5 Transforms of cumulative distribution functions

In this part, we introduce certain transforms and extensions of cumulative distribution functions, which are used in the classification calculation of the extreme value distributions.

5.1 Extended cumulative distribution function

Definition 5.1 Extended cumulative distribution function

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The extension \(\widetilde{F}\) of a c.d.f. \(F\) is the function

\begin{align*} \widetilde{F} \colon [-\infty ,+\infty ] \to [0,1] \end{align*}

given by

\begin{align*} \widetilde{F}(x) = \begin{cases} 0 & \; \text{ if } x = -\infty \\ F(x) & \; \text{ if } -\infty {\lt} x {\lt} +\infty \\ 1 & \; \text{ if } x = +\infty . \end{cases}\end{align*}

Lemma 5.2 Continuity points of extended c.d.f.

The extension \(\widetilde{F}\) of a c.d.f. \(F\) is continuous at \(x = -\infty \), \(x = + \infty \), and at any \(x \in (-\infty ,+\infty )\) where \(F\) is continuous.

Proof ▶

Since \(\lim _{x \to +\infty } F(x) = 1\) by properties of c.d.f.s and \(\widetilde{F}(+\infty ) = 1\) by definition of the extension, continuity at \(x = + \infty \) follows. Continuity at \(x=-\infty \) is similar.

Suppose \(F\) is continuous at \(x \in \mathbb {R}\). Then since \(\widetilde{F}\) coincides with \(F\) in a neighborhood of \(x\) (indeed on the open set \(\mathbb {R}\subsetneq [-\infty ,+\infty ]\)), the continuity of \(F\) at \(x \in \mathbb {R}\) implies continuity of \(\widetilde{F}\) at \(x\).

5.2 One over one minus cumulative distribution function

Definition 5.3 One over one minus cumulative distribution function

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The transform \(\frac{\mathbf{1}}{\mathbf{1}-\widetilde{F}}\) of a c.d.f. \(F\) is the function

\begin{align*} \frac{\mathbf{1}}{\mathbf{1}-\widetilde{F}} \colon [-\infty ,+\infty ] \to [1,+\infty ] \end{align*}

given by

\begin{align*} \frac{\mathbf{1}}{\mathbf{1}-\widetilde{F}}(x) = \begin{cases} 1 & \; \text{ if } \widetilde{F}(x) = 0 \\ \frac{1}{1-\widetilde{F}(x)} & \; \text{ if } 0 {\lt} \widetilde{F}(x) {\lt} 1 \\ +\infty & \; \text{ if } \widetilde{F}(x) = 1 , \end{cases}\end{align*}

where \(\widetilde{F} \colon [-\infty ,+\infty ] \to [0,1]\) is the extension of the c.d.f. \(F\).

Lemma 5.4 Continuity points of one over one minus c.d.f.

The transform \(\frac{\mathbf{1}}{\mathbf{1}-\widetilde{F}}\) of a c.d.f. \(F\) is continuous at \(x = -\infty \), \(x = + \infty \), and at any \(x \in (-\infty ,+\infty )\) where \(F\) is continuous.

Proof ▶

Since \(\lim _{x \to +\infty } \widetilde{F}(x) = \widetilde{F}(+\infty ) = 1\) by Lemma 5.2 and the continuous extension of \(p \mapsto \frac{1}{1-p}\) to a function \([0,1] \to [0,+\infty ]\) tends to \(+\infty \) as \(p \to 1\), we have

\begin{align*} \lim _{x \to +\infty } \frac{\mathbf{1}}{\mathbf{1}-\widetilde{F}}(x) \; = \; +\infty \; = \; \frac{\mathbf{1}}{\mathbf{1}-\widetilde{F}}(+\infty ) . \end{align*}

Therefore \(\frac{\mathbf{1}}{\mathbf{1}-\widetilde{F}}\) is continuous at \(+\infty \). Continuity at \(-\infty \) similarly follows from \(\lim _{x \to -\infty } \widetilde{F}(x) = \widetilde{F}(-\infty ) = 0\) and \(\frac{1}{1-p}\) tending to \(1\) as \(p \to 0\), which give

\begin{align*} \lim _{x \to -\infty } \frac{\mathbf{1}}{\mathbf{1}-\widetilde{F}}(x) \; = \; 1 \; = \; \frac{\mathbf{1}}{\mathbf{1}-\widetilde{F}}(-\infty ) . \end{align*}

