{2\sqrt{v}}\right)^{\frac{n}{2}-2}\Gamma(n-2)}{(2\sqrt{v})\,2^{\frac{n-1}{2}+\frac{n}{2}-2}\,\Gamma\left(\frac{n-1}{2}\right)\left(\Gamma\left(\frac{n}{2}-1\right)\right)^2}\mathbf1_{S}$$

$$=\frac{e^{-\frac{v}{2}}v^{\frac{n-4}{2}}(\sqrt{v}+u)^{\frac{n}{2}-2}(\sqrt{v}-u)^{\frac{n}{2}-2}\,\Gamma(n-2)}{2^{\frac{2n-3}{2}+\frac{n}{2}-2}\,(\sqrt{v})^{n-4}\,\Gamma\left(\frac{n-1}{2}\right)\left(\Gamma\left(\frac{n-2}{2}\right)\right)^2}\mathbf1_{S}$$

Now, using Legendre's duplication formula,

$\Gamma(n-2)=\frac{2^{n-3}}{\sqrt \pi}\Gamma\left(\frac{n-2}{2}\right)\Gamma\left(\frac{n-2}{2}+\frac{1}{2}\right)=\frac{2^{n-3}}{\sqrt \pi}\Gamma\left(\frac{n-2}{2}\right)\Gamma\left(\frac{n-1}{2}\right)$ where $n>2$.

So for $n>2$, $$f_{U,V}(u,v)=\frac{2^{n-3}\,e^{-\frac{v}{2}}(v-u^2)^{\frac{n}{2}-2}}{\sqrt \pi\,2^{\frac{3n-7}{2}}\,\Gamma\left(\frac{n}{2}-1\right)}\mathbf1_{S}$$

Marginal pdf of $U$ is then given by

$$f_U(u)=\frac{1}{2^{\frac{n-1}{2}}\sqrt \pi\,\Gamma\left(\frac{n}{2}-1\right)}\int_{u^2}^\infty e^{-\frac{v}{2}}(v-u^2)^{\frac{n}{2}-2}\,\mathrm{d}v$$

$$=\frac{e^{-\frac{u^2}{2}}}{2^{\frac{n-1}{2}}\sqrt \pi\,\Gamma\left(\frac{n}{2}-1\right)}\int_0^\infty e^{-\frac{t}{2}}\,t^{(\frac{n}{2}-1-1)}\,\mathrm{d}t$$

$$=\frac{1}{2^{\frac{n-1}{2}}\sqrt \pi\,\left(\frac{1}{2}\right)^{\frac{n}{2}-1}}e^{-\frac{u^2}{2}}$$

$$=\frac{1}{\sqrt{2\pi}}e^{-u^2/2}\,,u\in\mathbb{R}$$

$\endgroup$
2
$\begingroup$

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]] *)
$\endgroup$
1
$\begingroup$

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

Consider the Box-Muller transformation $Z=\sqrt{-2\ln U}\sin(2\pi V)$, where $U,V\sim U(0,1)$. We can show that $-\ln U\sim\operatorname{Exp}(1)$, i.e. $-2\ln U\sim\chi^2_2$. On the other hand, we can show that $\sin(2\pi V)$ has the location-scale arcsine distribution, which agrees with the distribution of $2\operatorname{B}(1/2,1/2)-1$. This means Box-Muller transformation is a special case of $(2Y-1)\sqrt{X}$ when $n=3$.

Related:

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.