3 Classification of extreme value distributions

3.1 Auxiliary classification I

Lemma 3.1 Second order differential equation for Q

Suppose that

\begin{align*} Q \colon \mathbb {R}\to \mathbb {R}\end{align*}

is differentiable and satisfies

\begin{align*} Q(0) = 0 \qquad \text{ and } \qquad Q’(0) = 1 \end{align*}

and

\begin{align*} Q(h+s) = Q(h) \alpha (s) + Q(s) \end{align*}

for some $\alpha \colon \mathbb {R}\to \mathbb {R}$ and every $s,h \in \mathbb {R}$. Then $Q$ is twice continuously differentiable and satisfies

\begin{align} Q”(s) = Q’(s) \, Q”(0) \qquad \text{ for every } s \in \mathbb {R}. \end{align}

Proof ▶

Note first that the equation implies (rearranging and dividing by $h$), for any $s$ and $h \ne 0$,

\begin{align*} \frac{Q(h+s) - Q(s)}{h} = \frac{Q(h)}{h} \alpha (s) . \end{align*}

Taking the limit as $h \to 0$ and using $Q(0)=0$ yields $Q'(s) = Q'(0) \, \alpha (s) = \alpha (s)$, where we also took into account $Q'(0) = 1$. Therefore necessarily $\alpha = Q'$, and the equation can be rewritten in the form

\begin{align*} Q(h+s) = Q(h) \, Q’(s) + Q(s) \qquad \text{ for } s,h \in \mathbb {R}. \end{align*}

Rearranging the equation, we find

\begin{align*} Q(h+s) - Q(s) = Q(h) \, Q’(s) \end{align*}

and interchanging the role of $s$ and $h$ also

\begin{align*} Q(h+s) - Q(h) = Q(s) \, Q’(h) . \end{align*}

Subtracting the last two equations yields

\begin{align*} Q(s) - Q(h) = Q(s) \, Q’(h) - Q(h) \, Q’(s) , \end{align*}

which by rearranging and dividing by $h \ne 0$ yields

\begin{align*} \frac{Q(h)}{h} \big( Q’(s) - 1 \big) = Q(s) \frac{Q'(h) - 1}{h} . \end{align*}

Taking the limit $h \to 0$, recalling $Q(0)=1$ and $Q'(0)=1$, gives the existence of the second derivative $Q''(0)$ and the equation

\begin{align*} Q’(s) - 1 = Q(s) Q”(0) . \end{align*}

Solving $Q'(s) = 1 + Q''(0) \, Q(s)$ and recalling that $Q$ is differentiable shows that $Q'$ is also differentiable, so $Q$ is indeed twice differentiable. Differentiating, we get the asserted equation

\begin{align*} Q”(s) = Q’(s) \, Q”(0) . \end{align*}

Since $Q'$ is differentiable and in particular continuous, this also shows that $Q''$ is continuous, i.e., that $Q$ is twice continuously differentiable.

Lemma 3.2 Solution for Q

Suppose that $Q \colon \mathbb {R}\to \mathbb {R}$ is twice continuously differentiable and $Q'$ is positive and $Q$ and satisfies $Q(0)=0$, $Q'(0) = 1$, and the equation concluded in Lemma 3.1 with $\gamma = Q''(0)$, i.e.,

\begin{align} \label{eq: Q eqn dd} Q”(s) = \gamma \, Q’(s) \qquad \text{ for every } s \in \mathbb {R}. \end{align}

Then $Q$ is given by

\begin{align*} Q(s) = \begin{cases} \frac{e^{\gamma s} - 1}{\gamma } & \; \text{ if $\gamma \ne 0$} \\ \; \; s & \; \text{ if $\gamma = 0$} \end{cases} \qquad \text{for $s \in \mathbb {R}$.} \end{align*}

Proof ▶

Since $Q$ is differentiable and $Q'(s){\gt}0$ for any $s \in \mathbb {R}$, we can write 2 as

\begin{align*} \frac{\mathrm{d}}{\mathrm{d}s} \log Q’(s) = \frac{Q''(s)}{Q'(s)} = \gamma . \end{align*}

Integrating and noting $\log Q'(0) = \log 1 = 0$, this yields $\log Q'(s) = \gamma s$ for all $s \in \mathbb {R}$, or equivalently $Q'(s) = e^{\gamma s}$. Integrating a second time and noting $Q(0) = 0$, this yields the desired formulas in the two cases $\gamma \ne 0$ and $\gamma = 0$, respectively, as follows.

If $\gamma \ne 0$, then we get

\begin{align*} Q(s) = \int _0^s e^{\gamma u} \, \mathrm{d}u = \frac{e^{\gamma s} - 1}{\gamma } . \end{align*}

If $\gamma = 0$, then we get

\begin{align*} Q(s) = \int _0^s e^0 \, \mathrm{d}u = s . \end{align*}

Theorem 3.3 Monotone functions are a.e. differentiable

✓

If $f : \mathbb {R}\to \mathbb {R}$ is nondecreasing, then the derivative $f'(x)$ exists at almost every $x \in \mathbb {R}$. In particular there exists points $x$ where $f'(x)$ exists.

