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hybrid model, there must be a simulator S� so that � IdealF S,Z ≈ ExecF H,Z Hence for every adversary A there exists a simulator S such that for every environment: � IdealF S, Z ≈ ExecQP ,A,Z Thus, QP securely realizes F�. 2 Another implication of the theorem is that we can immediately deduce security in a mu...
https://ocw.mit.edu/courses/6-897-selected-topics-in-cryptography-spring-2004/3b2e382e61c2740c5174cd29a3d51cd3_l8.pdf
Chapter  3:  Collecting  Data   population  is  a  collection  of  objects,  items,  humans/animals  (“units”)  about   A   which  information  is  sought.   A   sample  is  a  part  of  the  population  that  is  observed.   parameter  is  a  numerical ...
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sample  –  make  sure  you  know  what  you  can  calculate  and   what  you  can’t  calculate!  You  can’t  calculate  anything  from  the  population  if  you  only   have  a  sample.   sampling  frame  is  a  list  of  all  units  in  a  fin...
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sample  collector  (who  may  use   Judgment  sampling Bias  is  possible  with  these  sampling  methods.  To  avoid  bias,  sample   .   sampling) randomly  from  the  population.   quota   A   wi bei simple  random  sample  (SRS)  of  size  n  from  a ...
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SRS  from  each  one. to  d e.g.   o pulati o  statistics customers  stratified  by  race  (some  races  are  rarer  than  others)  involves  dividing  the  population  into  homogeneous   eful  when  you  want    This ons  as  well  as  on  the  wh ...
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3)  Draw  of  people  from  each  county   1-­‐in-­‐k  systematic  sampling  consists  of  selecting  every  kth  unit.  Useful  for  sampling   items  coming  off  assembly  lines.                                 MIT OpenCour...
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Lecture 7 Quantum Mechanical Measurements. Symmetries, conserved quantities, and the labeling of states Today’s Program: 1. Expectation values 2. Finding the momentum eigenfunctions and the dispersion relations for free particle. 3. Commutator and observables that commute 4. Symmetries and conserved quantities –...
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function space. That u x means that any wave function can be represented as a linear combination of n   : u x  x       C u x n n n Where the coefficients of the expansion are just as they are in the geometrical analogy projections of the function un direction given by the inner products between the o...
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1 Finding the eigenvectors and eigenvalues of operators, discuss the geometrical interpretation of eigenvectors and eigenvalues – scaling. Fourth Postulate (discrete non-degenerate): When the physical quantity a is measured on a system in the normalized state the probability P a n  of obtaining the non-degenera...
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�� x,t0    un *  x  x,t0  dx. ˆ n Cnun x  , where C x, t0    Substituting this into the equation above we get: Aˆ  x,t  x,t    * ˆ x x A   dx  ˆ A n * * n n   A C x dx    C u x  ˆ u     n n   n n C u x  C Au x dx  n   C u x  C a u x dx  n n n * * n...
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 Let’s consider an example: Particle flying in free space. We have previously approximated particles in free space with plane waves but in reality they are more like a linear superposition of multiple plane waves. Let’s consider a particle described by a wavefunction: 1 ik0xi0t  1 2ik0x4i0t   x, t  e ...
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k0e 3ik0x3i0t  1 2 2 k0e 2ik0x  1 2k0e 2 2 3ik0x3i0t  dx   1 4 So on average the momentum is equal to: k0 What is the probability for this particle to be flying left (in the opposite direction of x axis)? Here we need to find the probability that particle has negative momentum. Indeed from plane ...
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How precisely can we measure a physical quantity? – Before we can answer this question let’s consider a helpful mathematical concept: The Commutator operator and commutation relations The Commutator operator is defined as: Aˆ, Bˆ   AˆBˆ  BˆAˆ  Two observables Aˆ and Bˆ are said to commute if: AˆBˆ  BˆAˆ  A...
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x   3 3 x 2m x 3  i  3 3 x 2m x 3  0 Do Hˆ    2 2 2m x 2 and pˆ  i  x share eigenfunctions? The eigenfunctions for momentum are plane waves e ikx : i  e x ikx  k  eikx And from the previous lectures we remember that the plane waves are also eigenfunctions for the free space Hamilt...
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x  xˆ, pˆ    xi x  Then: xˆ, pˆ  i  i x    i  x 5 It turns out that this relationship is very significant as it Mathematics it means that x and p cannot be measured with absolute precision at the same time. In fact their uncertainties (or measurement...
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symmetries manifest themselves in equations?�Let us suppose that your system is symmetric with respect to translations in x that would imply that any physical property could not have an x dependence. In particular the energy would not have an explicit dependence on x thus: H x, p x  0    p  const dp dt  ...
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� 0 States are labeled by specific values of their properties, which do not change with time – these properties are called constants of motion. We learned that in QM physical properties are represented by operators and that the values of properties obtained in measurements are eigenvalues of the corresponding opera...
