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For a random variable $X\sim \text{Exp}(\lambda)$ ($\mathbb{E}[X] = \frac{1}{\lambda}$) I feel intuitively that $\mathbb{E}[X|X > x]$ should equal $x + \mathbb{E}[X]$ since by the memoryless property the distribution of $X|X > x$ is the same as that of $X$ but shifted to the right by $x$.
However, I'm struggling to use the memoryless property to give a concrete proof. Any help is much appreciated.
Thanks.
$\ldots$ by the memoryless property the distribution of $X|X > x$ is the same as that of $X$ but shifted to the right by $x$.
Let $f_X(t)$ denote the probability density function (pdf) of $X$. Then, the mathematical formulation for what you correctly state $-$ namely, the conditional pdf of $X$ given that $\{X > x\}$ is the same as that of $X$ but shifted to the right by $x$ $-$ is that $f_{X \mid X > x}(t) = f_X(t-x)$. Hence, $E[X\mid X > x]$, the expected value of $X$ given that $\{X > x\}$ is $$\begin{align} E[X\mid X > x] &= \int_{-\infty}^\infty t f_{X \mid X > x}(t)\,\mathrm dt\\ &= \int_{-\infty}^\infty t f_X(t-x)\,\mathrm dt\\ &= \int_{-\infty}^\infty (x+u) f_X(u)\,\mathrm du &\scriptstyle{\text{on substituting}~u = t-x}\\ &= x + E[X]. \end{align}$$ Note that we have not explicitly used the density of $X$ in the calculation, and don't even need to integrate explicitly if we simply remember that (i) the area under a pdf is $1$ and (ii) the definition of expected value of a continuous random variable in terms of its pdf.
For $x>0$, the event $\{X>x\}$ has probability $P\{X>x\}=1-F_X(x)=e^{-\lambda x} > 0$. Hence, $$ \newcommand{\E}{\mathbb{E}} \E[X\mid X> x] = \frac{\E[X\,I_{\{X>x\}}]}{P\{X>x\}} \, , $$ but $$ \E[X\,I_{\{X>x\}}] = \int_x^\infty t\,\lambda\,e^{-\lambda t}\,dt = (*) $$ (using Feynman's trick, vindicated by the Dominated Convergence Theorem, because it is fun) $$ (*) = -\lambda \int_x^\infty \frac{d}{d\lambda}(e^{-\lambda t}) \,dt = -\lambda \frac{d}{d\lambda} \int_x^\infty e^{-\lambda t} \,dt $$ $$ = -\lambda \frac{d}{d\lambda} \left(\frac{1}{\lambda} \int_x^\infty \lambda\,e^{-\lambda t} \,dt\right) = -\lambda\frac{d}{d\lambda}\left(\frac{1}{\lambda}\left(1 - F_X(x)\right)\right) $$ $$ = -\lambda\frac{d}{d\lambda}\left(\frac{e^{-\lambda x}}{\lambda}\right) = \left(\frac{1}{\lambda}+x\right)e^{-\lambda x} \, , $$ which gives the desired result $$ \E[X\mid X>x] = \frac{1}{\lambda}+x = \E[X] + x \, . $$
