A Generic Method for Density Forecasts Recalibration

Abstract

We address the calibration constraint of probability forecasting. We propose a generic method for recalibration, which allows us to enforce this constraint. It remains to be known the impact on forecast quality, measured by predictive distributions sharpness, or specific scores. We show that the impact on the Continuous Ranked Probability Score (CRPS) is weak under some hypotheses and that it is positive under more restrictive ones. We used this method on temperature ensemble forecasts and compared the quality of the recalibrated forecasts with that of the raw ensemble and of a more specific method, that is Ensemble Model Output Statistics (EMOS). Better results are shown with our recalibration rather than with EMOS in this case study.

8.5 Appendix

Here are gathered all the proofs concerning the results presented in the chapter. The first section is concerned by proofs of results in an infinite sample and the second by result in a finite sample.

Lemma 8.1

$$\begin{aligned} \textsc {E}_Y[\,L_{\tau }-L_{\tau _{c}}\,] = \sum _{e} p_e \int _{G_{e}^{-1}(\tau _{c})}^{G_{e}^{-1}(\tau )} (F_{e}(y)-\tau ) \text {d}y\,, \end{aligned}$$

with $\tau$,$\tau _c \in [0,1]$ and $p_e$ the frequency of appearance of the state e. Under the Assumption A.2.1, we prove Lemma 8.1.

Proof

We have:

$$\begin{aligned} \textsc {E}_Y[\,L_{\tau }-L_{\tau _{c}}\,] = \sum _e p_e \left( \textsc {E}_{Y}[\,L_{\tau }({{\varvec{Y}}},G_{e}^{-1}(\tau ))\,] - \textsc {E}_{Y}[\,L_{\tau }({{\varvec{Y}}},G_{e}^{-1}(\tau _c)\,] \right) . \end{aligned}$$

(8.8)

First, we only focus on a particular e. Thus, we are interested in:

$$\begin{aligned} \textsc {E}_{Y}[\,L_{\tau }({{\varvec{Y}}},G_e^{-1}(\tau ))\,] - \textsc {E}_{Y}[\,L_{\tau }({{\varvec{Y}}},G_e^{-1}(\tau _c)\,] \equiv \textsc {E}_{Y,e}[\,L_{\tau ,\tau _c}\,]. \end{aligned}$$

For ease of notation and comprehension, we suppress e in the notation since there is no confusion. Moreover, we suppose, for ease of notation again (and since we obtain the same result if we inverse the inequality) that $G^{-1}(\tau ) \le \,G^{-1}(\tau _c)$. So, we have:

$$\begin{aligned} \textsc {E}_{Y}[\,L_{\tau ,\tau _c}\,]&= \int _{-\infty }^{+\infty }\,\left( \,[y-G^{-1}(\tau )]\,\tau +[G^{-1}(\tau )-y]\,\mathbf {1}_{\{y\le G^{-1}(\tau )\}}\,\right) \,f_{Y}(y)\,\text {d}y\\&{\quad }-\int _{-\infty }^{+\infty }\,\left( \,[y-G^{-1}(\tau _c)]\,\tau + [G^{-1}(\tau _c)-y]\,\mathbf {1}_{\{y\le G^{-1}(\tau _c)\}}\,\right) \,f_{Y}(y)\,\text {d}y \\&=[G^{-1}(\tau _c)-G^{-1}(\tau )]\,\tau + [G^{-1}(\tau )-G^{-1}(\tau _c)]\,F\circ \,G^{-1}(\tau )\\&{\quad }- G^{-1}(\tau _c)\,[\,F\circ \,G^{-1}(\tau _c)-F\circ \,G^{-1}(\tau )\,] + \int _{y=G^{-1}(\tau )}^{G^{-1}(\tau _c)}\,\underbrace{y}_v\,\underbrace{f_{Y}(y)}_{u'}\, \text {d}y\,. \end{aligned}$$

Using integral by parts, we have:

$$\begin{aligned} \textsc {E}_{Y}[\,L_{\tau ,\tau _c}\,]&= [G^{-1}(\tau _c)-G^{-1}(\tau )]\,\tau + \int _{y=G^{-1}(\tau _c)}^{G^{-1}(\tau )}\,F(y)\,\text {d}y\\&= \int _{y=G^{-1}(\tau _c)}^{G^{-1}(\tau )}\,[\,F(y)-\tau \,]\,\text {d}y\,. \end{aligned}$$

Replacing it in (8.8) finishes the demonstration.   $\square$

1.1 8.5.1 Impact on Score: Conditions for Improvement

In this section, the reader can find the proofs of results mentioned in Sect. 8.3.1 of the chapter. We first demonstrate how to approximate the difference of $L_\tau$ expectation before showing that under some hypotheses, our correction improves systematically the quality of the forecasts.

1.1.1 8.5.1.1 Rewriting the Difference of $L_\tau$ Expectation

Under the Assumptions A.2.1, A.3.1.1–A.3.1.4 and using functional derivatives and the implicit function theorem, we prove (8.2).

Proof

Remember: Let H be a functional, h a function, $\alpha$ a scalar and $\delta$ an arbitrary function.

