How close to being memoryless can you make a distribution with bounded support? Related to Exponential-like distribution with support [0,1]  I wondered just how close to memorylessness a continuous distribution with bounded support can get. For a continuous variable to be memoryless, it has to be exponential, just as a discrete memoryless distributions must be geometric, so this is a defining feature of the exponential distribution. If the support is bounded, the distribution cannot be exponential so cannot be memoryless, but we may still be able to define a sense in which it comes "close" to being memoryless.
We say a continuous distribution is memoryless if for all $s, t \geq 0$ we have $$\Pr(X>t+s \mid X>t)=\Pr(X>s)$$.
Let's say that we have got "close" to being memoryless if, for example, the absolute value of $$\Pr(X>t+s \mid X>t) - \Pr(X>s)$$ is very small for any choice of $s, t$ and we might want to restrict it so that $X, s, t, s+t$ all lie between 0 and 1. One metric for "closeness to memorylessness" might be the least upper bound for that absolute value of the difference, but if another metric has been proposed before that's fine too.
So whichever sensible way we measure it, just how close to memorylessness can we get?
I suspect the answer is we can get arbitrarily close by using a truncated exponential distribution with mean increasingly nearer to zero. But for a fixed mean of $X$, e.g. $\mathbb{E}(X) = 0.1$, it's no longer intuitive (at least to me) that a truncated exponential would be optimal... does anyone have any suggestions? Is it something that has been researched?
 A: In terms of the CDF $F(t)$ or the survival function $S(t) = 1-F(t)$ you have
$$p(X>t+s|X>t) = \frac{S(t+s)}{S(t)}$$
You get this fraction to be constant for different $t$ and $s$ when $S(t)$ is an exponential function.
(And obviously the relation breaks when $t>1$ or $t+s>1$, because that exponential relation ends above 1. So you only have memoryless in some narrow sense)
Truncated exponential with point masses
We can have an exponential function for the survival function as follows
$$S(t) = \begin{cases}
1 &\quad & \text{for $t\leq0$}\\
a \exp(-bt) &\quad &\text{for $0<t\leq1$}\\
0 &\quad& \text{for $t>1$}
\end{cases}$$
This is a truncated exponential distribution with extra point masses at $t=0$ and $t=1$ (a mixture of a continuous and discrete distribution).
The most extreme case is when you have a single point mass at $t=1$, by setting $a=1$ and $b=0$, which is $S(t)=1$ for $t<1$ and $S(t)=0$ for $t\geq1$. Or when you have a single point mass at $t=0$, by setting $a=0$, in which case the definition of the conditional probability (which equals zero) becomes a vacuous truth.
Truncated exponential
At first I thought that the truncated exponential would satisfy as well. But in this case the survival function will be
$$S(t) = \begin{cases}
1 &\quad & \text{for $t\leq0$}\\
\frac{\exp(-bt) - \exp(-b)}{1-\exp(-b)}  &\quad &\text{for $0<t\leq1$}\\
0 &\quad& \text{for $t>1$}
\end{cases}$$
It is translated/shifted by a constant $\exp(-b)$ to ensure that $S(1)=0$ and continuous.
If the distribution needs to be continuous
In this case you can use the distribution with point masses and substitute the point masses with a continuous function and make them arbitrarily small.
You can also use the truncated exponential and make the constant $\exp(-b)$ arbitrarily small. In the extreme cases $b\to \infty$ you approach the situation with a point mass in $t=0$.