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��      t  i r  This type of differential equation is separable, i.e. we can look for a solution in the following form:  r   2      2m 1               r        i  t t    t  r 2  2m  V r  V r  t    r 1    t Note that the left side...
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 2      2m   r    E   uE  , 2  V   x   r  r E Then the solutions to time-dependent Schrodinger’s equation will have a form: 7 E t i     uE r E  t  uE r  e  E  r, t In general, since the Hamiltonian may have many eigenvalues and corresponding eige...
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�� x  E x      2 2 2m x2      E   uE     x   e x x e i i 2 mE 2 x  2mE 2 x  Using the energy as a “label” doesn’t completely and uniquely specify a state. What about momentum? – If momentum is a constant of motion then we can use it as an additional label to uniquely specify...
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uniquely and completely specifies the state. 8 Conservved Quantitiies Examplee II: Parity operator annd symmetrric potentialls Definitioon of a parityy operator: ˆˆ x    x What aree the eigenfuunctions and eigenvaluess of the parityy operator: ˆ    u x u x    ˆ ˆ    ˆ ...
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22 2  m2 x     2 2 2 2m x 1 2  2 2mm  m2 2  ˆ   x H x Let’s cheeck the commmutator:  ˆ     ˆ , Hˆ   x Hˆ x  Hˆ ˆ   2 2 2    x  ˆ  2  2mm x 2  m x 1 2 2       x   2 2   2  2m x 2       2m  2  x   mm2 x     ...
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 2mm 2 2 x 2   m2 x  x  0  1 2    1 2 m2 2   x  x    This meaans that one can always ffind a set of eigenfunctioons commonn to Hˆ and ˆ . In fact, llast lecture wwe have showwn that SHOO eigenfunctiions are alwaays even or oodd. 9 MIT OpenCourseWare http:...
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6.241 Dynamic Systems and Control Lecture 9: Transfer Functions Readings: DDV, Chapters 10, 11, 12 Emilio Frazzoli Aeronautics and Astronautics Massachusetts Institute of Technology March 2, 2011 E. Frazzoli (MIT) Lecture 9: Transfer Functions Mar 2, 2011 1 / 13 Asymptotic Stability (Preview) We have seen tha...
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can be written in the form: y [k] = CAk x[0] + C k−1 � � Ak−i−1Bu[i] + Du[t] � i=0 or y (t) = C exp(At)x(0) + C � t 0 exp(A(t − τ ))Bu(τ ) dτ + Du(t). However, the convolution integral (CT) and the sum in the DT equation are hard to interpret, and do not offer much insight. In order to gain a better underst...
https://ocw.mit.edu/courses/6-241j-dynamic-systems-and-control-spring-2011/3bf509c94d0977a81bf16f1d0e15c47e_MIT6_241JS11_lec09.pdf
ejωt + e−jωt ) = 2u0e σt cos(ωt), � and the input u is a “half” of a sinusoid with exponentially-changing amplitude. E. Frazzoli (MIT) Lecture 9: Transfer Functions Mar 2, 2011 4 / 13 Output response to elementary inputs (1/2) Recall that, y (t) = CeAt x(0) + C � t 0 e A(t−τ )Bu(τ ) dτ + Du(t). Plug in u(t)...
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will converge to zero. The steady state response can be written as: → 0, and the transient response yss = G (s)e st , G (s) ∈ Cny ×nu , where G (s) = C (sI − A)−1B + D is a complex matrix. The function G : s describes how the system transforms an input e → st G (s) is also known as the transfer function: it into...
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− i − 1: i=0 y [k] = CAk x[0] + C k−1 � Al Bu0z k−l−1 + Du0z k l=0 = CAk x[0] + Cz k−1 � k−1 � � (Az −1)i Bu0 + Du0z k . i=0 E. Frazzoli (MIT) Lecture 9: Transfer Functions Mar 2, 2011 8 / 13 Matrix geometric series Recall the formula for the sum of a geometric series: k−1 � i m = i=0 1 − mk 1 − m...
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(zI − A)−1Bu0 + C (zI − A)−1B + D u0z k . � � �� Transient response �� Steady−state response � � � � � If the system is asymptotically stable, the transient response will converge to zero. The steady state response can be written as: yss[k] = G (z)z k , G (z) ∈ C, where G (z) = C (zI − A)−1B + D is a comple...
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13 Models of continuous-time systems CT CT System CT x˙ (t) = Ax(t) + Bu(t) y (t) = Cx(t) + Du(t) ⎡ 1 . . . 0 0 . . . A = ⎢ ⎢ ⎣ . . . −a0 −a1 ⎤ 0 0 0 . . . . . . ⎥ ⎥ 1 ⎦ . . . −an−1 ⎤⎡ 0 . . . ⎥ B = ⎢ ⎥ ⎢ ⎦ ⎣ 0 1 � C = b0 b1 � . . . bn−1 D = d G (s) = C (sI − A)−1B + D G (s) = bn−1s n−1 ...