We can write the expression of the functional evaluated at $f+\delta \alpha$ as follow:

$$\begin{aligned} H[h+\delta \,\alpha ]= H[h] + \frac{\text {d}H[h+\delta \,\alpha ]}{\text {d}\alpha }|_{\alpha =0}\, \alpha + \frac{1}{2} \frac{\text {d}^{2} H[h+\delta \,\alpha ]}{\text {d}\alpha ^{2}}|_{\alpha =0}\,\alpha ^{2} + \cdots + \text {Rem}(\alpha )\,, \end{aligned}$$

with Rem($\alpha$) the remainder. Denote:

$$\begin{aligned} \varDelta PL [h]= & {} \sum _{e} p_e \int _{h_{e}^{-1}(\tau _{c})}^{h_{e}^{-1}(\tau )} (F_{e}(y)-\tau ) \text {d}y \nonumber \\ \nonumber \\= & {} \sum _{e} p_{e} \varDelta PL_{e} [h_{e}]\,. \end{aligned}$$

For ease of notation, denote $\varDelta PL_{e} [F_{e} + \delta _{e} \alpha ] \equiv \varDelta PL_{F,\delta ,e}$. Choosing $H= \varDelta PL_{e}$, $h=F_{e}$ and $\eta _e = \alpha \delta _e$ (even if we use $\alpha \delta _e$ in the development in order to use functional derivatives, directional derivatives and the implicit function theorem), we have:

$$\begin{aligned} \varDelta PL_{F,\delta ,e}&\sim \varDelta PL_{e} [F_{e}] + \frac{\text {d}\varDelta PL_{F,\delta ,e}}{\text {d}\alpha }|_{\alpha =0}\,\alpha + \frac{1}{2} \frac{\text {d}^{2} \varDelta PL_{F,\delta ,e}}{\text {d}\alpha ^{2}} \vert _{\alpha =0}\,\alpha ^{2} + \text {Rem}_{e}(\alpha )\\&= \left[ \frac{\partial \varDelta PL_{F,\delta ,e}}{\partial \alpha } |_{\alpha =0\,,\,\tau _{c}=\tau } + \frac{\partial \varDelta PL_{F,\delta ,e}}{\partial \tau _{c}} |_{\alpha =0\,,\,\tau _{c}=\tau } \frac{\text {d}\tau _{c}}{\text {d}\alpha } \right] \alpha \\&{\quad }+ \left[ \frac{\partial ^{2} \varDelta PL_{F,\delta ,e}}{\partial \alpha ^{2}} |_{\alpha =0\,,\,\tau _{c}=\tau } + 2 \frac{\partial ^{2} \varDelta PL_{F,\delta ,e}}{\partial \alpha \partial \tau _{c}} |_{\alpha =0\,,\,\tau _{c}=\tau } \frac{\text {d}\tau _{c}}{\text {d}\alpha } \right] \frac{\alpha ^{2}}{2}\\&{\quad }+ \left[ \frac{\partial ^{2} \varDelta PL_{F,\delta ,e}}{\partial \tau _{c}^{2}}|_{\alpha =0\,,\,\tau _{c}=\tau } \left( \frac{\text {d}\tau _{c}}{\text {d}\alpha }\right) ^{2} + \frac{\partial \varDelta PL_{F,\delta ,e}}{\partial \tau _{c}} |_{\alpha =0\,,\,\tau _{c}=\tau } \frac{\text {d}^{2} \tau _{c}}{\text {d}\alpha ^{2}} \right] \frac{\alpha ^{2}}{2}\\&{\quad }+ \text {Rem}_{e}(\alpha )\,. \end{aligned}$$

To calculate $\frac{\text {d}\tau _{c}}{\text {d}\alpha }$, we will use the equation which link $\tau _{c}$ and $\alpha$:

$$\begin{aligned} \sum _{e} p_{e} F_{e} \circ (F_{e} + \delta _{e} \alpha )^{-1} (\tau _{c}) = \tau . \end{aligned}$$

Using the implicit function theorem, we find:

$$\begin{aligned} \frac{\text {d}\tau _{c}}{\text {d}\alpha }= \sum _e p_e \delta _{e} \circ F_{e}^{-1}(\tau ) \end{aligned}$$

Now, we need to calculate partial derivatives:

$$\begin{aligned} \frac{\partial \varDelta PL_{F,\delta ,e}}{\partial \alpha } |_{\alpha =0\,,\,\tau _{c}=\tau }= & {} \frac{\partial \left( \int _{(F_{e}+\delta _{e} \alpha )^{-1}(\tau _{c})}^{(F_{e}+\delta _{e} \alpha )^{-1}(\tau )} (F_{e}(y)-\tau ) \text {d}y\right) }{\partial \alpha } |_{\alpha =0\,,\,\tau _{c}=\tau } =0\,;\\ \\ \frac{\partial \varDelta PL_{F,\delta ,e}}{\partial \tau _{c}} |_{\alpha =0\,,\,\tau _{c}=\tau }= & {} 0\,\,\,;\,\,\, \frac{\partial ^{2} \varDelta PL_{F,\delta ,e}}{\partial \tau _{c}^{2}} |_{\alpha =0\,,\,\tau _{c}=\tau } = -\frac{1}{f_{e} \circ F_{e}^{-1}(\tau )};\\ \\ \frac{\partial ^{2} \varDelta PL_{F,\delta ,e}}{\partial \alpha ^{2}} |_{\alpha =0\,,\,\tau _{c}=\tau }= & {} 0\,\,\,;\,\,\,\frac{\partial ^{2} \varDelta PL_{F,\delta ,e}}{\partial \alpha \partial \tau _{c}} |_{\alpha =0\,,\,\tau _{c}=\tau } = \frac{\delta _{e} \circ F_{e}^{-1}(\tau )}{f_{e} \circ F_{e}^{-1}(\tau )}\,. \end{aligned}$$

