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4.7 Regression ved beregning

Beregning af residualer

Vi vender tilbage til vores datasæt:

\(\begin {array}{ | c | c |} \hline x_i & y_i \\ \hline 1 & 2 \\ 3 & 5 \\ 5 & 4 \\ \hline \end {array}\)

Tidligere fandt vi ved lineær regression følgende model:

\[y=0{,}5x+2{,}17.\]

Ud fra vores model, kan vi nemt beregne residualerne. Vi husker at residualerne er afvigelsen mellem data og modellen. Altså kan vi finde det i’te residual ved:

\[e_i=y_i-\hat {y_i}\]

Her er:
\(e_i\) det i’te residual,
\(y_i\) den i’te y-værdi i data,
\(\hat {y_i}\) den i’te estimerede y-værdi ifølge modellen (altså den y-værdi man beregner med den estimerede model).

  • Eksempel 4.7.1
    Vi vil nu kigge på hvordan vi kan beregne residualerne vores data:

    \(\begin {array}{ | c | c |} \hline x_i & y_i \\ \hline 1 & 2 \\ 3 & 5 \\ 5 & 4 \\ \hline \end {array}\)

    Vi starter med at beregne \(\hat {y_1}\). Den første x-værdi er \(1\), så vi sætter \(1\) ind i forskriften.

    \[\hat {y_1}=0{,}5\cdot 1 + 2{,}166667=2,666667\]

    Tilsvarende regner vi resten af \(\hat {y_i}\)’erne

    .
    \(x_i\) \(y_i\) \(\hat {y_i}\)
    1 2 2,666667
    3 5 3,666667
    5 4 4,666667

    Vi regner så det første residual

    \[e_1=y_1-\hat {y_1}=(2-2{,}666667)=-0,666667,\]

    og på tilsvarende måde regner vi resten af residualerne:

    .
    \(x_i\) \(y_i\) \(\hat {y_i}\) \(e_i\)
    1 2 2,666667 -0,666667
    3 5 3,666667 1,333333
    5 4 4,666667 -0,666667

    Vi har hermed beregnet residualerne som altså ses i kolonnen til højre.

Mindste kvadraters metode

Der findes flere metoder til at beregne regressionsmodeller. Her skal vi se på en metode som hedder ”mindste kvadraters metode”.

Vi vil igen tage udgangspunkt i datasættet:

\(\begin {array}{ | c | c |} \hline x_i & y_i \\ \hline 1 & 2 \\ 3 & 5 \\ 5 & 4 \\ \hline \end {array}\)

Vi vil først se på ideen i metoden, og derefter hvordan man rent beregningsteknisk finder modellen. Vi starter med at tegne vores punkter ind sammen med en mulig regressionslinje (det er den vi skal bestemme) og residualerne (tegnet med blåt):

(-tikz- diagram)

Vi tegner nu kvadrater ud fra residualerne:

(-tikz- diagram)

Mindste kvadraters metode går ud på at bestemme den forskrift \(y=ax+b\), som gør de blå kvadrater mindst mulige. Da kvadraterne har sidelængde \(e_i\), og dermed areal \(e_i^2\), skal vi altså bestemme \(a\) og \(b\) så

\begin{equation} \label {eq:(1)} \Sigma _{i=1}^{n}e_i^2 \end{equation}

bliver mindst muligt.

Vi vil beregne residualerne r på tilsvarende måde som i eksempel 4.7.1, men denne gang er \(a\) og \(b\) ubekendte i forskriften \(\hat {y}=ax+b\). I vores første punkt er \(x_1=1\) og derfor sætter vi \(1\) ind på \(x\)’ets plads i forskriften:

\[\hat {y_1}=a\cdot 1 + b=a+b\]

Tilsvarende sætter regner vi resten af \(\hat {y_i}\)’erne

.
\(x_i\) \(y_i\) \(\hat {y_i}\)
1 2 \(a+b\)
3 5 \(3a+b\)
5 4 \(5a+b\)

Vi regner så de kvadratiske residualer:

\[e_1^2=(y_1-\hat {y_1})^2=(2-(a+b))^2=(-a-b+2)^2\]

og tilsvarende for alle de andre punkter.