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(z) = C (zI − A)−1B + D G (z) = bn−1z n−1 + . . . + b0 z n + an−1z n−1 + . . . + a0 + d E. Frazzoli (MIT) Lecture 9: Transfer Functions Mar 2, 2011 13 / 13 MIT OpenCourseWare http://ocw.mit.edu 6.241J / 16.338J Dynamic Systems and Control Spring 2011 For information about citing these materials or our Term...
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2.092/2.093 — Finite Element Analysis of Solids & Fluids I Fall ‘09 Lecture 4 - The Principle of Virtual Work Prof. K. J. Bathe MIT OpenCourseWare Su = Surface on which displacements are prescribed Sf = Surface on which loads are applied Su ∪ Sf = S ; Sf ∩ Su = ∅ Given the system geometry (V, Su, Sf ), loads (...
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y stresses (forces per unit area in the deformed geometry). II. τij nj = fi Sf on Sf Compatibility: ui = u S i u on Su and all displacements must be continuous. • • Stress-strain laws This is known as the differential formulation. Example Reading assignment: Section 3.3.4 • Equilibrium EA EA d2u + f B = 0...
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− Z L 0 dδu dx EA du dx dx + Z L 0 f Bδudx = 0 (A) The equation above becomes: Internal virtual work }| EA z Z L { dx = dδu dx du dx 0 External virtual work }| { f Bδudx z Z L + 0 Virtual work due to boundary forces z }| { (cid:12) (cid:12) Rδu (cid:12)L dx are the virtual strains, du where dδu on Su, since we do not k...
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εxx εyy εzz γxy γyz γzx εxx εyy εzz γxy γyz γzx ; εxx = ∂u ∂x ; εzz = ∂u ∂z We see that (B) is the generalized form of (A’). The principle of virtual work states that for any compatible virtual displacement field imposed on the body in its state of equilibrium, the total internal virtual work is 4 Lecture 4 The Princi...
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Metals and Insulators • Covalent bonds, weak U seen by e-, with EF being in mid-band area: free e-, metallic • Covalent or slightly ionic bonds, weak U to medium U, with EF near band edge – EF in or near kT of band edge: semimetal – EF in gap: semiconductor • More ionic bonds, large U, EF in very large gap, insulator...
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* + pe 2τh * mh p is analogous to n for holes, and so are τ and m* Note that in both photon stimulated promotion as well as thermal promotion, an equal number of holes and electrons are produced, i.e. n=p ©1999 E.A. Fitzgerald 5 Thermal Promotion of Carriers • We have already developed how electrons are promoted...
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1 E − 2 T dE e k b 3 * ⎛ 2m ⎞ 2 1 e ⎟⎟ 2 ⎜⎜ 2 2π ⎝ h ⎠ 3 2 E k b F T E − kb e g T NC EF −Eg n = N C e kbT 7 ©1999 E.A. Fitzgerald Density of Thermally Promoted of Carriers • A similar derivation can be done for holes, except the density of states for holes is used • Even though we know that n=p, ...
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in the ln term, the Fermi level sits about in the center of the band gap 3 kbT ⎛ ⎞ 2 (me p or n = ni = 2 ⎟ ⎜ 2 ⎝ 2πh ⎠ * )3 * mv 4 −E g e 2kbT ©1999 E.A. Fitzgerald 9 Law of Mass Action for Carrier Promotion 3 kbT ⎛ ⎞ 2 = np = 4 ni ⎟ ⎜ 2 ⎝ 2πh ⎠ (me * )3 * mh 2 E − kb e g T ; 2 = NC NV e ...
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∝ e 2kbT − Eg This can be a measurement for Eg For Si, Eg=1.1eV, and let μe and μh be approximately equal at 1000cm2/V-sec (very good Si!) σ~1010cm-3*1.602x10-19*1000cm2/V-sec=1.6x10-6 S/m, or a resistivity ρ of about 106 ohm-m max •One important note: No matter how pure Si is, the material will always be a poor...
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r e t p a h 4 C 4 Matrix estimation Over the past decade or so, matrices have entered the picture of high-dimensional statistics for several reasons. Perhaps the simplest explanation is that they are the most natural extension of vectors. While this is true, and we will see exam- ples where the extension from vectors t...
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{ } j i 82 ≤ r A = U DV ⊤ = λj ujvj⊤ , j=1 X , λ1, . . . , λr} r diagonal matrix with positive diagonal entries IR that are orthonormal and V is r} ∈ IRn that are also orthonormal. Moreover, { m × where D is a r U is a matrix with columns a matrix with columns it holds that u1, . . . , u { v1, . . . , vr} ∈ j uj , { AA...