Thus, we have:

$$\begin{aligned} \varDelta PL_{e} [F_{e} + \delta _{e} \alpha ]&\sim \left[ \left( \frac{\delta _{e} \circ F_{e}^{-1}(\tau )}{f_{e} \circ F_{e}^{-1}(\tau )} \right) \sum _e p_e \delta _{e} \circ F_{e}^{-1}(\tau )\right] \alpha ^2\\&{\quad }- \left[ \frac{(\sum _e p_e \delta _{e} \circ F_{e}^{-1}(\tau ))^{2}}{2 f_{e} \circ F_{e}^{-1}(\tau )}\right] \alpha ^{2} + \text {Rem}_{e}(\alpha )\,, \end{aligned}$$

and hence:

$$\begin{aligned} \varDelta PL [F + \delta \alpha ]&\sim \left( \sum _e\frac{p_e \delta _e (F_e^{-1}(\tau ))}{f_e (F_e^{-1}(\tau ))}\right) \left( \sum _e p_e \delta _e (F_e^{-1}(\tau ))\right) \times \alpha ^{2}\\&{\quad }- \left( \sum _e \frac{p_e}{2f_e (F_e^{-1}(\tau ))}\right) \left( \sum _e p_e \delta _e (F_e^{-1}(\tau ))\right) ^{2} \times \alpha ^{2}\\&{\quad }+ \sum _e p_e \text {Rem}_{e}(\alpha ). \end{aligned}$$

Now, let’s focus on the remainders. Following the Taylor–Lagrange inequality, if M such that $\left| \frac{\text {d}^{3} \varDelta PL_{F,\delta ,e}}{\text {d}\alpha ^{3}} \right| \le M$ exists, we have $\vert \,\text {Rem}_e(\alpha )\,\vert \le \frac{M |\alpha ^{3}|}{3!}$. Let’ s find conditions for the existence of M. The third derivative is:

$$\begin{aligned} \frac{\text {d}^{3} \varDelta PL_{F,\delta ,e}}{\text {d}\alpha ^{3}}&= \frac{\partial \varDelta PL_{F,\delta ,e}}{\partial \tau _{c}} \frac{\text {d}^{3} \tau _c}{\text {d}\alpha ^{3}}+ 3 \frac{\partial ^{2} \varDelta PL_{F,\delta ,e}}{\partial \tau _{c}\partial \alpha } \frac{\text {d}^{2} \tau _c}{\text {d}\alpha ^{2}} +3 \frac{\partial ^{2} \varDelta PL_{F,\delta ,e}}{\partial \tau _{c}^{2}} \frac{\text {d}^{2} \tau _c}{\text {d}\alpha ^{2}} \frac{\text {d}\tau _{c}}{\text {d}\alpha }\\&{\quad }+ \frac{\partial ^{3} \varDelta PL_{F,\delta ,e}}{\partial \alpha ^{3}} + 3 \frac{\partial ^{3} \varDelta PL_{F,\delta ,e}}{\partial \tau _{c}\partial \alpha ^{2}} \frac{\text {d}\tau _{c}}{\text {d}\alpha }+ 3 \frac{\partial ^{3} \varDelta PL_{F,\delta ,e}}{\partial \tau _{c}^{2}\partial \alpha } \left( \frac{\text {d}\tau _{c}}{\text {d}\alpha }\right) ^{2}\\&{\quad }+ \frac{\partial ^{3} \varDelta PL_{F,\delta ,e}}{\partial \tau _{c}^{3}} \left( \frac{\text {d}\tau _{c}}{\text {d}\alpha }\right) ^{3}\,. \end{aligned}$$

Let’s calculate the partial derivatives of order 3:

$$\begin{aligned} \frac{\partial ^{3} \varDelta PL_{F,\delta ,e}}{\partial \alpha ^{3}} |_{\alpha =0\,,\,\tau _{c}=\tau }&= 0 \,\,\,;\,\,\, \frac{\partial ^{3} \varDelta PL_{F,\delta ,e}}{\partial \tau _{c}^{3}} |_{\alpha =0\,,\,\tau _{c}=\tau } = 2 \frac{f_{e}^{'} \circ F_{e}^{-1}(\tau )}{f_{e} \circ F_{e}^{-1}(\tau )}\,;\\ \frac{\partial ^{3} \varDelta PL_{F,\delta ,e}}{\partial \tau _{c}^{2}\partial \alpha } |_{\alpha =0\,,\,\tau _{c}=\tau }&= -2 \frac{f_{e}^{'} \circ F_{e}^{-1}(\tau )}{f_{e} \circ F_{e}^{-1}(\tau )^{3}} (\delta _{e} \circ F_{e}^{-1}(\tau )) -2 \frac{\delta _{e}^{'} \circ F_{e}^{-1}(\tau )}{f_{e} \circ F_{e}^{-1}(\tau )^{2}}\,; \\ \frac{\partial ^{3} \varDelta PL_{F,\delta ,e}}{\partial \tau _{c}\partial \alpha ^{2}} |_{\alpha =0\,,\,\tau _{c}=\tau }&= \frac{f_{e}^{'} \circ F_{e}^{-1}(\tau )}{f_{e} \circ F_{e}^{-1}(\tau )^{3}} \left( \delta _{e} \circ F_{e}^{-1}(\tau )\right) ^{2}\\&{\quad }- 2 \frac{\delta _{e}^{'} \circ F_{e}^{-1}(\tau )}{f_{e} \circ F_{e}^{-1}(\tau )^{2}} \left( \delta _{e} \circ F_{e}^{-1}(\tau )\right) . \end{aligned}$$

Moreover, we have:

$$\begin{aligned} \frac{d^{2} \tau _c}{d \alpha ^{2}} = \sum _e p_e \left( \frac{2 \delta _{e}^{'} \circ F_{e}^{-1}(\tau ) - f_{e}^{'} \circ F_{e}^{-1}(\tau )}{f_{e} \circ F_{e}^{-1}(\tau )} \right) \delta _{e} \circ F_{e}^{-1}(\tau ). \end{aligned}$$

