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by: Johnson Kozey


Johnson Kozey
GPA 3.5


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This 78 page Class Notes was uploaded by Johnson Kozey on Thursday October 22, 2015. The Class Notes belongs to MS 10 at University of California Santa Barbara taught by Staff in Fall. Since its upload, it has received 20 views. For similar materials see /class/227074/ms-10-university-of-california-santa-barbara in Military Science at University of California Santa Barbara.


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Date Created: 10/22/15
UCSB ECON 134A Introductory Finance Lecture 10 Risk and Return The CAPM Corresponds to Ch 10 in textbook Spring quarter 2009 Instructor Ragnar Arnason Portfolios A set collection of individual assets is called Portfolio We are interested in the Expected Return and Variance andor Standard Deviation of portfolios Turns out that the Covariance and Correlation between assets in the portfolio are important Expected value ofx Va rxEx Ex2 I W 972 moo x Standard deviation SDx tVarOc Covariance ofx and y covxy y Eo N N xn aoon y N 1 Note covxy measures realationship between x and y covxygt0 3 positive relationship covxylt0 gt negative relationship Note covxyecgt cgt0 Correlation between X and y calX y oovx y 9 var x var y SDX SD Note corxye 1 1 corxy0 gt no relationship between x and y corxygtO 3 positive relationship corxyltO 3 negative relationship corxy1gt exact linear positive relationship corxy1gt exact linear negative relationship How to calculate these things Examples Con den Two risky assets stocks and bonds Three possible states of the world recession normal and boom There is a 13 chance of each state of the world Rate of Return Stoc Fund Bond nd Expected Return Stock Fund Bond Fund Rate of Squared Rate of Squared Return Deviation Return Deviation Recession 7 00324 17 00100 Normal 12 00001 7 00000 Boom 28 00289 3 00100 Expected return 1100 700 Variance 00205 00067 Standard Deviation 143 82 Expected Return Stock Fund Bond Fund Rate of Squared Rate of Squared Scenario Return Deviation Return Dev39atio Ens gtlt 7 12 g 28 ErS11 Variance Stock Fund Bond Fund Rate of Squared Rate of Squared Scenario Return Deviation Return Deviation 7 112 0324 Variance Stock Fund Bond Fund Rate of Squared Rate of Squared Scenario Return Deviation Return Deviation 0289 0205 Standard Deviation Stock Fund Bond Fund Rate of Squared Rate of Squared Scenario Return Deviation Return Deviation 143 x00205 Cova ance 001813 Correlation C0vS B C0709 C0rS B W 0998 Form a portfolio Arbitrarily select 50 stocks 50 bonds Expected return and variance ofthe Portfolio Rate of Return ocun out n Portfolio squared deviation 50 00016 95 00000 125 00012 90 00010 How did we get these numbers Rate of Return Scenario ndfund Portfolio squared deviation gt a 50 00016 i 95 00000 39 39 39 39 125 00012 The expected rate of return on the portfolio is the weighted average of the expected returns on the securities in the portfolio E032 WBEFBWSEFS 50X1150gtlt7 9 Portfolios Rate of Return Scenario Stockfund Bondfund Portfolio squared deviation 50 00016 95 00000 125 00012 i 039 0 067 00010 3 00 ESta2121165115d I atib f 14 31 3 The variance of the rate of return in the portfolio is varP 2 w w coves B 25 00205 25 00067 205 05 0117 000095 Crucial Observation The risk ofthe portfolio as measured by the varP or SDP is less than for either asset The expected return however is between the two individual returns These are the benefits of asset diversification This is the fundamental insight of asset portfolio theory General theory Assets i12l r return of asseti w weight of asset i in Portfolio W wi r i1 If equal weights gt EU Variance and Standard deviation much more complicated Case 1 2 assets VarltPgt2gt 2 W1 W2 covltr1r2gt SDP JVarU Case 2 N assets VarP 2 WT Qw Var r1 C 0121 1 r2 C0vr1rN w1 C 012r2 r1 Varr2 w2 VarP I C0vrNr1 VarrN wN SDP JVarP For any two assets ram w comm 30m 304 SDP S w1 