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a n × n PSD matrix A, we have λmax (A) = max x⊤Ax . x ∈S n−1 Norms and inner product Let A = aij} { in the following notation. and B = { be two real matrices. Their size will be implicit bij} Vector norms The simplest way to treat a matrix is to deal with it as if it were a vector. In particular, we can extend ℓq norms...
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called the Nuclear norm (or trace norm) of A. kop is called the operator norm (or A k q = k spectral norm) of A. ∞ ∞ We are going to employ these norms to assess the proximity to our matrix of interest. While the interpretation of vector norms is clear by extension from the vector case, the meaning of “ kop is small” i...
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the regression function in this case X = x] is a function from IRd to IRT . Clearly, f can be estimated f (x) = IE[Y independently for each coordinate, using the tools that we have developed in the previous chapter. However, we will see that in several interesting scenar- ios, some structure is shared across coordinate...
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for each of the n days are already known to headquarters and are stored in d. In this case, it may be reasonable to assume that the a matrix X same subset of variables has an impact of the sales for each of the franchise, though the magnitude of this impact may differ from franchise to franchise. As a result, one may as...
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than in [LPTVDG11]. Indeed, rather than exploiting sparsity, observe that such a matrix Θ∗ has rank k. This is the kind of structure that we will be predominantly using in this chapter. Rather than assuming that the columns of Θ∗ share the same sparsity pattern, it may be more appropriate to assume that the matrix Θ∗ i...
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: 2 = u⊤ 2 Akin to the sub-Gaussian sequence model, we have a direct observation model where we observe the parameter of interest with additive noise. This |2 = 1. w u 2 | | | X X ⊤ u =n 4.2. Multivariate regression 86 |0 is small. enables us to use thresholding methods for estimating Θ∗ when However, this also follow...
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1 kop ≤ δ. A − 4σ log(12)(d ∨ T ) + 2σ 2 log(1/δ) p p d Proof. This proof follows the same steps as Problem 1.4. Let 1 and net for that we can always choose − , it holds u N2 be a 1/4-net for N1 be a 1/4- It follows from Lemma 1.18 . Moreover, for any T − S 12d and 1 |N | ≤ 2 |N | ≤ S 1, v T 1 2T 1. 1 − − d ∈ S ∈ S u⊤A...
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∈ N2.Together with the above display, it yields t 8σ2 kop > t A 12d+T exp IP ≤ − × 2 ∼ subG(σ2) for any x ∈ δ ≤ k (cid:0) (cid:1) (cid:0) (cid:1) or f 4σ log(12)(d t ≥ ∨ T ) + 2σ 2 log(1/δ) . p p The following theorem holds. Theorem 4.3. Consider the multivariate linear regression model (4.1) under the assumption ORT o...
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A 1 − ≥ implies that S Note that it follows from Weyl’s inequality that | > τ λj| ≤ } ⊂ { ¯ Next define the oracle Θ = ˆ λj − . 3τ } | S λj ujvj⊤ and note that and Sc λj| ⊂ { j : j : | j λj| ≤ k F kop ≤ τ . It . A ˆΘsvt k Θ∗ 2 F k − ∈ ˆ2 Θsvt k P ≤ ¯ Θ k − 2 F + 2 ¯ Θ k − Θ∗ 2 F k (4.4) Using Cauchy-Schwarz, we control ...
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| ∈ Plugging the above two displays in (4.4), we get ˆΘsvt k − Θ∗ 2 F ≤ k 144 τ 2 + 2 λj | Sc | jX∈ S jX ∈ 2) and on Sc, Since on S, τ 2 = min(τ 2, | ˆΘsvt k λj| Θ∗ 2 F ≤ k − 432 3 min(τ 2, 2), it yields, λj| | | 2 λj| ≤ min(τ 2, j X rank(Θ∗) 2) λj | | 432 τ 2 ≤ j=1 X = 432 rank(Θ∗)τ 2 . In the next subsection, we exte...
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σ rank(Θ∗) n d ∨ (cid:16) T + log(1/δ) . (cid:17) Proof. We begin as usual by noting that Y X ˆ Θrk 2 F + 2nτ 2 rank(Θrk) ˆ k − which is equivalent to k Y − ≤ k XΘ∗ 2 kF + 2nτ 2 rank(Θ∗) , XΘ∗ X ˆ r Θ k 2 F ≤ Next, by Young’s inequality, we have ˆ E, Θ h X rk X − − k 2 k Θ∗ i − 2nτ rank(Θ ) + 2nτ rank(Θ∗) . ˆ rk 2 2 2 ...
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k − Note that rank(N ) equality, we get ≤ E, U h 2 = i Φ⊤E, N/ h Φ⊤E k ≤ k ≤ rank(N ) k 2 op Φ⊤E N 2 kF i k 2 N 1 2 k op k 2 N F k k Φ⊤E rank(Θrk) + rank(Θ∗) . 2 op ˆ k k Next, note that Lemma 4.2 yields ≤ k (cid:2) Φ⊤E 2 op ≤ nτ 2 rank(Θrk) + rank(Θ∗) . nτ 2 so that (cid:3) k ˆ k E, U h 2 i ≤ Together with (4.5), this...