Since $\eta _e$, its first, second and third derivatives are finite in $F_e^{-1}(\tau )$, it is also the case for $\delta _e$ and the partial derivatives are finite. Furthermore, $f_e$, $\delta _e$ and their derivatives are bounded (since $\eta _e$ and their derivatives are bounded), which implies that the second derivatives of $\varDelta PL_{e} [F_{e} + \delta _{e} \alpha ]$ are also bounded. Thus, under these conditions, M exists. Then, we can write $\frac{\text {d}^{3} \varDelta PL_{F,\delta ,e}}{\text {d}\alpha ^{3}} = M_1 \delta _e^3$ and hence $\left| \text {Rem}_e(\alpha ) \right| \le \frac{|M_1| |\alpha \delta _e|^3}{3!}$ which implies that $\lim \,\,\frac{\text {Rem}_e(\alpha )}{(\alpha \delta _e)^2} = 0$, $\alpha \delta _e \rightarrow 0$, which shows that $\text {Rem}_e(\alpha )$ is negligible compared to $\frac{\text {d}^{2} \varDelta PL_{F,\delta ,e}}{\text {d}\alpha ^{2}}$.

Moreover, since $\forall \,\,e \in E$ the functions $F_e$ are $C^{3}$ and the functions $f_e$ and their derivatives are bounded by a constant which doesn’t depend on e, $\forall \,\, e \in E$, the development is valid for all directions and thus, since $\eta _e = G_e - F_e$, we have:

$$\begin{aligned} \textsc {E}_Y[\,L_{\tau }-L_{\tau _{c}}\,]&\sim \left( \sum _e\frac{p_e \eta _e (F_e^{-1}(\tau ))}{f_e (F_e^{-1}(\tau ))}\right) \left( \sum _e p_e \eta _e (F_e^{-1}(\tau ))\right) \nonumber \\&{\quad }- \left( \sum _e \frac{p_e}{2f_e (F_e^{-1}(\tau ))}\right) \left( \sum _e p_e \eta _e (F_e^{-1}(\tau ))\right) ^{2}\\&{\quad }\text {as}\,\, max\,\, \eta _e \rightarrow 0. \end{aligned}$$

To finish the demonstration, remark that Lemma 8.1 proves that:

$$\begin{aligned} \varDelta PL [G] = \textsc {E}_Y[\,L_{\tau }-L_{\tau _{c}}\,]. \end{aligned}$$

   $\square$

1.1.2 8.5.1.2 Systematic Improvement of the Quality

Under the Assumption A.3.1.5 or A.3.1.6, if $\exists \,\,\nu \ge 0$ (sufficiently small)$\, \, \forall \,\, e \in E$ $\,\, \forall \,\, y \in \mathbf {R};|\eta _e(y)| \le \nu$, we show (8.3) and (8.4):

Proof

Prove (8.3) is equivalent to show that $\varDelta PL [G]$ is positive, and if we rewrite:

$$\begin{aligned} \varDelta PL [G] \sim (2\textsc {E}[f^{-1}\eta ]- \textsc {E}[f^{-1}]\textsc {E}[\eta ])\textsc {E}[\eta ]\,, \end{aligned}$$

it is clear that the Assumption A.3.1.6 ensures the positivity of $\varDelta PL [G]$.

However, we need more argumentation to understand the complete utility of the Assumption A.3.1.5. Let’s look at one of the two worst cases: only two states of the world, the correlation coefficient $\rho =-1$, $\eta > 0$ (the other case is when $\rho =1$ and $\eta < 0$) and at each bound of the support of $\delta$ and $f^{-1}$, there is half of the probability mass. We also consider that the ratios between max and min of the supports are equal. If we define $max_e=M$ and $min_e= \frac{M}{r}$, one has the following equation:

$$\begin{aligned} \frac{1}{2}= \frac{2(r^{2}+1)}{(r+1)^{2}}-1. \end{aligned}$$

Solving this equation in r produces the expected result concerning the ratio between max and min values of $\eta$ and $f^{-1}$.

Now, let’s prove (8.4). According to (8.1), we have:

$$\begin{aligned} \textsc {E}_Y[\,CRPS_{G, C \circ G}\,]= & {} 2 \int _{-\infty }^{+\infty } \left( \int _0^1 L_\tau (y, G^{-1}(\tau ))- L_\tau (y, G^{-1} \circ C^{-1} (\tau )) \text {d}\tau \right) f_Y (y) \text {d}y. \end{aligned}$$

We can rewrite:

$$\begin{aligned} \textsc {E}_Y[\,CRPS_{G, C \circ G}\,]&= 2 \int _{-\infty }^{+\infty } \int _0^1 L_\tau (y, G^{-1}(\tau )) \, f_Y (y)\, \text {d}\tau \, \text {d}y \\&{\quad }- 2 \int _{-\infty }^{+\infty } \int _0^1 L_\tau (y, G^{-1} \circ C^{-1} (\tau )) \,f_Y (y) \, \text {d}\, \tau \text {d}y \, , \end{aligned}$$

and using the Fubini–Tonelli theorem, one obtains:

$$\begin{aligned} \textsc {E}_Y[\,CRPS_{G, C \circ G}\,]= & {} 2 \int _0^1 \textsc {E}_Y[\,L_{\tau }-L_{\tau _{c}}\,] \text {d}\tau \nonumber \\\ge & {} 0. \end{aligned}$$

(8.9)

   $\square$

1.2 8.5.2 Impact on Score: Bounds on Degradation

Under the Assumptions A.2.1, A.3.2.1–A.3.2.6 we prove (8.6) and (8.7).