SDr1 w2 SDr2 w1 SDr1 w2 SDI 2 only if corr1r21 So combining assets with less than perfect correlation in a portfolio leads to reduction in standard deviation ie risk The efficient set Consider A certain amount to invest A portfolio of two assets Calculate the expected return and standard deviation for all combination of these assets Plot these combinations in EPSDPspace The resulting curve is the efficient set Example The efficient set I39 Efficient I50 I a 39 frontier E 50 E 5 40 G K E 5 30 2 1 20 m 5 3 10 ii 00 000 020 040 060 080 100 120 140 Standard deviatinn of partfolin 160 The efficient set depends 0 C rx39yp AA return bonds If p 10 no risk reduction is possible If p 10 complete risk reduction is possible Illustrative calculations EXCEL The Efficient Set for Many Securities Portfolios may consist of many assets We can still calculate a set of risk returns of various combinations of assets The calculations however can become quite demanding The number of variance terms number of assets The number of covariance terms N2N Then have to calculate a huge number of asset combinations Number X 1 X number of intervals eg 10 But feasible with modern computers some freeware available for this write your own programs The efficient set is now dense notjust one curve return Individual Assets L r 0P The efficient frontier still exists the upper boundary of the set return Individual Assets L Minimum Variance ortfolio r 0P Diversification and Portfolio Risk Diversification can substantially reduce the variability of returns without an equivalent reduction in expected returns 0 This reduction in risk arises because large deviations in the returns of one asset compared to the expected tend to be offset by lesser deviations from another 0 This happens unless the assets are perfectly correlated cor1 Overview of Neuronal Dynamics Basic Features of Nervous System Organization Bipolar Mitral Pyramidal Motor cell cell cell neuron retina olfactory bulb cortex spinal cord Cephalic ganglia Longitudinal nerve cord Transverse nerve cord Tentacles Base nerve mt c elegans in hydra 302 neurons 32 chemosensory 7000 synaptic connections 1011 neurons 1015 synapses extracellular action potential 100 V 20 msec Intracellular Recording 2 I I I m I 139 I 1 qlmnwjlumnbrimu inlmuullulazr ummuullulelr zlu1ivzm puEm iuls 2D m d 2 sec mailing pnmnlial l39 30 111 l 111quot Fast Spiking Interneuron Response to Current Injection Brain Slice Intracellular Recording encoding mean ring rate 1 HAT nt ffE i El t tld1 6131M JI In H linglhur 1quotIquot391H39 WW 391 Hiquot 39 Thalamocortical Neurons Have Two Firing Modes Neuronal Dynamres elecmcal and enernres1 srgnals bropnysres ofsmgle neurons interacuons between neurons Encodmg how are sensory sumuh motor ntxol com s memory reward rn d represented m these srgnals Neural Campmnonn cadmg ayrnnnes nnnnnnnnnnnn Us Eye Fixation Head Direction Cells Neural Signal action potential Encoding mean firing rate encodes motor command eye position or computed head direction in laboratory frame of reference Dynamics Persistent Neural ActiVity S W Frequency 5 O W E N A New Form of Neural Dynamics During Acoustic Sequence Production zebra nch Mu f 4 339 Jr1 Uh LIE IFI39IE39 SI Bird Song Production Neural signal action potential Encoding bursts of action potentials signify articulator positions Dynamics burst sequence with synchronized transitions Tnmm 255 xespunse Intrinslcally bursting neuron LimaX Procerebral Lobe TRAVELING WAVE OF NEURAL ACTIVITY BANDS OF LABELED NEURONS AFTER LEARNING RIGHT PC LOBE Central Pattern Generators Olfactory Systems Neural Signal action potential subthreshold membrane potential Encoding CPGs muscle tension for rhythmic contractions olfaction Dynamics Oscillation 7 Hz 100 Hz 15 How to measure the openloop gain 0 V 10 ki How to measure the offset voltage of 10 ki 10 ki D Md How to measure the CMRR of highgain am 1 ki 1 Mi Md wa lMi f highgain ampli ers In the circuit to the right apply an AC input signal as