how to derive this property of exponential distribution?

Question

Let $X$ be a real random variable with exponential distribution on a measure space with probability measure $\Bbb P$. Let $\Bbb E$ be the expected value. Then \begin{equation}\Bbb P(X>s+t|X>s)=\Bbb P(X>t).\end{equation}

How can we derive the following property? \begin{equation}\Bbb E(X-t|X>t) = \Bbb E(X).\end{equation}

Answer 1

Let $f_X$ be the density function of $X$. We have (by memorylessness) for all real $x,a$ \begin{equation}\Bbb P(X>x\mid X>a)=\Bbb P(X>x-a).\end{equation}

Thus, by the Lebesgue Differentiation Theorem, $f_{X\mid X>a}(x)=f_X(x-a)$ for all $a$ and Lebesgue almost all $x$. It follows that (taken from the first answer on Conditional expectation of exponential random variable by Dilip Sarwate)

\begin{equation}\begin{split} \Bbb E(X\mid X > x) &:= \int_{-\infty}^\infty t f_{X \mid X > x}(t)\,\mathrm dt\\ &= \int_{-\infty}^\infty t f_X(t-x)\,\mathrm dt\\ &= \int_{-\infty}^\infty (x+u) f_X(u)\,\mathrm du \quad{\text{on substituting}~u = t-x}\\ &= x + E(X). \end{split}\end{equation}

This is equivalent to your claim by linearity of the expected value.