Proof

adding and substracting $\textsc {E}_Y[\,L_{\tau }(Y,G^{-1}(\tau _c))\,]$ to $\textsc {E}_Y[L_{\,\widehat{\tau }_{c}}-L_{\tau }\,]$, we obtain:

$$\begin{aligned} \textsc {E}_Y[L_{\,\widehat{\tau }_{c}}-L_{\tau }\,]&= \textsc {E}_Y[\,L_{\tau }(Y,G^{-1}(Q_{\tau }))\,] - \textsc {E}_Y[\,L_{\tau }(Y,G^{-1}(\tau _c))\,]\\&{\quad } +\,\textsc {E}_Y[\,L_{\tau }(Y,G^{-1}(\tau _c))\,] - \textsc {E}_Y[\,L_{\tau }(Y,G^{-1}(\tau ))\,]\,, \end{aligned}$$

and finally:

$$\begin{aligned} \textsc {E}_Y[L_{\,\widehat{\tau }_{c}}-L_{\tau }\,] = \textsc {E}_{Y,e}[\,L_{\tau }(Y,G_{e}^{-1}(Q_{\tau }))\,] - \textsc {E}_{Y,e}[\,L_{\tau }(Y,G_{e}^{-1}(\tau _c))\,] - \textsc {E}_Y[\,L_{\tau }-L_{\tau _{c}}\,]. \end{aligned}$$

To begin with, we treat the third term on the right side. We have:

$$\begin{aligned} \textsc {E}_{Y,e}[\,L_{\tau ,\tau _{c}}\,] = \int _{y=G_{e}^{-1}(\tau _c)}^{G_{e}^{-1}(\tau )}\,[\,F_{e}(y)-\tau \,]\,\text {d}y \,. \end{aligned}$$

Using the change of variable $y=G_{e}^{-1}(z)$ and taking the absolute value, we find:

$$\begin{aligned} \left| \textsc {E}_{Y,e}[\,L_{\tau ,\tau _{c}}\,] \right| = \left| \int _{z=\tau _c}^{\tau }\,(F_e \circ G_{e}^{-1}(z)-\tau )\,\frac{1}{g_{e}(G_{e}^{-1}(z))}\,\text {d}z \right| \,. \end{aligned}$$

Now, one needs to distinguish two cases.

If $\tau \,>\,\tau _c$, one has:

$$\begin{aligned} \left| \,\textsc {E}_{Y,e}[\,L_{\tau ,\tau _{c}}\,]\,\right|= & {} \int _{z=\tau _c}^{\tau }\,\left| \,(F_e \circ G_{e}^{-1}(z)-\tau )\,\frac{1}{g_{e}(G_{e}^{-1}(z))}\,\right| \,\text {d}z\\ \\ \\\le & {} \int _{z=\tau _c}^{\tau }\,\left| \,(F_e \circ G_{e}^{-1}(z)-\tau )\,\right| \,\xi \,\,\text {d}z\,. \end{aligned}$$

Since $|\,F_{e}(z)- G_{e}(z)\,| \le \varepsilon ,$ $\forall z \in \mathbf {R},\,\forall e \in E$, one obtains $|\,F_e \circ G_{e}^{-1}(z)-z\,| \le \, \varepsilon ,$ $\forall z \in [0,1],\,\forall e \in E$ and then:

• if $z=\,\tau$, one has $\left| \,F_e \circ G_{e}^{-1}(\tau )-\tau \,\right| \le \varepsilon$,

• if $z=\,\tau _c$, $\left| \,F_e \circ G_{e}^{-1}(\tau _c)-\tau )\,\right| = \left| \,F_e \circ G_{e}^{-1}(\tau _c)-\tau _{c} + \tau _{c}-\tau \,\right|$.

Moreover, one has:

$$\begin{aligned} \left| \,\tau _{c}-\tau \,\right|= & {} \left| \,\sum _e\,p_e\,\left( \,\tau _{c}-F_{e} \circ G_{e}^{-1}(\tau _{c})\,\right) \,\right| \\\le & {} \sum _e\,p_e\,\left| \,F_{e} \circ G_{e}^{-1}(\tau _{c})-\tau _c \,\right| \\\le & {} \varepsilon \,, \end{aligned}$$

and finally:

$$\begin{aligned} \left| \,F_{e} \circ G_{e}^{-1}(\tau _{c}-\tau )\,\right|\le & {} \left| \,F_{e} \circ G_{e}^{-1}(\tau _{c})-\tau _{c}\,\right| + \left| \,\tau _{c}-\tau \,\right| \\\le & {} 2 \,\varepsilon \,. \end{aligned}$$

One deduces, when $\tau \,>\,\tau _c$:

$$\begin{aligned} \left| \,\textsc {E}_{Y,e}[\,L_{\tau ,\tau _{c}}\,]\,\right| \le \,2\,(\,\tau -\,\tau _c\,)\,\varepsilon \,\xi . \end{aligned}$$

When $\tau \,<\,\tau _c$, one obtains:

$$\begin{aligned} \left| \,\textsc {E}_{Y,e}[\,L_{\tau ,\tau _{c}}\,]\,\right| \le \int _{z=\tau }^{\tau _c}\,\left| \,(F_e \circ G_{e}^{-1}(z)-\tau )\,\right| \,\xi \,\,\text {d}z\,,\\ \end{aligned}$$

and using the same arguments as previously:

$$\begin{aligned} \left| \,\textsc {E}_{Y,e}[\,L_{\tau ,\tau _{c}}\,]\,\right| \le \,2\,(\,\tau _c-\,\tau \,)\,\varepsilon \,\xi . \end{aligned}$$

