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I am reading an article in which the author states that given two i.i.d random variables $X,Y\sim\mathcal{N}(0,1)$, we have $\mathbb{P}(\frac{X}{Y}\le t)=\mathbb{P}(\frac{X}{|Y|}\le t)$ since the random varaibles $\frac{X}{Y}$ and $\frac{X}{|Y|}$ are identically distributed by the symmetry of the standard normal distribution.

But I don't really understand why the random variables $\frac{X}{Y}$ and $\frac{X}{|Y|}$ are identically distributed. Can you provide a reason for why these two random variables are identically distributed?

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  • $\begingroup$ Even without the symmetry argument, one can derive both distributions to show that they have the standard Cauchy distribution. This has already been done in multiple posts here. $\endgroup$ – StubbornAtom Feb 26 '18 at 15:51
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    $\begingroup$ You should have waited before cross-posting on Math S.E. $\endgroup$ – StubbornAtom Feb 26 '18 at 16:23
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    $\begingroup$ @StubbornAtom the difference between answers from mathematicians and statisticians (or physicists/engineers to which I associate more closely) is certainly amusing and worthwhile for meta-discussion. $\endgroup$ – Martijn Weterings Feb 26 '18 at 17:23
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    $\begingroup$ Please do not post the same question on multiple sites. Each community should have an honest shot at answering without anybody's time being wasted. $\endgroup$ – D.W. Feb 27 '18 at 1:32
  • $\begingroup$ Previously asked: stats.stackexchange.com/questions/243849/… $\endgroup$ – StubbornAtom Apr 14 '18 at 9:43
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There are different way to see this.

1

The top side (y positive) is a (point symmetric) mirror image of the bottom side (y negative). So you can express $P(\frac{x}{y}|y<0)$ by $P(\frac{x}{y}|y>0)$ and similarly $P(\frac{x}{|y|}|y<0)$ by $P(\frac{x}{|y|}|y>0)$

2

  • $\frac{x}{y}=b$ occurs when $\frac{x}{y}=b$ for $y>0$ and $\frac{x}{y}=b$ for $y<0$
  • $\frac{x}{|y|}=b$ occurs when $\frac{x}{y}=b$ for $y>0$ and $\frac{x}{y}=-b$ for $y<0$

thus effectively you flip the bottom half from $\frac{x}{y}=b$ to $\frac{x}{y}=-b$ but this half is symmetric in $P(\frac{x}{y}|y<0) = P(-\frac{x}{y}|y<0)$

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a conversion to cyclic coordinates may also work. $P(\frac{x}{y}=a) \propto 2 P(\theta=tan^{-1}(a))$ and also $P(\frac{x}{|y|}=a) \propto 2 P(\theta=tan^{-1}(a))$

image: 1000 points x,y distributed i.i.d. N(0,1)

1000 points x,y distributed i.i.d. N(0,1)

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Setting $B=\frac{X}{|X|}$ and $C=\frac{Y}{|Y|}$ we have $B,C,|X|,|Y|$ independent.

Observe that $\frac{X}{Y}=\frac{B}{C}\frac{|X|}{|Y|}$ and $\frac{X}{|Y|}=B\frac{|X|}{|Y|}$.

It is not difficult to prove that $\frac{B}{C}$ and $B$ have the same distribution.

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If you look at a Cartesian plane, then given any t, the points where $\frac{x}{y}$ = t will be a line with slope $\frac{1}{t}$. That is, the equation of the line will be x=ty. The points where $\frac{x}{|y|}$ = t will be that line with the bottom half reflected horizontally. That is, for negative y, the equation will become x= -ty. So by including that absolute value, we are, for negative y values, going from looking for P(x=2y0, y=y0) to looking for P(x= -2y0, y=y0). But because the normal distribution is symmetric, the probabilities are the same.

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