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For which distribution same pdf is generated for given random variable

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1

$begingroup$

For which of the distribution same pdf is generated for random variable X and 1/X.
Is it F(2,2)

statistics

share|cite|improve this question

edited May 27 at 8:13

ten do

asked May 27 at 8:08

ten doten do

335

$endgroup$

• 
  
  
  

  
  5
  

  
  
  
  
  
  

  
  $begingroup$
  It is unclear from what you are actually asking.
  $endgroup$
  – 1123581321
  May 27 at 8:10
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @1123581321 Suppose $X$ had density function $f(x)$ and $Y=frac1X$ with density function $g(y)$. The question wants to consider cases where $g(y) = f(y)$
  $endgroup$
  – Henry
  May 27 at 9:03
  

  
  
  
  

add a comment | 

1

$begingroup$

For which of the distribution same pdf is generated for random variable X and 1/X.
Is it F(2,2)

statistics

share|cite|improve this question

edited May 27 at 8:13

ten do

asked May 27 at 8:08

ten doten do

335

$endgroup$

• 
  
  
  

  
  5
  

  
  
  
  
  
  

  
  $begingroup$
  It is unclear from what you are actually asking.
  $endgroup$
  – 1123581321
  May 27 at 8:10
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @1123581321 Suppose $X$ had density function $f(x)$ and $Y=frac1X$ with density function $g(y)$. The question wants to consider cases where $g(y) = f(y)$
  $endgroup$
  – Henry
  May 27 at 9:03
  

  
  
  
  

add a comment | 

$begingroup$

For which of the distribution same pdf is generated for random variable X and 1/X.
Is it F(2,2)

statistics

share|cite|improve this question

edited May 27 at 8:13

ten do

asked May 27 at 8:08

ten doten do

335

$endgroup$

share|cite|improve this question

edited May 27 at 8:13

ten do

asked May 27 at 8:08

ten doten do

335

share|cite|improve this question

edited May 27 at 8:13

ten do

asked May 27 at 8:08

ten doten do

335

asked May 27 at 8:08

ten doten do

335

asked May 27 at 8:08

ten doten do

335

2 Answers
2

active

oldest

votes

5

$begingroup$

Let $U,V$ be iid random variables with $P(U=0)=0$.

Then if $X$ is defined as $fracUV$ it will have the same distribution as $frac1X=fracVU$.

If moreover $X$ has a PDF then $frac1X$ will also have the same PDF.

Special case: $U$ has chi-squared distribution. Then $X$ has $F$-distribution.

share|cite|improve this answer

answered May 27 at 9:00

drhabdrhab

106k546137

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  It would be interesting if all solutions could be decomposed into a ratio like this
  $endgroup$
  – Henry
  May 27 at 9:23
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Henry Indeed. Uptil now I did not manage to prove that.
  $endgroup$
  – drhab
  May 27 at 9:30
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Rahul : Ah. Slow about an obvious. <sigh> Thanks.
  $endgroup$
  – Eric Towers
  May 27 at 17:33
  

  
  
  
  

add a comment | 

5

$begingroup$

You are looking for a solution to $f(x) = dfracf(1/x)x^2$

So take any non-negative function $g(x)$ on $[-1,1]$ where $k= intlimits_-1^1 g(x), dx$ is positive and finite

then a solution will be

• $f(x) = dfracg(x)2k text when -1 le x le 1$


• $f(x) = dfracg(1/x)2kx^2 text when x lt -1 text or x gt 1$

and I think all solutions will essentially be of this form

One solution is $f(x)=dfrac1(1+x)^2$ for $x gt 0$ and this is indeed an $F(2,2)$ distribution

but there are many others, including the Cauchy density $f(x)=dfrac1pi(1+x^2)$ on $x in mathbb R$

Another simple illustration, with $g(x)=1$ and so $k=2$, has $f(x)=frac14$ when $x in [-1,1]$ and $f(x)=frac14x^2$ otherwise.

share|cite|improve this answer

edited May 27 at 18:05

answered May 27 at 8:59

HenryHenry

103k483172

$endgroup$

add a comment | 

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5

$begingroup$

Let $U,V$ be iid random variables with $P(U=0)=0$.

Then if $X$ is defined as $fracUV$ it will have the same distribution as $frac1X=fracVU$.

If moreover $X$ has a PDF then $frac1X$ will also have the same PDF.

Special case: $U$ has chi-squared distribution. Then $X$ has $F$-distribution.

share|cite|improve this answer

answered May 27 at 9:00

drhabdrhab

106k546137

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  It would be interesting if all solutions could be decomposed into a ratio like this
  $endgroup$
  – Henry
  May 27 at 9:23
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Henry Indeed. Uptil now I did not manage to prove that.
  $endgroup$
  – drhab
  May 27 at 9:30
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Rahul : Ah. Slow about an obvious. <sigh> Thanks.
  $endgroup$
  – Eric Towers
  May 27 at 17:33
  

  
  
  
  

add a comment | 

5

$begingroup$

Let $U,V$ be iid random variables with $P(U=0)=0$.