Hence, one concludes that:

$$\begin{aligned} \left| \,\textsc {E}_{Y,e}[\,L_{\tau ,\tau _{c}}\,]\,\right| \le \,2\,\left| \,\tau -\,\tau _c\,\right| \,\varepsilon \,\xi . \end{aligned}$$

To finish, replacing $\textsc {E}_{Y,e}[\,L_{\tau ,\tau _{c}}\,]$ in (8.8), we have:

$$\begin{aligned} \left| \,\textsc {E}_Y[\,L_{\tau }-L_{\tau _{c}}\,]\,\right| \le 2\,\varepsilon ^{2}\,\xi . \end{aligned}$$

Now let’s focus on the remainder on the right side. First, we only focus on a particular e. Thus, we are interested in:

$$\begin{aligned} \textsc {E}_{Y}[\,L_{\tau }(Y,G_e^{-1}(Q_{\tau }))\,] - \textsc {E}_{Y}[\,L_{\tau }(Y,G_e^{-1}(\tau _c))\,] \equiv \textsc {E}_{Y,e}[L_{\,\widehat{\tau }_{c}}-L_{\tau _c}\,]. \end{aligned}$$

For ease of notation and comprehension, we suppress e in the notation since there is no confusion. So, we have:

$$\begin{aligned} \textsc {E}_{Y}[L_{\,\widehat{\tau }_{c}}-L_{\tau _c}\,]&= \left( \frac{1}{2}-\tau \right) \,\textsc {E}_{Y} \left[ G^{-1}(Q_\tau )- G^{-1}(\tau _c)\right] \nonumber \\&{\quad }+\frac{1}{2}\,\textsc {E}_{Y} \left[ \,|Y-G^{-1}(Q_\tau )| - |Y-G^{-1}(\tau _c)|\right] . \end{aligned}$$

We find:

$$\begin{aligned} \left| \,\textsc {E}_{Y}[L_{\,\widehat{\tau }_{c}}-L_{\tau _c}\,]\,\right|&\le \left| \frac{1}{2}\,\textsc {E}_{Y}\left[ \,|Y-G^{-1}(Q_\tau )| - |Y-G^{-1}(\tau _c)|-G^{-1}(Q_\tau ) + G^{-1}(\tau _c)\,\right] \right| \nonumber \\&{\quad }+ (1-\tau )\left| \,\textsc {E}_{Y} \left[ G^{-1}(Q_\tau ) - G^{-1}(\tau _c)\right] \,\right| . \end{aligned}$$

Let’s focus on the second term on the right side. Using a Taylor series approximation around $\tau _c \in [0,1]$ and the Taylor–Lagrange formula for the remainder, one has:

$$\begin{aligned} G^{-1}(Q_\tau ) = G^{-1}(\tau _c) + \frac{1}{g(G^{-1}(\tau _c))}\,(Q_{\tau }-\tau _c) + \frac{g^{'}(\gamma )}{g(\gamma )^{3}}\,\frac{(Q_{\tau }-\tau _c)^{2}}{2}\,, \end{aligned}$$

with $\gamma = \tau _c +(Q_{\tau }-\tau _c)\,\theta$, and $0< \theta < 1$.

And so

$$\begin{aligned} (1-\tau )\left| \,\textsc {E}_{Y} \left[ G^{-1}(Q_\tau ) - G^{-1}(\tau _c)\right] \,\right| \le \,\frac{(1-\tau ) \,\alpha \,\xi ^{3}}{2}\,\,\frac{\lambda }{n}. \end{aligned}$$

Now, one can study the first term on the right side. Some useful remarks before the next: one can easily see that the study of such a function can be restricted to a study on the interval $I_{y}:= ]-\infty ,G^{-1}(\tau _c)]$, since we can find results on the interval $[G^{-1}(\tau _c),\infty [$ using the same arguments.

Let’s define $G^{-1}(Q_\tau ) \equiv Z_\tau$, $G^{-1}_{\tau _c} \equiv G^{-1}({\tau _c})$ and $f_{Y}^{\,\,G^{-1}_{\tau _c}} \equiv f_{Y}(G^{-1}(\tau _c))$, for ease of notation.

Thus, we are interested in calculating:

$$\begin{aligned} \frac{1}{2}\, \int _{y=-\infty }^{G^{-1}_{\tau _c}} f_{Y}(y)\,\underbrace{(\,\textsc {E}_{Z_\tau }[\,|G^{-1}_{\tau _c} - Z_\tau | \,+ |Z_\tau -y|\,]-G^{-1}_{\tau _c}+y)}_{ = \textsc {E}_{Z_\tau }[\,|Z_\tau -y| - Z_\tau \,]+y}\,\text {d}y\,\nonumber .\\ \end{aligned}$$

(8.10)

However, the function studied in the integral is complicated to work with. So, one will prefer to use its integral version, that is,

$$\begin{aligned} \textsc {E}_{Z_\tau } [\,|Z_\tau -y| - Z_\tau \,]+y= & {} \int _{u=-\infty }^{y} \frac{\text {d}}{\text {d}u}\,(\textsc {E}_{Z_\tau }[\,|Z_\tau -u| - Z_\tau \,]+u)\,\text {d}u\,. \end{aligned}$$

For the bounds of the integral, the upper one is obvious. To justify the lower one, it is important to note that $\lim \,\,\textsc {E}_{Z_\tau } [\,|Z_\tau -y| - Z_\tau \,]+y=0$, $y\rightarrow -\infty$.