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(Θ) Y XΘ 2 F k ≤ can be solved efficiently. To that end, let Y = X(X⊤X)†X⊤Y denote the orthog- onal projection of Y onto the image space of X: this is a linear operator from IRd T . By the Pythagorean theorem, we get for any Θ T into IRn IRd T , × × × ¯ ∈ k ¯ Next consider the SVD of Y: − k Y XΘ 2 F = Y Y¯ 2 k − kF + Y¯ ...
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F over matrices of rank at most k ¯ 7→ k XΘ Y ˆ ˆ squares but this is not necessary for our results. Remark 4.5. While the rank penalized estimator can be computed efficiently, it is worth pointing out that a convex relaxation for the rank penalty can also be used. The estimator by nuclear norm penalization Θ is defined t...
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empirical covariance matrix ˆΣ defined by ≻ ∈ (cid:2) (cid:3) n ˆΣ = 1 n XiX i⊤ . Using the tools of Chapter 1, we c i=1 X an prove the following result. Theorem 4.6. Let X1, . . . , Xn be n i.i.d. sub-Gaussian random vectors such that IE[XX ⊤] = Σ and X Σ subGd( k ∼ kop). Then d + log(1/δ) n ∨ ˆΣ k Σ kop . k Σ kop − r ...
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− I dkopk − Σ1/2 kop be a 1/4-net for Let Lemma 4.2 that N d − 1 such that |N | ≤ 12d. It follows from the proof of So that for any t ≥ ˆ IP Σ k (cid:0) It holds, Idkop > t − ≤ (cid:1) IP x⊤ ˆ(Σ x,y X ∈N (cid:0) Id)y > t/2 . (4.6) − (cid:1) ˆ x⊤(Σ Id)y = − n1 n i=1 X (cid:8) Using polarization, we also have (Xi⊤x)(Xi⊤y...
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2 − − IE[Z 2 ] − 1/2 , (cid:16) (cid:0) where in the last inequality, we u(cid:1)s(cid:1)(cid:3) subG d(1), we have Z+, Z (cid:17) hwarz. Next(cid:1),(cid:1)(cid:3)s ince X subG(2), and it follows from Lemma 1.12 that (cid:0) (cid:2) (cid:2) ∼ Z 2 + − IE[Z 2 +] ∼ − ∼ subE(32) , and Z 2 − − IE[Z 2 ] − ∼ subE(32) Therefo...
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17) h i ( ) . 144d exp n 2 2 t 32 t 32 t 32 ≥ 2d n 2 n (cid:16) es our proof. This conclud log(144) + log(1/δ) log(144) + log(1/δ) 2d n ∨ (cid:17) (cid:16) 2 n (4.7) (0, 1) if ∈ 1/2 (cid:17) Theorem 4.6 indicates that for fixed d, the empirical covariance matrix is a consistent estimator of Σ (in any norm as they are al...
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12) Var(X ⊤u) d = (cid:12) (cid:12) δ. . Σ k kop r (cid:16) d + log(1/δ) n d + log(1/δ) n ∨ (cid:17) with probability 1 − The above fact is useful in the Markowitz theory of portfolio section for IRd such that example [Mar52], where a portfolio of assets is a vector u |1 = 1 and the risk of a portfolio is given by the ...
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7 points in two dimension. This image has become quite popular as it shows that gene expression levels can recover the structure induced by geographic clustering. How is it possible to “compress” half a million dimensions into only two? The answer is that the data is intrinsically low dimensional. In this case, a plaus...
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matrix Σ covariance model if it is of the form ∈ IRd × d is said to satisfy the spiked Σ = θvv⊤ + Id , where θ > 0 and v ∈ S d − 1. The vector v is called the spike. 4.4. Principal component analysis 95 Courtesy of Macmillan Publishers Ltd. Used with permission. Figure 4.1. Projection onto two dimensions of 1, 387 poi...
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(Davis-Kahan sin(θ) theorem). Let Σ satisfy the spiked covari- ance model and let Σ be any PSD estimator of Σ. Let v˜ denote the largest ˜ eigenvector of Σ. Then we have ˜ εv˜ | − v 2 2 ≤ | 2 sin2 ∠(v˜, v) 8 ˜ Σ Σ 2 . θ2 k − kop ≤ min 1 ∈{± ε } (cid:0) (cid:1) 1, it holds under the spiked covariance model Proof. Note t...