Then if $X$ is defined as $fracUV$ it will have the same distribution as $frac1X=fracVU$.

If moreover $X$ has a PDF then $frac1X$ will also have the same PDF.

Special case: $U$ has chi-squared distribution. Then $X$ has $F$-distribution.

share|cite|improve this answer

answered May 27 at 9:00

drhabdrhab

106k546137

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  It would be interesting if all solutions could be decomposed into a ratio like this
  $endgroup$
  – Henry
  May 27 at 9:23
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Henry Indeed. Uptil now I did not manage to prove that.
  $endgroup$
  – drhab
  May 27 at 9:30
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Rahul : Ah. Slow about an obvious. <sigh> Thanks.
  $endgroup$
  – Eric Towers
  May 27 at 17:33
  

  
  
  
  

add a comment | 

$begingroup$

Let $U,V$ be iid random variables with $P(U=0)=0$.

Then if $X$ is defined as $fracUV$ it will have the same distribution as $frac1X=fracVU$.

If moreover $X$ has a PDF then $frac1X$ will also have the same PDF.

Special case: $U$ has chi-squared distribution. Then $X$ has $F$-distribution.

share|cite|improve this answer

answered May 27 at 9:00

drhabdrhab

106k546137

$endgroup$

share|cite|improve this answer

answered May 27 at 9:00

drhabdrhab

106k546137

answered May 27 at 9:00

drhabdrhab

106k546137

answered May 27 at 9:00

drhabdrhab

106k546137

5

$begingroup$

You are looking for a solution to $f(x) = dfracf(1/x)x^2$

So take any non-negative function $g(x)$ on $[-1,1]$ where $k= intlimits_-1^1 g(x), dx$ is positive and finite

then a solution will be

• $f(x) = dfracg(x)2k text when -1 le x le 1$


• $f(x) = dfracg(1/x)2kx^2 text when x lt -1 text or x gt 1$

and I think all solutions will essentially be of this form

One solution is $f(x)=dfrac1(1+x)^2$ for $x gt 0$ and this is indeed an $F(2,2)$ distribution

but there are many others, including the Cauchy density $f(x)=dfrac1pi(1+x^2)$ on $x in mathbb R$

Another simple illustration, with $g(x)=1$ and so $k=2$, has $f(x)=frac14$ when $x in [-1,1]$ and $f(x)=frac14x^2$ otherwise.

share|cite|improve this answer

edited May 27 at 18:05

answered May 27 at 8:59

HenryHenry

103k483172

$endgroup$

add a comment | 

5

$begingroup$

You are looking for a solution to $f(x) = dfracf(1/x)x^2$

So take any non-negative function $g(x)$ on $[-1,1]$ where $k= intlimits_-1^1 g(x), dx$ is positive and finite

then a solution will be

• $f(x) = dfracg(x)2k text when -1 le x le 1$


• $f(x) = dfracg(1/x)2kx^2 text when x lt -1 text or x gt 1$

and I think all solutions will essentially be of this form

One solution is $f(x)=dfrac1(1+x)^2$ for $x gt 0$ and this is indeed an $F(2,2)$ distribution

but there are many others, including the Cauchy density $f(x)=dfrac1pi(1+x^2)$ on $x in mathbb R$

Another simple illustration, with $g(x)=1$ and so $k=2$, has $f(x)=frac14$ when $x in [-1,1]$ and $f(x)=frac14x^2$ otherwise.

share|cite|improve this answer

edited May 27 at 18:05

answered May 27 at 8:59

HenryHenry

103k483172

$endgroup$

add a comment | 

$begingroup$

You are looking for a solution to $f(x) = dfracf(1/x)x^2$

So take any non-negative function $g(x)$ on $[-1,1]$ where $k= intlimits_-1^1 g(x), dx$ is positive and finite

then a solution will be

• $f(x) = dfracg(x)2k text when -1 le x le 1$


• $f(x) = dfracg(1/x)2kx^2 text when x lt -1 text or x gt 1$

and I think all solutions will essentially be of this form

One solution is $f(x)=dfrac1(1+x)^2$ for $x gt 0$ and this is indeed an $F(2,2)$ distribution

but there are many others, including the Cauchy density $f(x)=dfrac1pi(1+x^2)$ on $x in mathbb R$

Another simple illustration, with $g(x)=1$ and so $k=2$, has $f(x)=frac14$ when $x in [-1,1]$ and $f(x)=frac14x^2$ otherwise.

share|cite|improve this answer

edited May 27 at 18:05

answered May 27 at 8:59

HenryHenry

103k483172

$endgroup$

share|cite|improve this answer

edited May 27 at 18:05

answered May 27 at 8:59

HenryHenry

103k483172

answered May 27 at 8:59

HenryHenry

103k483172

answered May 27 at 8:59

HenryHenry

103k483172