Expected remaining lifetime of exponentially distributed variable

Question

If we express the expected remaining lifetime of the exponentially distributed variable X = Exp(1), given that X>1 as

$$E[X−1|X>1]$$

Then given memorylessness we get

$E[Y-1] = 1/1 - 1 = 0$

Answer 1

No: memorylessness implies $E[X-c|X>c]=E[X]=\lambda$ (here $\lambda=1$) for any $c>0$. We can verify the exponential distribution is memoryless:$$f_X(x|X>c)=\frac{f_X(x)}{P(X>c)}=\frac{\lambda e^{-\lambda x}}{e^{-\lambda c}}=\lambda e^{\lambda(c-x)}$$for all $x\ge c$, i.e. $X-c|X>c\sim\operatorname{Exp}(\lambda)$.