Bound the probability - $P(|1 - e^{-(x - y)}| \geq \delta) $ if $P(|x-y| \geq \epsilon) \leq \kappa $.

Question

Is the quantity $|1 - e^{-(x - y)}|$ bounded if $|x-y|$ is bounded? I think answer is obviously yes. But can we bound the following probability - $P(|1 - e^{-(x - y)}| \geq \delta) $ if $P(|x-y| \geq \epsilon) \leq \kappa $ ?

Answer 1

Assuming we are interested in $0<\delta<1$, we have $$P(|1-e^{y-x}|\le\delta)=P\left(\ln(1-\delta)\le y-x\le \ln(1+\delta) \right)\\ \ge P\left(-\delta\le y-x\le \frac\delta{1+\delta} \right)\\ \ge P\left(|y-x|\le \frac\delta{1+\delta} \right), $$ where I have used the second set of inequalities given in Intuition behind logarithm inequality: $1 - \frac1x \leq \log x \leq x-1$.

Now if $P(|x-y|\le \epsilon)\ge 1-\kappa_\epsilon$, we get $$P(|1-e^{y-x}|\le\delta)\ge 1-\kappa_{\delta/(1+\delta)}.$$