Suppose \(F\) is continuous at \(x \in \mathbb {R}\), and recall from Lemma 5.2 that \(\widetilde{F}\) is then also continuous at \(x\). Now \(\frac{\mathbf{1}}{\mathbf{1}-\widetilde{F}}\) is a composition of the continuous function \(p \mapsto \frac{1}{1-p} \colon [0,1] \to [0,+\infty ]\) with \(\widetilde{F}\), and as such also becomes continuous at \(x\).

5.3 One over negative logarithm cumulative distribution function

Definition 5.5 One over negative logarithm cumulative distribution function

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The transform \(\frac{\mathbf{1}}{\widetilde{\log } \big( 1 / \widetilde{F} \big)}\) of a c.d.f. \(F\) is the function

\begin{align*} \frac{\mathbf{1}}{\widetilde{\log } \big( 1 / \widetilde{F} \big)} \colon [-\infty ,+\infty ] \to [0,+\infty ] \end{align*}

given by

\begin{align*} \frac{\mathbf{1}}{\widetilde{\log } \big( 1 / \widetilde{F} \big)}(x) = \begin{cases} 0 & \; \text{ if } \widetilde{F}(x) = 0 \\ \frac{1}{\log \big( 1/\widetilde{F}(x) \big)} & \; \text{ if } 0 {\lt} \widetilde{F}(x) {\lt} 1 \\ +\infty & \; \text{ if } \widetilde{F}(x) = 1 , \end{cases}\end{align*}

where \(\widetilde{F} \colon [-\infty ,+\infty ] \to [0,1]\) is the extension of the c.d.f. \(F\).

Lemma 5.6 Continuity points of one over negative logarithm c.d.f.

The transform \(\frac{\mathbf{1}}{\widetilde{\log } \big( 1 / \widetilde{F} \big)}\) of a c.d.f. \(F\) is continuous at \(x = -\infty \), \(x = + \infty \), and at any \(x \in (-\infty ,+\infty )\) where \(F\) is continuous.

Proof ▶

Since \(\lim _{x \to +\infty } \widetilde{F}(x) = \widetilde{F}(+\infty ) = 1\) by Lemma 5.2 and the continuous extension of \(p \mapsto \frac{1}{\log (1/p)}\) to a function \([0,1] \to [0,+\infty ]\) tends to \(+\infty \) as \(p \to 1\), we have

\begin{align*} \lim _{x \to +\infty } \frac{\mathbf{1}}{\widetilde{\log } \big( 1 / \widetilde{F} \big)}(x) \; = \; +\infty \; = \; \frac{\mathbf{1}}{\widetilde{\log } \big( 1 / \widetilde{F} \big)}(+\infty ) . \end{align*}

Therefore \(\frac{\mathbf{1}}{\widetilde{\log } \big( 1 / \widetilde{F} \big)}\) is continuous at \(+\infty \). Continuity at \(-\infty \) similarly follows from \(\lim _{x \to -\infty } \widetilde{F}(x) = \widetilde{F}(-\infty ) = 0\) and \(\frac{1}{\log (1/p)}\) tending to \(0\) as \(p \to 0\), which give

\begin{align*} \lim _{x \to -\infty } \frac{\mathbf{1}}{\widetilde{\log } \big( 1 / \widetilde{F} \big)}(x) \; = \; 0 \; = \; \frac{\mathbf{1}}{\widetilde{\log } \big( 1 / \widetilde{F} \big)}(-\infty ) . \end{align*}

Suppose \(F\) is continuous at \(x \in \mathbb {R}\), and recall from Lemma 5.2 that \(\widetilde{F}\) is then also continuous at \(x\). Now \(\frac{\mathbf{1}}{\widetilde{\log } \big( 1 / \widetilde{F} \big)}\) is a composition of the continuous function \(p \mapsto \frac{1}{\log (1/p)} \colon [0,1] \to [0,+\infty ]\) with \(\widetilde{F}\), and as such also becomes continuous at \(x\).