= Pareto tail conditional scaling
{c}
{title2=$X\mid X\geq u\sim\operatorname{Pareto}(\alpha,u),\quad u\geq\beta$}
Conditioning a <Pareto distribution> on exceeding a threshold changes its minimum while preserving its <shape parameter>. For $u\geq\beta$ and $x\geq u$, the definition of <conditional probability> gives
$$
\mathbb P(X>x\mid X\geq u)=\frac{(\beta/x)^\alpha}{(\beta/u)^\alpha}=\left(\frac{u}{x}\right)^\alpha.
$$
Equivalently, for $r\geq1$,
$$
\mathbb P(X/u>r\mid X\geq u)=r^{-\alpha}.
$$
The ratio to the threshold has the same <Pareto distribution> at every admissible threshold.<Random: Probability, Mathematical Statistics, Stochastic Processes by Kyle Siegrist/The Pareto Distribution>[Random: Probability, Mathematical Statistics, Stochastic Processes by Kyle Siegrist; The General Pareto Distribution, Related Distributions: closure under right-tail conditioning]{ref} For the chosen $\alpha=2$, $\beta=10$, only $1/4$ of outcomes exceed 20, but among those outcomes the <probability> of exceeding 40 is again $1/4$. When $\alpha>1$, the same identity gives <conditional expectation> $\mathbb E[X\mid X\geq u]=\alpha u/(\alpha-1)$.
This multiplicative scaling differs from the additive <memoryless property> of an <exponential distribution>. Indeed $Y=\log(X/\beta)$ has <exponential distribution> of rate $\alpha$, because $\mathbb P(Y>y)=e^{-\alpha y}$ for $y\geq0$. Thus logarithms turn multiplication into addition and explain the scaling property. Conversely, if $U$ has <uniform distribution> on $(0,1)$, <inverse transform sampling> gives $X=\beta(1-U)^{-1/\alpha}$.
The excess $Z=X-u$ conditional on $X\geq u$ has <survival function> $(1+z/u)^{-\alpha}$ for $z\geq0$. In the <generalized Pareto distribution> convention $\mathbb P(Z>z)=(1+\xi z/\sigma)^{-1/\xi}$, this means $\xi=1/\alpha$ and $\sigma=u/\alpha$. Here the excess starts at zero; the original <Pareto distribution> starts at $\beta>0$.
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