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Question

Suppose that a random variable $X$ has the property that, for any $a > 1$, the conditional distribution of $\frac 1 a X$ given that $X > a$ is the same as the distribution of $X$. Specify all possible distributions of $X$.

My thoughts

I do know that $f_{X\mid Y}(x\mid y) = \frac {f_{X, Y}(x,\ y)} {f_Y(y)}$ and I suppose I have to make use of this relationship, but I have no idea how to even begin - When I find conditional distributions, I am usually given a specific value, not something like $X > a$. Any proposed solutions with intuitive explanations will be greatly appreciated :)

Ethan Mark
  • 2,167

1 Answers1

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The condition in the question can be written as $$ \mathsf{P}(X>ax\mid X>a)=\mathsf{P}(X>x), $$ assuming that $\mathsf{P}(X>a)>0$ for all $a>1$. First, it is clear that $\mathsf{P}(X\le x)=0$ for all $x\le 1$. For $x>1$, the equation becomes (using Bayes' rule): $$ \mathsf{P}(X>ax)=\mathsf{P}(X>x)\mathsf{P}(X>a), $$ which is equivalent to $$ \bar{F}(ax)=\bar{F}(x)\bar{F}(a), $$ where $\bar{F}$ is the complementary cdf of $X$. This is Cauchy’s power equation, and its solution is given by $\bar{F}(x)=x^c$ for an arbitrary constant $c<0$ ($\because \lim_{x\to\infty}\bar{F}(x)=0$).

  • The distribution of a random variable is $P_X(A)=\mathsf{P}(X\in A)$, where $A$ is a "nice" set. In particular, $P_X((x,\infty))=\mathsf{P}(X>x)$.
    • –  Apr 23 '21 at 06:31
  • Given (1), $P_{X/a}((x,\infty))=\mathsf{P}(X/a>x)=\mathsf{P}(X>ax)$.
    • –  Apr 23 '21 at 06:33
  • When $x<1$, ${X>a}\cap {X>ax}={X>a}$.
    • –  Apr 23 '21 at 06:35