Proof ▶

(The proof is already in Mathlib.)

Lemma 3.4 Solution for E

Suppose that $E \colon (0,\infty ) \to \mathbb {R}$ is nondecreasing and nonconstant function which satisfies $E(1) = 0$ and

\begin{align*} E(\lambda \sigma ) = E(\lambda ) A(\sigma ) + E(\sigma ) \end{align*}

for some $A \colon (0,\infty ) \to (0,\infty )$ and all $\lambda , \sigma {\gt} 0$. Then, denoting $c = E'(1)$ and $\gamma = \frac{1}{E'(1)} \frac{\mathrm{d}^2}{\mathrm{d}{s}^2} E(e^s) \big|_{s=0}$, for all $\lambda \in \mathbb {R}$ we have

\begin{align*} E(\lambda ) = \begin{cases} c \big( \lambda ^\gamma - 1 \big) & \; \; \text{ if $\gamma \ne 0$.} \\ c \, \log (\lambda ) & \; \; \text{ if $\gamma = 0$.} \end{cases}\end{align*}

Proof ▶

Denote $H(s) = E(e^s)$ for $s \in \mathbb {R}$. Then $H(0) = E(1) = 0$ and $H$ is also nondecreasing and nonconstant. By Theorem 3.3, there exists some $s_0 \in \mathbb {R}$ such that the derivative $H'(s_0)$ exists.

The equation for $E$ yields an equation for $H$,

\begin{align*} H(h+s) = H(h) A(e^{s}) + H(s) \qquad \text{ for any } s, h \in \mathbb {R}. \end{align*}

We can rearrange this equation and divide by $h$, and we get for any $s, h \in \mathbb {R}$, $h \ne 0$,

\begin{align*} \frac{H(h+s) - H(s)}{h} = \frac{H(h)}{h} A(e^{s}) . \end{align*}

If $s = s_0$, then the LHS tends to $H'(s_0)$ as $h \to 0$. The RHS must therefore also have a limit as $s \to 0$, and observing that $\frac{H(h)}{h} = \frac{H(h) - H(0)}{h}$, that limit is $H'(0) A(e^{s_0})$, which shows that the derivative $H'(0)$ exists. Then applying the same equation at general $s \in \mathbb {R}$ shows that $H'(s) = H'(0) A(e^{s})$ must exist, so $H \colon \mathbb {R}\to \mathbb {R}$ is in fact everywhere differentiable.

Note that we must have $H'(0) {\gt} 0$, because if $H'(0) = 0$ then by the above equation $H'(s) = 0$ for every $s$ and then $H$ is a constant function, which is a contradiction. Denote $c = H'(0) = \frac{\mathrm{d}}{\mathrm{d}s} E(e^s) \big|_{s=0} = E'(1)$.

Also the equation and positivity of the function $A$ give $H'(s){\gt}0$ for all $s$.

Now consider $Q(s) = \frac{H(s)}{c}$. This $H$ is obviously also differentiable, since $H$ is. The equation for $H$ yields the following equation for $Q$,

\begin{align*} Q(h+s) = Q(h) \alpha (s) + Q(s) \qquad \text{ for any } s,h \in \mathbb {R}, \end{align*}

where $\alpha (s) = A(e^{s})$. By Lemma 3.1 $Q$ is then twice continuously differentiable and satisfies

\begin{align} Q’(s) Q” (0) = Q” (s) \qquad \text{ for all } s \in \mathbb {R}. \end{align}

By Lemma 3.2 we then get a formula for $Q$, which involves

\begin{align*} \gamma = Q”(0) = \frac{H''(0)}{c} = \frac{1}{c} \frac{\mathrm{d}^2}{\mathrm{d}{s}^2} E(e^s) \big|_{s = 0} . \end{align*}

If $\gamma \ne 0$ then the formula reads $Q(s) = \frac{e^{\gamma s} - 1}{\gamma }$. Now just tracing the definitions, we get the asserted formula

\begin{align*} E(\lambda ) = H(\log \lambda ) = c \, Q(\log \lambda ) = c \, \big( e^{\gamma \, \log (\lambda )} - 1 \big) = c \, \big( \lambda ^{\gamma } - 1 \big) . \end{align*}

If $\gamma = 0$ then the formula reads $Q(s) = s$. Now just tracing the definitions, we get instead

\begin{align*} E(\lambda ) = H(\log \lambda ) = c \, Q(\log \lambda ) = c \, \log (\lambda ) , \end{align*}

and the proof is complete.

3.2 Inverting to recover a cumulative distribution function