indicated and measure the ampli er output voltage and the voltage at the junction of the two 10 k9 resistors call this VX The ampli er gain is then 101 times VoutVX You may want to reduce all the resistor values by a factor of 10 from the values given Note also this or some similar network needs to be in place when you are doing general testing opamps dont bias up correctly unless feedback is applied highgain ampli ers The DC output voltage will be about 200 times the input offset voltage pli ers Ifthe resistors are wellmatched take the time to nd resistors accurate to 1 or better before testing then the circuit gain will be about 2000ACM AD vvv IAIA HIV Examples of OP AMP Circuits A very simple one but still with good performance Costantcurrent sources are used for both the differential stage and the following commonemitter stage The circuit could be easily quot ippedquot by swapping PNPs for NPNs and Viceversa Here we have simplified the circuit by eliminating the constantcurrent loads This will reduce the gain and make the amplifier harder to bias right Note also that the constantcurrent source mirror for the differential pair uses emitter degeneration resistors to increase its output impedance This ampli er is considerably more complex The commonemitter stage now uses a darlington which increases its input impedance thereby increasing the voltage gain of the first differential stage The ouput stage also uses a darlington thereby increasing the load impedance on the commonemitter 39 stage increasing its gain The diode stack used for setting the output stage bias can be replaced as indicated by a quotVbe multiplierquot which could be adjustable by making the resistor quot3Rquot a trimpot This illustrates some more tricks Emitter followers are used in the current mirrors to eliminate basecurrent errors and a Darlington input stage is used to increase the ampli er differential input impedance and to decrease the amplifier input bias current base current Reading Capacitor Codes Large capacitors have the value printed plainly on them such as 10uF Ten Micro Farads but smaller disk types along with plastic lm types often have just two or three numbers on them For example the biasdecoupling capacitors that we use in ECE 2 are often 01uF ceramic capacitors and those are marked quot104Kquot What does that mean The rst ke point is that the unit of capacitance is usually taken to be pico Farads 1039 2 F The three numbers represent a code that is somewhat similar to the resistor code The first two are the 1St and 2nd signi cant digits and the third is a multiplier Most of the time the last digit tells you how many zeros to write after the first two digits but the standard EIA standard RS198 has some peculiarities Multiplier this times the Third digit rst two digits gives you the value in PicoFarads 0 1 1 10 2 100 3 1000 4 10000 5 100000 6 not used 7 not used 8 01 9 1 So the quot104Kquot device means 104 zeroes in pF so 100000pF which is the same as 100nF or 01 uF Similarly a capacitor marked 103 means a 10000pF capacitor or 001 uF The last letter is a tolerance code as shown in the table below 80 quotKquot means 10 Prof R York Gammaray Bursts from the Birth of Magnetars Eliot Quataert UC Berkeley Collaborators Niccolo Bucclan ni Brian Home Todd Thompson Jon Arena Phil Chang Overview Diversity onoung Neutron Stars 1 amp B Out ows from Magnetized Rotating Proto NSs What sets M E P in the rst 10 sec A Magnetar in a Bottle interaction w the host star What ls the origln of the observed collimation of GRBs Emphasize Central Engine Physics not GRB Phenomenology Goal E 10 ergs in collimated l 1103 material in 30 s a SN The Diversity of Neutron Stars Thermal Emission RotationPowered Pulsars if T5 397 39quot 39 39 is Crab P33 msB4x101ZG 1 quot my x Dec 04 Flare From Sgr180620 1 39 Magnetars Strong B Fields 3 101440156 2 35255 l inferred for some NSs Chandra Xra spindown Typlcal pulsars have flares B 10111013G quiescentflux P 30 ms 1s 10 of