Gaussian Mixture Modeling and PCA – Lecture Notes

= \lambda _i^{-\frac{1}{2}}\) . Then \(\Omega \Omega = \Lambda \! ^{-1}\) , and \(\Omega ^T = \Omega\) . \[\begin{align*} \Sigma ^{-1} = V \Lambda \! ^{-1} V^{-1} & = V\Omega ^T \Omega V^T = (\Omega V^T)^T [...] 184,24}{\frac1{\sqrt{\vphantom{\pi ^d}\smash {(2\pi )^d |\Sigma |}}}} \cdot e^{\smash {-\frac{1}{2} \textcolor[RGB]{0,155,170}{\big((p-\mu )^T\Sigma ^{-1}(p-\mu ) \big)}} }\end{align*}\] 1-dimensional normal [...] tion to factorize \(\Sigma\) into: \[\Sigma = V \Lambda V^{-1} \quad \text{ and, therefore, } \quad \Sigma ^{-1} = V \Lambda ^{\! -1} V^{-1}\] where \(V\) is orthonormal and contains the eigenvectors \(\cong\) …

Evaluation-Winter20202021-Grafiken

mehr Rückmeldungen 1,00 2,00 3,00 4,00 5,00 6,00 1,00 1,50 2,00 2,50 3,00 3,50 4,00 4,50 5,00 5,50 6,00 B ew er tu ng D oz en t* in in sg es am t Bewertung Veranstaltung insgesamt Noten: 1=sehr gut, ..., 6=sehr [...] e ausgesprochen hat. Übersichten 1,00 1,50 2,00 2,50 3,00 3,50 4,00 4,50 5,00 5,50 6,00 0 10 20 30 40 50 60 70 80 90 100 B ew er tu ng B et re uu ng / D oz en t* in 1= se hr g ut , . .., 6 = s eh r sc [...] 2020/2021: Vorlesungen 1,00 2,00 3,00 4,00 5,00 6,00 1,00 2,00 3,00 4,00 5,00 6,00 B ew er tu ng d er B et re uu ng in sg es am t Bewertung der Veranstaltung insgesamt Noten: 1=sehr gut, ..., 6=sehr schlecht …

eine Wahrscheinlichkeit für die Klasse C für M als P (C|Am1 ∩ . . . ∩ Amk ) = P (C) ∏l j=1 P (Amj |C) P (C) ∏l j=1 P (Amj |C) + (1− P (C)) ∏l j=1 P (Amj |C) Dabei entspricht C dem Komplementär-Ereignis [...] (daher naive Bayes). Sei eine E-Mail M mit den Attribut-Werten Am1 , . . . , Aml mit {m1, . . . ,ml} ⊆ {1, . . . , k} gegeben (z.B. durch auf [0; 1] normierte Häufigkeit vorher gelernter Worte wj). Mit Hilfe [...] gegebenen Beispiel M dann die wahrschein- lichste Klasse prediction(M) = arg max C P (C|Am1 ∩ . . . ∩ Aml ). Aufgabe 10 Punkte 1. Implementieren Sie einen Operator NaiveBayes, der nach dem Satz von Bayes eine Menge …

constants C, ν ≥ 1, 0 < σ ≤ β, σ0 ≤ U ≤ C2 √ nβ,C2 = (4 √ ν log C1) −1, C1 = max(C 1 ν , e). Let C3 = C2 1 16ν and A a constant. Then we have: E ( sup g∈G | n∑ i=1 εig(Xi)| ) ≤ A( √ νnσ2 0 log ( C1β σ0 ) + 2νU [...] constants C, ν ≥ 1, 0 < σ ≤ β, σ0 ≤ M ≤ C2 √ nβ,C2 = (4 √ ν logC1)−1, C1 = max(C 1 ν , e). Let C3 = C2 1 16ν and A be a constant. Then we have: E ( sup g∈G | n∑ i=1 εig(Xi)| ) ≤ A (√ νnσ2 0 log ( C1β σ0 ) + 2νM [...] above by a−1 n n− 1 2ωn(m∗bn). If Assumptions (A1)-(A3) are fulfilled for bn, we get with that ωn(m∗bn) = O( √ bn(log bn)) and finally ||fn(x)− E(fn(x))||∞ = O(a−1 n n− 1 2 (log bn) 1 2 b 1 2 n ). With …

Estimations of the Roman Life Expectancy Using Ulpian´s Table

12) provides an approximation formula, which is, with our symbols: 0 1 0 1 0 1 0 10 1 r kk u mk e 2r 1 1 k k (u m k(u m) 2 r kk k e a u 1 r ! "! " & '& ' ! "( )& ' & ' & '& ' ( )& ' ( ) %# % % # + # + # %% [...] present value of a payment of 1 in m years and an interest rate of r, using simple interest, can be calculated by 0 1 0 11 1 r m 1 r m - % # + # . If r m# is small 0 1 n 1 r m m 1 % #9 $ is therefore an ap [...] over the whole span of life is 0 1 0 10 1 0 1 0 10 1 f l x 0 x 20 *l x f l 20 l x x 20 l(20) 3 2 24 4 $5 4 # . 46 . The life expectancy after the mortality change is: 0 1 0 1 200 * * *e(x) l x dx l x dx 0 …