.
\(x_i\) \(y_i\) \(\hat {y_i}\) \(e_i^2\)
1 2 \(a+b\) \((-a-b+2)^2\)
3 5 \(3a+b\) \((-3a-b+5)^2\)
5 4 \(5a+b\) \((-5a-b+4)^2\)

Vi finder summen af alle de kvadratiske residualer

\begin{align*} \Sigma _{i=1}^{3}e_i^2 & = e_1^2+e_2^2+ e_3^2 \\ & = (-a-b+2)^2+(-3a-b+5)^2+(-5a-b+4)^2 \end{align*}

Vi ser nu at summen af kvadratiske residualer afhænger af \(a\) og \(b\), og vi kan derfor se den som en funktion af \(a\) og \(b\):

\[f(a,b)= (-a-b+2)^2+(-3a-b+5)^2+(-5a-b+4)^2,\]

hvor \(f(a,b)\) altså betegner summen af de kvadratiske residualer. Åhhh bare vi dog havde en metode til at minimere en funktion i to variable... nej hov vent... Det har vi jo!!! Den lærte vi i afsnittet om ekstrema for partielle afledede. Ifølge sætning 4.6.1 skal vi sætte de partielle aflede lig nul. Vi starter med at bestemme de partielle afledede (og vi husker at bruge reglen for sammensatte funktioner).

\begin{align*} \frac {\partial }{\partial a}f(a,b) & = 2(-a-b+2) (-1)+2(-3a-b+5)(-3) +2(-5a-b+4)(-5)\\ & = 2a+2b-4+18a+6b-30+50a+10b-40\\ & = 70a+18b-74 \end{align*}

\begin{align*} \frac {\partial }{\partial b}f(a,b) & = 2(-a-b+2) (-1)+2(-3a-b+5)(-1) +2(-5a-b+4)(-1)\\ & = 2a+2b-4+6a+2b-10+10a+2b-8\\ & = 18a+6b-22 \end{align*}

Vi sætter nu de partielle afledede lig med nul:

\[70a+18b-74=0\quad 18a+6b-22=0\]

Vi mangler altså bare at løse to ligninger med to ubekendte. Vi isolerer \(b\) i den første ligning:

\begin{align*} 70a+18b-74 & =0\\ b & = -3{,}88889a+4{,}111111 \end{align*} og sætter ind på \(b\)’s plads i den anden ligning

\begin{align*} 18a+6b-22 & = 0\\ 18a+6(-3{,}88889a+4{,}111111)-22 & = 0\\ a & = 0{,}5. \end{align*} Vi kan nu finde \(b\):

\begin{align*} b & = -3{,}88889a+4{,}111111\\ & = -3{,}88889\cdot 0{,}5+4{,}111111\\ & = 2{,}17. \end{align*} Vi har nu fundet både \(a\) og \(b\) og derfor har vi fundet forskriften

\[y=0{,}5x+2{,}17.\]

Den kritiske læser vil måske være bekymret over, om vi kan være sikker på, at vi har minimeret de kvadratiske afvigelser. At de partielle afledede giver nul, er ikke nogen garanti for et minimum. Vi ved dog at hvis en funktion har et minimum, så er de partielle afledede nul. Da det er oplagt at de kvadratiske afvigelser må have et minimum (hvorfor er det oplagt?), og da der kun en løsning, hvor de partielle afledede er nul, må det være minimum vi har fundet.

Øvelse 4.7.1

Betragt data

\(\begin {array}{ | c | c |} \hline x_i & y_i \\ \hline 1 & 4 \\ 2 & 2 \\ 4 & 5 \\ \hline \end {array}\)

Du skal nu bestemme en lineær model for data. Jeg har delt det op i 3 bidder, så det bliver nemmere at identificere, hvad der er gået galt, når det er gået galt (for det gør det).

  • a) Bestem funktionen \(f(a,b)\) som beskriver summen af de kvadratiske residualer.

  • b) Bestem de partielle afledte for \(f\)

  • c) Bestem den lineære model.

Løsning 4.7.1

  • a) \(f(a,b)=(-a-b+4)^2+(-2a-b+2)^2+(-4a-b+5)^2\)

  • b) \(\frac {\partial }{\partial a}f(a,b)=42a + 14b - 56\)
    \(\frac {\partial }{\partial b}f(a,b) = 14a + 6b - 22\)

  • c) \(y=0{,}5x+2{,}5\)

Øvelse 4.7.2

Betragt data:

\(\begin {array}{ | c | c |} \hline x_i & y_i \\ \hline 0 & 0 \\ 1 & 1 \\ 2 & 4 \\ 3 & 5 \\ \hline \end {array}\)

Bestem ved beregning:

  • a) funktionen \(f(a,b)\) som beskriver summen af de kvadratiske residualer.

  • b) de partielle afledte for \(f\)

  • c) en lineær model for data

  • d) residualerne

Løsning 4.7.2

  • a) \(f(a,b)=(-b)^2+(-a-b+1)^2+(-2a-b+4)^2+(-3a-b+5)^2\)

  • b) \(\frac {\partial }{\partial a}f(a,b)= 28a + 12b - 48\)
    \(\frac {\partial }{\partial b}f(a,b) = 12a + 8b - 20\)

  • c) \(y=1{,}8x-0{,}2\)

  • d)
    \(\begin {array}{ | c | c | c |} \hline x _i & y_i & e_i \\ \hline 0 & 0 & 0{,}2 \\ 1 & 1 & -0{,}6 \\ 2 & 4 & 0{,}6 \\ 3 & 5 & -0{,}2 \\ \hline \end {array}\)