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is the largest eigenvector ˜of Σ and in the last one, we used the fact that the matrix v˜v˜⊤ vv⊤ has rank at most 2. − Next, we have that v˜v˜⊤ k − vv⊤ k 2 F = 2(1 − (v⊤v˜)2) = 2 sin2(∠(v˜, v)) . Therefore, we have proved that 2 ∠ θ sin ( (v˜, v)) ˜ 2 Σ ≤ k Σ kop sin(∠(v˜, v)) , − so that sin(∠(v˜, v)) 2 ˜ θ ≤ k Σ Σ − ...
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+ log(1/δ) n ∨ (cid:17) This result justifies the use of the empirical covariance matrix Σ as a re- placement for the true covariance matrix Σ when performing PCA in low di- n, mensions, that is when d the above result is uninformative. As before, we resort to sparsity to overcome this limitation. n. In the high-dimensi...
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� and X Theorem 4.10. Let X1, . . . , n i.i.d. copies of a sub-Gaussian random subGd( vector X kop). Assume k further that Σ = θvv⊤ + Id satisfies the spiked covariance model for v such d/2. Then, the k-sparse largest eigenvector vˆ of the empirical that ≤ covariance matrix satisfies, ∈ |0 = k ∼ Σ v (cid:3) (cid:2) | min...
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, d } ⊂ { Σ(S), vˆ(S)vˆ(S)⊤ vv = Σ(S) v(S)v(S)⊤ IRd, x(S) | × | IR| × − − − − ∈ ∈ S ˆ ⊤ i h S | v⊤Σv vˆ⊤Σvˆ − ≤ k ˆ Σ(S) Σ(S) kopk − vˆ(S)vˆ(S)⊤ − v(S)v(S)⊤ k1 . Following the same steps as in the proof of Theorem 4.8, we get now that min 1 ∈{± ε } εvˆ | − v 2 2 ≤ | 2 sin2 ∠(vˆ, v) 8 θ2 ≤ S : sup S | | =2k ˆ Σ(S) k Σ(S...
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by exp n 2 ( − t 32 t 32 ∧ 2 (cid:17) ) + 2k log(144) + k log ed 2k (cid:0) (cid:1)i (cid:16) Choosing now t such that h C t ≥ r k log(ed/k) + log(1/δ) n ∨ k log(ed/k) + log(1/δ) n , for large enough C ensures that the desired bound holds with probability at least 1 δ. − 4.5. Problem set 99 4.5 PROBLEM SET Problem 4.1...
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there exists a matrix n some estimator Θ and ˆ × n matrix P such that P M = XΘ for ˆ ˆ 1 n k X ˆΘ XΘ∗ 2 kF . − σ2 rank(Θ∗) n T ) (d ∨ with probability .99. 3. Comment on the above results in light of the results obtain in Section 4.2. Problem 4.3. be the any solution to the minimization problem ˆ Consider the multivari...
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cid:76)(cid:82)(cid:81)(cid:68)(cid:79)(cid:3)(cid:54)(cid:87)(cid:68)(cid:87)(cid:76)(cid:86)(cid:87)(cid:76)(cid:70)(cid:86) Spring 2015 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms(cid:17)
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MIT OpenCourseWare http://ocw.mit.edu 18.969 Topics in Geometry: Mirror Symmetry Spring 2009 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms. MIRROR SYMMETRY: LECTURE 8 DENIS AUROUX Last time: 18.06 Linear Algebra. Today: 18.02 Multivariable Calculus. / 18.04 C...
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� = 1, � If ei is a basis of H 2( ˇ X, Z), ei in the K¨ complexified K¨ahler moduli space: if [B + iω] = � e∗ i B + iω. ahler cone, we obtain coordinates on the tˇiei, let ˇqi = exp(2πitˇi), tˇi = � Example. Returning to our example, ˇq = exp(2πi T 2 B + iω). Conjecture 1 (Mirror Symmetry). Let f : X → (D∗)S ...
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(since 2πiqˇi ∂qˇi ei ∈ H 1,1 etc.). �p = � �m(p) ∂ Xˇ ∂ i ∂q , Xˇ , Ω) and the RHS to a (1, 1)­ ∂ = ∂tˇi = 1 � 2 DENIS AUROUX Remark. A more grown-up version of mirror symmetry would give you an equiv- T X) with its usual product structure and H ∗( ˇX, C) alence between H ∗(X, with the quantum twisted pro...
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x0 : . . . : x4) �→ (ξax0 : x1 : . . . : x4) gives Xψ = Xξφ, so let z = (5ξ)−5 . Then z 0, i.e. ψ → ∞, gives a toric degeneration of Xψ to {x0x1x2x3x4 = 0}. This is maximally unipotent, as the monodromy on H 3 is given by → ∼ (4) ⎜ ⎜ ⎝ ⎛ 1 1 0 0 0 1 1 0 0 0 1 1 0 0 0 1 ⎞ ⎟ ⎟ ⎠ so it is LCSL. We want to compute th...
