[proofplan] For the memoryless property, we compute $\mathbb{P}(X \ge x + z \mid X \ge x)$ directly using the survival function of the exponential, $\mathbb{P}(X \ge t) = e^{-\lambda t}$. The ratio simplifies to $e^{-\lambda z}$ by the exponential law of exponents. For uniqueness, we show that any continuous distribution satisfying the memoryless property must have survival function $G(x) = e^{-\lambda x}$ by converting the functional equation $G(x+z) = G(x)G(z)$ into a Cauchy equation. [/proofplan] [step:Verify the memoryless property using the exponential survival function] Let $X \sim \operatorname{Exp}(\lambda)$ with $\lambda > 0$. The survival function is $\mathbb{P}(X \ge t) = e^{-\lambda t}$ for $t \ge 0$. For $x, z \ge 0$, \begin{align*} \mathbb{P}(X \ge x + z \mid X \ge x) &= \frac{\mathbb{P}(X \ge x + z \text{ and } X \ge x)}{\mathbb{P}(X \ge x)}. \end{align*} Since $\{X \ge x + z\} \subset \{X \ge x\}$ (as $x + z \ge x$), the numerator is $\mathbb{P}(X \ge x + z)$. Therefore \begin{align*} \mathbb{P}(X \ge x + z \mid X \ge x) = \frac{\mathbb{P}(X \ge x + z)}{\mathbb{P}(X \ge x)} = \frac{e^{-\lambda(x+z)}}{e^{-\lambda x}} = e^{-\lambda z} = \mathbb{P}(X \ge z). \end{align*} [/step] [step:Show uniqueness by reducing the memoryless property to a Cauchy equation] Suppose $X$ is a continuous non-negative random variable (not identically zero) satisfying \begin{align*} \mathbb{P}(X \ge x + z \mid X \ge x) = \mathbb{P}(X \ge z) \quad \text{for all } x, z \ge 0. \end{align*} Define $G : [0, \infty) \to [0, 1]$ by $G(x) = \mathbb{P}(X \ge x)$. The memoryless property is equivalent to \begin{align*} G(x + z) = G(x)\, G(z) \quad \text{for all } x, z \ge 0. \end{align*} Note that $G(0) = \mathbb{P}(X \ge 0) = 1$, and $G$ is non-increasing with $G(x) \to 0$ as $x \to \infty$ (since $X$ is not identically zero and has a continuous distribution). In particular, $G(x) > 0$ for all $x \ge 0$: if $G(x_0) = 0$ for some $x_0 > 0$, then for all $x \ge x_0$ we would have $G(x) \le G(x_0) = 0$, and for any $x < x_0$ we could write $G(x_0) = G(x) G(x_0 - x) = 0$, forcing $G(x) = 0$ for all $x > 0$. But then $\mathbb{P}(X = 0) = 1$, contradicting continuity of the distribution. [/step] [step:Solve the functional equation to obtain $G(x) = e^{-\lambda x}$] Since $G(x) > 0$ for all $x \ge 0$, define $\varphi(x) = -\log G(x)$, so $\varphi : [0, \infty) \to [0, \infty)$ with $\varphi(0) = 0$. The multiplicative equation becomes additive: \begin{align*} \varphi(x + z) = \varphi(x) + \varphi(z) \quad \text{for all } x, z \ge 0. \end{align*} This is Cauchy's functional equation on $[0, \infty)$. Since $G$ is non-increasing (as a survival function), $\varphi$ is non-decreasing. A non-decreasing solution of Cauchy's equation on $[0, \infty)$ must be linear: $\varphi(x) = \lambda x$ for some constant $\lambda \ge 0$. Since $G(x) \to 0$ as $x \to \infty$, we need $\lambda > 0$. Therefore $G(x) = e^{-\varphi(x)} = e^{-\lambda x}$, which is the survival function of $\operatorname{Exp}(\lambda)$. Since the survival function uniquely determines a continuous distribution, $X \sim \operatorname{Exp}(\lambda)$. [/step]