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I have to demonstrate the following Probability/Statistics result:

"Find a family of random variables $X$, having pdf $f$, such that $X$ and $Y=f(X)$ have the same distribution."

I have tried to find a proof to this result and I found out that, in the case of pdf $f$ strictly increasing, the result is true if $f$ is the identity function.

I have to find a characterization even in the following 2 cases:

  1. $f$ strictly decreasing;
  2. $f$ stictly increseasing for $x\leq m$ and stricly decreasing for $x \geq m$ (where $m$ is the mode of the distribution).

Any ideas and/or suggestions?

Jose Avilez
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met.91
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    Just a remark. The identity function is not a valid PDF, since it can take negative values and doesn't integrate to $1$. $$f(x) = \begin{cases} x & x\in[0,\sqrt{2}] \ 0 & \text{otherwise} \end{cases}$$ would do though. – Leander Tilsted Kristensen Jun 09 '21 at 22:27
  • @Leander Tilsted Kristensen Thanks for the remark. I already realized that the identity function was not a valid pdf and I had to put some restrictions to the domain in order to have the correct properties for the pdf. Thanks for having told me that! – met.91 Jun 09 '21 at 22:27
  • Interesting problem! I'll spend more time on it tomorrow. In the meantime, one potential way to move forward is to note that for r.v. $X$ with cdf $F_X(x)$, $F_X(X) \sim \text{Uniform}(0, 1)$, that is, plugging any r.v. into its own cdf results in a standard uniform distribution. – Jack Gallagher Jun 10 '21 at 01:30
  • @Jack Gallagher Thanks for your interesting comment! – met.91 Jun 10 '21 at 17:32

2 Answers2

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My previous answer contained a fatal flaw in logic, caused by not taking the absolute value of the determinant of the Jacobian in the change of variables theorem. The following is a more precise answer.

Let $X$ have pdf $f_X(x)$, $Y$ have pdf $f_Y(y)$, and $Y=g(X)$.

By the change of variables theorem (sketches of the proof of which can be found in a previous MSE post Intuitive proof of multivariable changing of variables formula (jacobian) without using mapping and/or measure theory?, we have

$f_Y(y)=f_X(g^{-1}(y))\cdot \left| D_y g^{-1}(y) \right| $

Now, for this question, we are interested in the case where $f_X=f_Y=g$. Call this $f$.

Then the change of variables equation becomes

$f(y)=f\left(f^{-1}(y)\right)\cdot\left|D_y f^{-1}(y) \right|$

and we must now assume $f$ is monotonic since otherwise $f^{-1}$ does not exist as a function. Applying the chain rule, we obtain:

$f(y)=f\left(f^{-1}(y)\right) \cdot \left| \frac{1}{(D_y f) \left( f^{-1}(y) \right) } \right| $

Due to monotonicity,

$f(y)=y \cdot \left| \frac{1}{(D_yf)\left( f^{-1}(y) \right) } \right| $

It can easily be shown that $f(y)=ay^n$, $n=\pm1$, $a\geq0$ are sufficient to fulfill this criterion, but I'm not sure how to prove they are necessary.

Wikipedia's article on Probability density functions gives an interesting formula I hadn't seen in my prob theory courses, which deals with nonmonotonic transformation functions:

$$f_Y(y)=\sum_{k=1}^{n(y)}\left|D_y g_k^{-1}(y)\right|\cdot f_X(g_k^{-1}(y))$$

where $n(y)$ is the number of $x$ such that $g(x)=y$ (essentially this is just combining the strictly increasing and strictly decreasing cases into one function). In your third case, where the density is unimodal, $n=2$ and the formula becomes:

$f(y)=\left|D_yf_1^{-1}(y)\right|\cdot f\left(f_1^{-1}(y)\right)+\left|D_yf_2^{-1}(y)\right|\cdot f\left(f_2^{-1}(y)\right)$

where $f_1, f_2$ are respectively your strictly increasing and strictly decreasing densities on their proper intervals (denoted here with indicator functions $I$):

$f_1(y)=p\frac{y-\mathcal{l}}{m-\mathcal{l}}I_{(\mathcal{l}, m]}(y)$

$f_2(y)=p\frac{1}{y-m+1}I_{(m, n]}(y)$

where $0<\mathcal{l}<m<n<\infty$ and $p=\frac{1}{\frac{m-\mathcal{l}}{2}+\ln{(n-m+1)}}$.

Since

$\left|D_yf_1^{-1}(y)\right|=\left|\frac{m-\mathcal{l}}{p}\right|=\frac{m-\mathcal{l}}{p}$ and

$\left|D_yf_2^{-1}(y)\right|=\left|-\frac{p}{y^2}\right|=\frac{p}{y^2}$

Our density becomes

$f(y)=(y-\mathcal{l})I_{(\mathcal{l}, m]}(y)+\frac{p^2}{y^2(y-m+1)}I_{(m, n]}(y)$

