Distributions having this specific property

Question

Are there any distributions $S$ for which we have the property that $$P(S \leq x-y|S \leq x) \text{ is independent of x} \quad \quad (1)$$ for $x,y \geq 0$. I know that for the 'complementary' property that is $$P(S \geq x+y|S \geq x) \text{ is independent of x}$$ exponential distribution is an example with this property but I am not aware of any distribution with property (1). Any help will be appreciated.

Answer 1

No, it is not going to be possible.

$P(S \leq x-y\mid S \leq x) = \dfrac{F(x-y)}{F(x)}$ using the CDF and

if this is independent of $x$ then it is some function of $y$, let's say $g(y)$.

So $F(x-y)=g(y)F(x)$ and $F(x)=g(y)F(x+y)$ and $F(x+y)=g(y)F(x+2y)$ etc.

which means $F(x+ny)=\dfrac{F(x)}{g(y)^n}$.

• If $g(y)>1$ then $F(x+ny)$ would be decreasing as $n$ increases, but it is a CDF
• If $g(y)=1$ then $F(x+ny)$ would be constant or periodic as $n$ increases, but it is a CDF
• If $0<g(y)<1$ then there would be an $n>\frac{\log(F(x))}{\log(g(x))}$ for which $F(x+ny)>1$, but it is a CDF