Indeed, one has:

$$\begin{aligned} \textsc {E}_{Z_\tau } [\,|Z_\tau -y| - Z_\tau \,]+y&= \int _{z=-\infty }^{y}(y-z)\,h(z)\,dz + \int _{z=y}^{\infty }(z-y)\,h(z)\,\text {d}z \nonumber \\&{\quad }+ \int _{z=-\infty }^{\infty } (y-z)\, h (z)\,\text {d}z \nonumber \\&=\int _{z=-\infty }^{y}2(y-z)\,h(z)\,\text {d}z \nonumber \\&= 2y\,H(y) - \int _{z=-\infty }^{y}2 z\,h(z)\,\text {d}z\,, \end{aligned}$$

with h and H the p.d.f and the c.d.f of the variable $Z_t$. If the variable $Z_\tau$ has a finite mean, $\lim ,h(y)=0$, $y\rightarrow -\infty$, and thus it is clear that the choice of $-\infty$ for the lower bound of the integral is the good one.

At this stage, it is not easy to see the usefulness of the transformation, but it will be after the following calculus:

$$\begin{aligned} \frac{d}{du}(\textsc {E}_{Z_\tau } [\,|Z_\tau -u| - Z_\tau \,]+u)&= 1 + \frac{d}{du}\left( \int _{z=-\infty }^{u} (u-z)\,h(z)\,\text {d}z \,\right) \nonumber \\&{\quad }+ \frac{d}{du}\left( \int _{z=u}^{\infty } (z-u)\,h(z)\,\text {d}z\,\right) . \end{aligned}$$

Finally, we have:

$$\begin{aligned} \frac{d}{du}(\textsc {E}_{Z_\tau } [\,|Z_\tau -u| - Z_\tau \,]+u)&= \int _{z=-\infty }^{u} h(z)\,dz - \int _{z=u}^{\infty } h(z)\,\text {d}z +1\\&= H(u)-(\,1-H(u)\,)+1\\&= 2\,H(u). \end{aligned}$$

Now, it is clear that this transformation could help us for the calculus of (8.10) since it is equivalent to study:

$$\begin{aligned} \int _{y=-\infty }^{G^{-1}_{\tau _c}}\,f_{Y}(y)\,\left( \,\int _{u=-\infty }^{y} H(u)\,du\,\right) \,\text {d}y \equiv \text {Half Int}. \end{aligned}$$

A difficulty remains, though. Indeed, $f_{Y}$ in unknown, and in consequence, not easy to work with. That’s why, at first, one will use $f_{Y}^{\,\,G^{-1}_{\tau _c}}$ for our calculus, and then we will study the impact of such a manipulation.

Let’s start with the first task. Using an integral by part on Half Int:

$$\begin{aligned} \int _{y=-\infty }^{G^{-1}_{\tau _c}}\,\underbrace{f_{Y}^{\,\,G^{-1}_{\tau _c}}}_{u'}\,\underbrace{\left( \,\int _{u=-\infty }^{y} H(u)\,du\,\right) }_v\,\text {d}y\,. \end{aligned}$$

One obtains:

$$\begin{aligned} \text {Half Int}&=\left[ yf_{Y}^{\,\,G^{-1}_{\tau _c}}\left( \,\int _{u=-\infty }^{y}H(u)\,du\,\right) \,\right] _{y=-\infty }^{G^{-1}_{\tau _c}} -\int _{y=-\infty }^{G^{-1}_{\tau _c}} y f_{Y}^{\,\,G^{-1}_{\tau _c}}\,H(y)\,\text {d}y\\&=\int _{u=-\infty }^{G^{-1}_{\tau _c}} f_{Y}^{\,\,G^{-1}_{\tau _c}}\, \underbrace{[\,G^{-1}_{\tau _c}-u\,]}_{u'}\, \underbrace{H(u)}_v \,\text {d}u\\&=\left[ \,f_{Y}^{\,\,G^{-1}_{\tau _c}}\,\left( \,u\,G^{-1}_{\tau _c}-\frac{u^{2}}{2}\,\right) \,H(u)\,\right] _{u=-\infty }^{G^{-1}_{\tau _c}}\\&{\quad }- \int _{u=-\infty }^{G^{-1}_{\tau _c}} f_{Y}^{\,\,G^{-1}_{\tau _c}}\,\left( \,u\,G^{-1}_{\tau _c}-\frac{u^{2}}{2}\,\right) h(u)\,\text {d}u\,. \end{aligned}$$

Since $\left( \,u\,G^{-1}_{\tau _c}-\frac{u^{2}}{2}\,\right) = \left( \,\frac{(u-G^{-1}_{\tau _c})^{2}}{2}-\frac{(G^{-1}_{\tau _c})^{2}}{2}\right)$, we have:

$$\begin{aligned} \text {Half Int} = f_{Y}^{\,\,G^{-1}_{\tau _c}}\,\left( \int _{u=-\infty }^{G^{-1}_{\tau _c}} \frac{(u-G^{-1}_{\tau _c})^{2}}{2}\,h(u)\,\text {d}u\right) . \end{aligned}$$