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x3 We want to extend it to z =� should be given by the root of fψ which tends to 0 as ψ → ∞. We need to show that there is only one such value (giving us a simple degeneration rather than a branched covering). Explicitly, set x3 = (ψx0x1x2)1/4y: (6) i.e. (7) fψ = 0 ⇔ x0 5 + x1 5 + x 5 2 + (ψx0x1x2)5/4 y 5 + 1 − ...
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Ωψ be the 3-form on Xψ characterized uniquely by (8) Ωψ ∧ dfψ = 5ψdx0 ∧ dx1 ∧ dx2 ∧ dx3 ∂fψ at each point of Xψ. At a point where ∂x3 and = 0, ( x0, x1, x2) are local coordinates, (9) Ωψ = 5ψdx0 ∧ dx1 ∧ dx2 ∂fψ ∂x3 = 5ψdx0 ∧ dx1 ∧ dx2 4 − 5ψx0x1x2 5x3 Defining it in terms of other coordinates, we get the sa...
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5ψdx3 fψ � dx0dx1dx2 where fψ has a unique pole at x3. The residue is precisely 5ψ (∂f /∂x3) , giving us (13) = � 5ψ T0 (∂f /∂x3) dx0dx1dx2 = � T0 Ωψ � DENIS AUROUX � T0 Ωψ = 1 2πi � (5ψ)−1(x5 0 + x5 dx0dx1dx2dx3 x0x1x2x3 dx0dx1dx2dx3 2 + x5 1 + x5 � 1 − (5ψ)−1 x0 3 + 1) − x0x1x2x3 5 + x2 5 + ...
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+ x1 + x2 + x (n!)5 . 5 5 3 1 2 4 So (14) (15) � Ωψ = −(2πi)3 ∞ � n)! (5 5(5 ψ)5n (n!) T0 In terms of z = (5ψ)−5, the period is proportional to n=0 (16) Set an = (5n)! Then (n!)5 . φ0(z) = ∞ (5n)! n � z (n!)5 n=0 (17) (n + 1)4 an+1 = � d (5n + 5)! (n!)5(n + 1) � cnzn) = Setting Θ = z dz : ...
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Electric field lines in the space surrounding a charge distribution show: PRS02 1. Directions of the forces that exist in space at all times. 2. Only directions in which static charges would accelerate when at points on those lines 3. Only directions in which moving charges would accelerate when at points on ...
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⎣ 3/ 2 ˆi ˆi E-Field of Two Equal Charges PRS02 G 1. E = ˆj 2k qs e ⎡ 2 d ⎤ ⎢ s + ⎥ 4 ⎦ ⎣ 2 3/ 2 There are a several ways to see this. For example, consider d→0. Then, G E → ke 2q ˆj 2s which is what we want (sitting above a point charge with charge 2 q) E-Field of Five Equal Charges PRS02 Six equa...
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than 1/r2 3) More slowly than 1/r2 4) Who knows? PRS02 E-Field of a Dipole (2) It falls off more rapidly We know this must be a case by thinking about what a dipole looks like from a large distance. To first order, it isn’t there (net charge is 0), so the E- Field must decrease faster. PRS02 An electric di...
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3.012 Fund of Mat Sci: Bonding – Lecture 1 bis WAVE MECHANICS Photo courtesy of Malene Thyssen, www.mtfoto.dk/malene/ 3.012 Fundamentals of Materials Science: Bonding - Nicola Marzari (MIT, Fall 2005) Last Time 1. Players: particles (protons and neutrons in the nuclei, electrons) and electromagnetic fields (photons)...
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000, p. 495, figure 14.2. 3.012 Fundamentals of Materials Science: Bonding - Nicola Marzari (MIT, Fall 2005) The total energy of the system • Kinetic energy K • Potential energy V 3.012 Fundamentals of Materials Science: Bonding - Nicola Marzari (MIT, Fall 2005) Polar Representation Diagram of the Argand plane remove...
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1926…) 3.012 Fundamentals of Materials Science: Bonding - Nicola Marzari (MIT, Fall 2005) When is a particle like a wave ? Wavelength • momentum = Planck Image of the double-slit experiment removed for copyright reasons. See the simulation at http://www.kfunigraz.ac.at/imawww/vqm/movies.html: "Samples from Visual Q...
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3.012 Fundamentals of Materials Science: Bonding - Nicola Marzari (MIT, Fall 2005) Stationary Schrödinger’s Equation (I) − 2 h 2 m 2 Ψ∇ r r * ),( ),( trVtr + Ψ r ),( tr = i h r ),( tr Ψ∂ t ∂ 3.012 Fundamentals of Materials Science: Bonding - Nicola Marzari (MIT, Fall 2005) Stationary Schrödinger’s Equation (II) − ⎡ ⎢...
