= Return level
{title2=$z_T=F_{\rm annual}^{-1}(1-1/T)$}
A <return level> answers the inverse question to a <return period>: which threshold has the specified rarity? Let $F(z)=F_{\rm annual}(z)=\mathbb P(M_j\leq z)$ be the common <cumulative distribution function> of annual <sample maxima> in a model with <stationarity>. For a continuous, strictly increasing $F$ and $T>1$, the $T$-year <return level> is
$$
z_T=F^{-1}\!\left(1-\frac1T\right),\qquad
F(z_T)=1-\frac1T,\qquad \mathbb P(M_j>z_T)=\frac1T.
$$
It has the same units as $M_j$, whereas $T$ is a count of annual blocks. <Independence> between years is unnecessary for this one-year <quantile> definition; it is needed for the geometric waiting and multi-year formulas under <return period>. For a <discrete probability distribution>, use the generalized <quantile function> $F^{-1}(q)=\inf\{z:F(z)\geq q\}$. Then $\mathbb P(M_j>z_T)\leq1/T$, with equality not guaranteed at jumps; an exact requested exceedance <probability> may be unattainable.
For a concrete standard model, choose the maxima convention of the <Gumbel distribution>, with location $\mu$ and scale $b>0$:
$$
F(z)=\exp\!\left[-\exp\!\left(-\frac{z-\mu}{b}\right)\right],\qquad
z_T=\mu-b\log\!\left[-\log\!\left(1-\frac1T\right)\right].
$$
The second formula follows by taking logarithms twice and solving $F(z_T)=1-1/T$. The maxima <cumulative distribution function> and its inverse are given in https://www.itl.nist.gov/div898/handbook/eda/section3/eda366g.htm[NIST: extreme value type I distribution, maximum case]. With the deliberately chosen toy values $\mu=10$ and $b=2$ in arbitrary units, $z_{10}=14.501$, $z_{100}=19.200$, and $z_{1000}=23.815$, rounded to three decimals. In particular, $19.200$ is approximately the annual <quantile> at cumulative <probability> $0.99$; it is not a prediction of the largest observation in the next 100 years. Under <independence>, the <probability> that all those annual <sample maxima> stay below the exact $z_{100}$ is $0.99^{100}\approx0.366$.
The time block matters. If a toy year contains $m$ <independent> daily blocks, each with exceedance <probability> $q$ at one fixed threshold, its annual exceedance <probability> is $p=1-(1-q)^m$, giving annual <return period> $1/[1-(1-q)^m]$. The reciprocal $1/q$ counts daily blocks. Dependence among daily exceedances requires their joint law instead; a threshold crossed on several successive days still produces only one exceedance indicator for that year's <sample maximum>. The <peaks-over-threshold method> and <declustering of extremes> address a different choice of extreme-event sampling.
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