Complementary of memoryless property for Exponential random variables

Question

Suppose $S$ is an exponential random variable with parameter $\lambda$. We know from the memoryless property of exponential $$P(S \geq x+y|S \geq x)=P(S \geq y)$$ But can we say anything about the conditional probability $$P(S \leq x-y|S \leq x)?$$

Answer 1

Well, you can compute it: $$P( S\le x-y\mid S\le x) = \frac{1-e^{-\lambda(x-y)}}{1-e^{-\lambda x}}.$$

What this is saying is that conditional on $S\le x,$ $x-S$ has a reversed and truncated exponential distribution. This is no surprise: that's how conditional distributions of this sort always work: the distribution of $X$ conditional on $X\in [a,b]$ is just given by the PDF of $X,$ truncated to that interval and rescaled to be normalized.

Just to be explicit, $$ f_{S\mid S\le x} (s) = \frac{\lambda e^{-\lambda s}}{\int_0^x \lambda e^{-\lambda s}ds } = \frac{\lambda e^{-\lambda s}}{1-e^{-\lambda x}}=\frac{\lambda e^{\lambda(x-s)}}{e^{\lambda x}-1}$$ for $s\in [0,x].$

How memorylessness manifests in the case where we're conditioning on $S\ge x$ is that the conditional distribution happens to be the original distribution, shifted up by $x$ $$ f_{S\mid S\ge x}(s) = \frac{\lambda e^{-\lambda s}}{\int_x^\infty \lambda e^{-\lambda s}ds } = \lambda e^{-\lambda(s-x)}$$ for $s\in[x,\infty).$

So in general, conditioning in either direction introduces some scaling and truncation that depends on $x.$ Just in the very particular case of conditioning an exponential to be larger than $x$ this scaling and truncation takes the exact form of shifting the variable to the right by $x.$ There doesn't seem to be much distinguishing the case where we condition an exponential to be smaller than $x$ from the general case.