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System Identification 6.435 SET 6 – Parametrized model structures – One-step predictor – Identifiability Munther A. Dahleh Lecture 6 6.435, System Identification 1 Prof. Munther A. Dahleh Models of LTI Systems • A complete model u = input y = output e = noise (with PDF). Lecture 6 6.435, System Identification 2 Prof...
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• Linear Regression (a function of past data) • Prediction error Lecture 6 6.435, System Identification 9 Prof. Munther A. Dahleh Examples …. ARMAX – ARMAX (Autoregressive moving average with exogenous input) • Description • Standard model • More general, includes ARX model structure. Lecture 6 6.435, System Identifi...
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Describes the model structure Define It follows Predictor: Lecture 6 6.435, System Identification 18 Prof. Munther A. Dahleh For a given • Notice: can be stable even though is not! • Lecture 6 6.435, System Identification 19 Prof. Munther A. Dahleh Predictor Models Def: A predictor model is a linear time-invariant ...
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. is the map is one particular model Lecture 6 6.435, System Identification 24 Prof. Munther A. Dahleh Example: ARX model structure stable Lecture 6 6.435, System Identification 25 Prof. Munther A. Dahleh General Structure Lecture 6 6.435, System Identification 26 Prof. Munther A. Dahleh You need to be differenti...
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(local or global) if it is identifiable (local or global) for all Lecture 6 6.435, System Identification 31 Prof. Munther A. Dahleh Central question II: Is the identified parameter equal to the “true parameters” ? Parametrized structure: true system Case I: Case II: Define: Let for some . If is identifiable at , ...
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Harvard-MIT Division of Health Sciences and Technology HST.951J: Medical Decision Support, Fall 2005 Instructors: Professor Lucila Ohno-Machado and Professor Staal Vinterbo 6.873/HST.951 Medical Decision Support Spring 2004 Evaluation Lucila Ohno-Machado Outline Calibration and Discrimination • AUCs • H-L stat...
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a graft lesion vessel treated ostial experience unscheduled case lab device antagonists dissection post rotablator atherectomy angiojet max pre stenosis max post stenosis no reflow Cases Women Age > 74yrs Acute MI Primary Shock Study Population Development Set 1/97-2/99 Validation Set 3/99-12/9...
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.40 0.30 0.20 0.10 0.00 LR Score aNN ROC Area 0.840 LR: Score: 0.855 aNN: 0.835 ROC = 0.50 0.00 0.20 0.40 0.60 0.80 1.00 1 - Specificity Risk Score of Death: BWH Experience Unadjusted Overall Mortality Rate = 2.1% 3000 2500 s e s a C f o r e b m u N 2000 1500 1000 500 0 Number of Cases ...
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e.g., 0.34) “True” Estimate 0.3 0.2 0.5 0.1 0 0 1 0 • In practice, classification into category 0 or 1 is based on Thresholded Results (e.g., if output or probability > 0.5 then consider “positive”) – Threshold is arbitrary threshold normal Disease True Negative (TN) True Positive (TP) FN FP 0 e...
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50 5 40 50 50 50 nl TN disease TP 0.0 FN FP 0.6 1.0 Sensitivity = 30/50 = .6 Specificity = 1 threshold nl D “nl” “D” 50 20 70 30 0 50 30 50 nl TN disease TP FN 0.0 0.7 1.0 4 . 0 d l o h s e r h T 6 . 0 d l o h s e r h T 7 . 0 d l o h s e r h T “nl” “D” 40 60 nl D 40 0 10 50 ...
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1 y t i v i t i s n e S Area under ROC: 0.5 0 1 - Specificity 1 Perfect discrimination 1 y t i v i t i s n e S 0 1 - Specificity 1 Perfect discrimination 1 y t i v i t i s n e S Area under ROC: 1 0 1 - Specificity 1 1 y t i v i t i s n e S ROC curve Area = 0.86 0 1 - Specificity 1 What ...
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8 0.2 0.5 0.7 0.9 concordant discordant concordant concordant concordant All possible pairs 0-1 Systems’ estimates for • Healthy 0.3 0.2 0.5 0.1 0.7 • Sick 0.8 0.2 0.5 0.7 0.9 concordant tie concordant concordant concordant C - index • Concordant 18 • Discordant 4 • Ties 3 C -index = Concordant...
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• “If the system is good in discrimination, calibration can be fixed” Calibration • System can reliably estimate probability of – a diagnosis – a prognosis • Probability is close to the “real” probability What is the “real” probability? • Binary events are YES/NO (0/1) i.e., probabilities are 0 or 1 for a given ...
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r e p s e u l a v d e v r e s b o f o g v A 0 Avg of estimates per group 1 Goodness-of-fit Sort systems’ estimates, group, sum, chi-square Estimated Observed 0.1 0.2 0.2 0.3 0.5 0.5 0.7 0.7 0.8 0.9 sum of group = 0.5 sum of group = 1.3 sum of group = 3.1 χ2 = Σ [(observed - estimated)2/estim...
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