Now, using the change of variable $G(u)= z$, a Taylor series approximation around $\tau _c$ and the Taylor–Lagrange formula, one has the following approximation for Half Int:

$$\begin{aligned} \frac{f_{Y}^{\,\,G^{-1}_{\tau _c}}}{2}\int _{z=0}^{\tau _c}\left[ \frac{1}{g(G^{-1}_{\tau _c})^{2}}(z-\tau _c)^{2} + \frac{g^{'}(\gamma )}{g(G^{-1}_{\tau _c})\,g(\gamma )^{3}}(z-\tau _c)^{3}+\frac{g^{'}(\gamma )^{2}}{4g(\gamma )^{6}}(z-\tau _c)^{4}\right] \phi (y)\text {d}y\,, \end{aligned}$$

with $\phi$ the p.d.f of the random variable $Q_{\tau }$. Using the Jensen inequality and since $0\le z \le \tau _c$, we find:

$$\begin{aligned} |\text {Half Int}|&\le \frac{f_{Y}^{\,\,G^{-1}_{\tau _c}}}{2} \left[ \,\frac{\xi ^{2}}{2} \frac{\lambda }{n} + \frac{\alpha \xi ^{4}}{2}\frac{\lambda }{n} + \frac{\alpha ^{2} \xi ^{6}}{8}\frac{\lambda }{n}\,\right] \\&\le \frac{1}{2}\,\frac{C_{int}\,\lambda }{n} . \end{aligned}$$

Since $\frac{\lambda }{n}$, which is the variance of the random variable $Q_{\tau }$, is decreasing with n, let’s study:

$$\begin{aligned} \varDelta _{\int f} \,\equiv \left| \, \int _{u=-\infty }^{G^{-1}_{\tau _c}}(\,f_{Y}(y) - f_{Y}^{\,\,G^{-1}_{\tau _c}}\,)\,\left( \,\int _{u=-\infty }^{y} H(u)\,\text {d}u\,\right) \,\text {d}y \,\right| . \end{aligned}$$

Since one supports the hypothesis that $f_{Y}^{'}$ is bounded, using the mean value theorem, one has:

$$\begin{aligned} \varDelta _{\int f}\,&\le \int _{y=-\infty }^{G^{-1}_{\tau _c}}|\,f_{Y}(y) - f_{Y}^{\,\,G^{-1}_{\tau _c}}\,|\,\left( \,\int _{u=-\infty }^{y} H(u)\,\text {d}u\,\right) \,\text {d}y\\&\le \int _{y=-\infty }^{G^{-1}_{\tau _c}}M \,\underbrace{ (\,G^{-1}_{\tau _c}- y \,) }_{u'}\,\underbrace{\left( \,\int _{u=-\infty }^{y} H(u)\,\text {d}u\,\right) }_v\,\text {d}y\,, \end{aligned}$$

and thus,

$$\begin{aligned} \varDelta _{\int f}&\le M \left( \left[ \left( y\,G^{-1}_{\tau _c} - \frac{y^{2}}{2}\right) \,\int _{u=-\infty }^{y} H(u)\,\text {d}u \right] _{y=-\infty }^{G^{-1}_{\tau _c}} - \int _{y=-\infty }^{G^{-1}_{\tau _c}} \left( y\,G^{-1}_{\tau _c} - \frac{y^{2}}{2}\right) H(y)\text {d}y\right) \\&= M\left( \frac{(G^{-1}_{\tau _c})^{2}}{2}\,\int _{u=-\infty }^{G^{-1}_{\tau _c}} H(u)\,\text {d}u + \int _{u=-\infty }^{G^{-1}_{\tau _c}}\left( \,\frac{(u-G^{-1}_{\tau _c})^{2}}{2}-\frac{(G^{-1}_{\tau _c})^{2}}{2}\right) H(u)\,\text {d}u\,\right) \nonumber \\&= M\,\int _{u=-\infty }^{G^{-1}_{\tau _c}} \,\underbrace{H(u)}_v \,\underbrace{\frac{(u-G^{-1}_{\tau _c})^{2}}{2}}_{u'}\,\text {d}u\\&= M \,\left( \,\left[ \,\frac{(u-G^{-1}_{\tau _c})^{3}}{6}\,H(u)\,\right] _{u=-\infty }^{G^{-1}_{\tau _c}} - \int _{u=-\infty }^{G^{-1}_{\tau _c}} \frac{(u-G^{-1}_{\tau _c})^{3}}{6}\,h(u)\,\text {d}u\,\right) . \end{aligned}$$

Finally, we obtain with the same change of variable and Taylor approximation as previously:

$$\begin{aligned} \varDelta _{\int f} \,\le & {} \, \frac{M}{6}\,\int _{z=0}^{\tau _c}\left[ \,\frac{1}{g(G^{-1}_{\tau _c}))}(\tau _c-z)+ \frac{g^{'}(\gamma )}{2g(\gamma )^{3}}(\tau _c-z)^{2}\,\right] ^{3}\,\phi (z)\,\text {d}z\\ \\ \\\le & {} \,\frac{M}{6}\,\left[ \frac{\xi ^3}{2} \frac{\lambda }{n} + \frac{3 \xi ^3 \alpha }{4} \frac{\lambda }{n} + \frac{3 \xi ^3 \alpha ^2}{8} \frac{\lambda }{n} + \frac{\xi ^3 \alpha ^3}{16} \frac{\lambda }{n}\right] \\ \\\le & {} \,\frac{1}{2}\,\frac{C_s\,\lambda }{n}. \end{aligned}$$

Thus, one has $\left| \,\textsc {E}_{Y,e}[L_{\,\widehat{\tau }_{c}}-L_{\tau _c}\,]\,\right| \le \,\frac{(C_{int} + C_s)\lambda }{n}$. Since $C_{int}$ and $C_s$ do not depend on e, this result remains meaningful when we are interested in the conditional expectation with respect to the random variable E and so $\left| \,\textsc {E}_Y[L_{\,\widehat{\tau }_{c}}-L_{\tau }\,]\,\right| \le \, 2\,\varepsilon ^{2}\,\xi + \frac{C\,\lambda }{n}$.

Moreover, using (8.9), we prove (8.7).    $\square$