Jack Gallagher
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  • I worked a lot on this result in order to find a characterization for the cases of strictly decreasing and the other one in the last days, but I didn't find any class of valid functions that can work. Your answer confirms what I thought: there's not a criteria for the other 2 cases. Thanks, @Jack Gallagher! – met.91 Jun 13 '21 at 22:35
  • How did you come to the conclusion that $f$ must satisfy $f(f(x))=x$ ??. I mean if we transform a random variable $X$ with PDF $f$ by a strictly monotone differentiable function $g$, we would get $$f_{g(X)}(y) = f_X(g^{-1}(y)) | \frac{d}{dy} g^{-1}(y) |,$$ so it seems like $f$ should rather satisfy $$f(y) = f(f^{-1}(y))|\frac{d}{dy} f^{-1}(y)| = y|\frac{d}{dy} f^{-1}(y)|$$ – Leander Tilsted Kristensen Jun 16 '21 at 19:02
  • @LeanderTilstedKristensen I initially had a proof using that formula and got $f(x) = x$ as the only solution. But you're absolutely right that that first step of mine seems like it's on quite shaky ground. I'll review my argument – Jack Gallagher Jun 17 '21 at 00:54
  • Ah, I found my error in that first proof--I had written the transformation formula as $f_Y(y)=f_X(y) D_y f^{-1}(x)$, as I read the $|\cdot|$ in my notes (we covered the more general 2D case in my Intro Prob class 3 years ago) as just taking the determinant of the Jacobian, not also taking the absolute value. That explains but certainly doesn't excuse the faulty logic I used in reducing met.91's problem to f(f(x))=f(x). – Jack Gallagher Jun 17 '21 at 01:43
  • @met.91 I've updated my answer. Barring any errors, I believe I found such an $f$ for the unimodal case. – Jack Gallagher Jun 23 '21 at 21:06
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I'm here to submit another result that I've just found out.

Considering $f$ strictly decreasing, I know (from previous demonstrated results) that the following equation is valid:

$$1-F\left(f^{-1}(y)\right)=F(y)$$

where we know that $y=f(x)$ so that the equation becomes

$$ 1 - F(x)=F\big(f(x)\big)$$

where $F$ is the CDF and $f$ is the PDF of the random variable $X$, such that

$$1-F(x)=F\big(F'(x)\big)\text.$$

The solution is $$F(x)=\frac{1}{2}+a^{2}\cdot\ln\left(\frac{x}{a}\right)\text.$$

Derivating the expression of $F(x)$ we obtain that $$ f(x)=\frac{a^{2}}{x}\text.$$

If we consider the following

$$ f(x)= \begin{cases} \frac{a^{2}}{x}, \, &{x \in \left[1, e^{\frac{1}{a^{2}}}\right] } \\ 0, \, &\text{otherwise} \end{cases} $$

Can it be considered as a valid PDF?

  1. $f(x)$ is $\geq 0$, for $x \in \left[1, e^{\frac{1}{a^{2}}}\right] $;
  2. the integral of $f(x)$ between $1$ and $e^{\frac{1}{a^{2}}}$ is equal to $1$.
Mohsen Shahriari
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met.91
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  • You are correct and I am wrong. My logic was faulty due to missing an absolute value sign in the transformation. I'll edit my answer to fix its faulty logic over the coming days. – Jack Gallagher Jun 17 '21 at 01:44
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    This is nice, and almost correct. The problem is that if $$f_X(x) = \begin{cases} \frac{a^2}{x} &, x \in [1,e^{1/a^2}] \ 0 &, \text{otherwise} \end{cases}$$ , then $Y=f_X(X)$ has PDF $$f_Y(y) = \begin{cases} \frac{a^2}{y} &, y \in [a^2e^{-1/a^2},a^2] \ 0 &, \text{otherwise} \end{cases},$$ so the PDF's are not defined on the same interval. – Leander Tilsted Kristensen Jun 17 '21 at 09:58
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    Setting $a=1$ i found that $$f(x) = \begin{cases} \frac{1}{x} &, x \in [e^{-1/2},e^{1/2}] \ 0 &,\text{otherwise} \end{cases}$$ does the trick, since $x\mapsto \frac{1}{x}$ maps the interval $[e^{-1/2},e^{1/2}]$ into itself. – Leander Tilsted Kristensen Jun 17 '21 at 10:03
  • @Leander Tilsted Kristensen Thanks for your interesting comments. Very nice observations! This problem that I'm trying to solve and demonstrate is soooo interesting, but so tough at the same time. – met.91 Jun 17 '21 at 10:25
  • @LeanderTilstedKristensen If we consider $$ f(x)= \begin{cases} \frac{a^{2}}{x}, , &{x \in \Bigr[e^{\frac{-1}{2a^{2}}}, e^{\frac{1}{2a^{2}}}\Bigl] } \ 0, , &{otherwise} \end{cases} $$

    does this function generalize your $f(x)$? This should work as a valid PDF and the interval $\Bigr[e^{\frac{-1}{2a^{2}}}, e^{\frac{1}{2a^{2}}}\Bigl]$ should go, through $f(x)$, to itself. Is this right?

     – met.91 Jun 17 '21 at 11:59
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    @met.91 not quite, but close. I believe the general formula should be $$f(x) = \begin{cases} \frac{a^2}{x} &, x \in [ae^{\frac{-1}{2a^2}},ae^{\frac{1}{2a^2}}] \ 0 &,\text{otherwise} \end{cases}.$$ I found this interval by noticing that $x\mapsto a^2/x$ maps an interval $[b,c]$ to $[\frac{a^2}{c},\frac{a^2}{b}]$. So we need $b = \frac{a^2}{c}$ and $c=\frac{a^2}{b}$ or equavalently $a^2 = bc$. And also, since the integral must be $1$ we must have $\log(\frac{c}{b}) = \frac{1}{a^2}$. – Leander Tilsted Kristensen Jun 17 '21 at 21:37
  • @Leander Tilsted Kristensen That interval should be correct. If we use those values and we calculate the limits of $F(x)$ (the CDF of the random variable $X$), the limit to $a e^{\frac{-1}{2a^{2}}}$ is $0$ and the one to $a e^{\frac{1}{2a^{2}}}$ goes to $1$ and $F(x)$ should be a correct CDF. Is this right? – met.91 Jun 18 '21 at 00:26