closest pair 1. p1, ... , pn points of P in random order p1 p2 Closest Pair: Randomized Algorithm Compute ‖p− q‖ of closest pair 1. p1, ... , pn points of P in random order 2. α = ‖p1 − p2‖ p1 p2 Closest [...] first i points Xi = 1I{αi−1<αi} indicator variable running time proportional to R = 1 + ∑n i=3(1 + iXi) E[R] = E[1 + ∑n i=3(1 + i · Xi)] ≤ n + ∑n i=3 i · E[Xi] = n + ∑n i=3 i · Pr[Xi = 1] ≤ n + ∑n i=3 i · [...] first i points Xi = 1I{αi−1<αi} indicator variable running time proportional to R = 1 + ∑n i=3(1 + iXi) E[R] = E[1 + ∑n i=3(1 + i · Xi)] ≤ n + ∑n i=3 i · E[Xi] = n + ∑n i=3 i · Pr[Xi = 1] ≤ n + ∑n i=3 i · …

Training Support Vector Machines – Lecture Notes

f(x_1)-y_1)-(f(x_2)-y_2)}{\left\lVert \smash {x_1-x_2}\mathstrut \right\rVert ^2} = \lambda _2 + y_2\frac{(f(x_1)-y_1)-(f(x_2)-y_2)}{\left\langle x_1,x_1\mathstrut \right\rangle -2\left\langle x_1,x_2 [...] by \eqref{eqn:svm-lambda0}}} b & +\sum \nolimits _{i=1}^{N} \lambda _i \\ & = \tfrac {1}{2} & \hfill \sum \nolimits _{i=1}^{N} \sum \nolimits _{j=1}^{N} \lambda _i\lambda _j y_i y_j \left\langle \vec{x_i} [...] Equation K3 becomes (for any two \(i\) , we simply use 1, 2 here): \[\begin{align*} \sum \nolimits _{i=1}^{N} \lambda _i y_i & = 0 \\ \lambda _1 y_1 + \lambda _2 y_2 & = - \sum \nolimits _{i=3}^{N} \lambda …

Training Support Vector Machines – Lecture Notes

f(x_1)-y_1)-(f(x_2)-y_2)}{\left\lVert \smash {x_1-x_2}\mathstrut \right\rVert ^2} = \lambda _2 + y_2\frac{(f(x_1)-y_1)-(f(x_2)-y_2)}{\left\langle x_1,x_1\mathstrut \right\rangle -2\left\langle x_1,x_2 [...] by \eqref{eqn:svm-lambda0}}} b & +\sum \nolimits _{i=1}^{N} \lambda _i \\ & = \tfrac {1}{2} & \hfill \sum \nolimits _{i=1}^{N} \sum \nolimits _{j=1}^{N} \lambda _i\lambda _j y_i y_j \left\langle \vec{x_i} [...] Equation K3 becomes (for any two \(i\) , we simply use 1, 2 here): \[\begin{align*} \sum \nolimits _{i=1}^{N} \lambda _i y_i & = 0 \\ \lambda _1 y_1 + \lambda _2 y_2 & = - \sum \nolimits _{i=3}^{N} \lambda …

with corresponding transmembrane receptors (S1P1, S1P2, S1P3). (64) 2.5.3.1 HDL-induced cell signaling mediated by scavenger receptor 1 HDL binding to SR-BI is leading to the recruitment of SRC to the [...] ........................................ тп 6.1.1.4 LIPG knockdown results in significantly reduced cell migration ............................. тп 6.1.1.5 LIPG knockdown results in decreased expression [...] ..... млт 6.1.5. HDL signaling properties are impaired by LIPG deficiency in MDA-MB-468 cell ммм 6.1.5.1. HDL-induced SR-BI protein expression is impaired in LIPG knockdown cells . ммм 6.1.5.2 HDL-induced …

000) · 1 1.14 · 1.1 4 − 1 0.1 + 10, 000 1.14 = −30, 000− 22, 000 · 1 1.14 · 1.1 4 − 1 0.1 + 10, 000 1.14 = −92, 907 C0B = −40, 000− (9, 000 + 0.3 · 35, 000) · 1 1.14 · 1.1 4 − 1 0.1 + 15, 000 1.14 = −40 [...] −40, 000− 19, 500 · 1 1.14 · 1.1 4 − 1 0.1 + 15, 000 1.14 = −91, 567 48 The equivalent annuities are: aA = −92, 907 · 1.14 · 0.1 1.14 − 1 = −29, 309 aB = −91, 567 · 1.14 · 0.1 1.14 − 1 = −28, 887. To calculate [...] the NPVs equal: −30, 000− (8, 000 + 0.4 · x) · 1 1.14 · 1.1 4 − 1 0.1 + 10, 000 1.14 = −40, 000− (9, 000 + 0.3 · x) · 1 1.14 · 1.1 4 − 1 0.1 + 15, 000 1.14 Solving for x gives: xcrit = 30, 774 km. In this …