NSs Origin of NS Diversity B Q Presumably tied to diversity in progenitors SN explosions Magnetars predicted by Thompson amp Duncan NS magnetic fields generatedmodi ed by dynamo during the birth of the NS Duncan amp Thumpsun 1992 Planetary Dynamo Sims Slow rotation P gtgt 1cm quot15 BDipole 1012 G Rapid rotation P 1cm ms B 1015 G Dipole Current Magnetars P 510 sec Spindown in 10 yrs or less 0 E g m i m m 9 gt o m x m o o m Progenitor quot7EI391IIIOI39 15 M o 39 Progenmtcn 12x107 wsl Collapses ml Ban we amp Shack neutrino F arma ou coolingv 39 Explosion 2 1 sec mmimmim W 1 0 Egg quotWHITE DWARFquot Ml wy m RotationallyDriven SN Erot w 2 X 1052PIS2 ergs E A c S C a 1mm Thompson Quataert amp Burrows 2005 Burrows et aIIZO O7 uniform strong seed eld ala Leblanc ampWilson 1970 Early Spindown of Magnetars o Vacuum Dipole pulsars E N 1049Pn lng5 ergs 81 SpindownTime TE 5 1 P13183132 hrs 0 Suggested as possible oean engine for GRBs cg Uoov 1992 Thompson 1994 During KelvinHelmholtz cooling phase of a young NS rst 10 sec strong mass loss from neutrino driven wind modi es spindown amp sets how relativistic the out ow is not a vacuum problem GR Aligned Dipole Stellar Spindown Sims Strong Bfield forces matter to corot eite extracts much more angular momentum Cotdtation lasts until the Alfven Point 5 RA B 1 2 7 U 87139 2 RA 3 J iMRiQ gtgt iMRfvs Light Cylinder RAlt cQw 68 km P ms 00 05 Buccian tini et al 2006 P 1 ms B 1 015 G Equatorial Wind Solutions 1D neutrinoheated MHD Winds nonrel vy 39 microphysics Given 8 0 LV Solve for E Strong dependence on 9 because of centrifugal flinging cent expansion 4000 6000 of NS atmosphere 95 1 Metzger et ai 2007 B31D14G ThermallyDriven Enhanced Mass Loss As PNS cools outflow evolves from NR to Rel with 6 gt1 Enhanced Early SpindoWn NR Wind E X M li3 00 05 Buccian tini et al 2006 P 1 ms B 1015 G a 2i Enhanced Early SpindoWn NR Wind E X M li3 Rv Lastv Closed Field Line 00 05 Buccian tini et al 2006 P 1 ms B 1015 G a 2i Enhanced Early SpindoWn NR Wind E X M li3 Rv Lastv Closed Field Line Vacuum Dipole Rv RL Light Cylinder Sims More Open Field Lines Ry 0305 R1 for 0 0120 7 RL 2 E B2P 4 7 X RY 00 05 Buccian tini et al 2006 P 1 ms Bz1015 G a 2i Evolutionary Calculations o P 12 ms Magnetar naturally produces an outflow with 10513952 ergs amp 0 1103 in 1030 sec as required for RBs dEdt 1051 ergs 5391 1000 W A A A A A I I A l 39 m P 1 ms TIme Slnce Core Collapse 5 initial Metzger et al 2007 suf cient energy may be extracted in few sec to modify nucleosynthesis in the outgoing SN shock a at light cylinder l 10 Time Since Core Collapse s Collimated Outflows from GRBs EiSO 105354 ergs gammarays Early afterglow observatiorjsprovided evidence for jet breaks indicative of beaming mm 2 g i Fraii 21 ai Latetime radioobservations probe total energy in late time SedovTaylor phrase E 105 52 ergs irvi m z EM e a i ae39mmrai 2n Ideal Relativistic Winds do not Efficiently SelfCollimate Dip magnetic eld Rel Winds Little Collimation Energy Flux Primarily Equatorial a 4 F51msB1Dl Ga20 Bipolar Morphology of the Crab on Large amp Small Scales A Magnetar in a Bottle Magnetar Inflates a Bubble of Rel Plasma and Magnetic Field Behind the Outgoing SN Shock Nebula reaches MagnetoHydro Equil SN Shock A Magnetar in a Bottle V ltlt C Magnetar Inflates a Bubble of Rel Plasma and Magnetic Field Behind the Outgoing SN Shock Nebula reaches MagnetoHydro Equil SN Shock 3 4 5 Cylindrical RadiusScale Height H Based on Begelman amp Li 1992 stronger B gt H i gt Higher Pressure on Axis 2D axisymmetric thin shell calcs Et0tt from evolutionary calcs 35 M5un progenitor T 390 Mn 2 Hyperenergetic SN 2D axisymmetric thin shell calcs 35 Msun progenitor E 2D axisymmetric thin shell calcs Et0tt from evolutionary calcs 35 Msun progenitor O 2 4 6 810 RUDBCW 2D axisymmetric thin shell calcs Etot 1 from ew31uticgtnary ca 35 Msun progenitor Tm105 d Hyp renergetit SN M109 cm 02496810 02468 We Uquot Mmgun l 1 2 3 F2009 Cm Emag 1 39 Emag i Emag 7 4 Etm Etm Etm 1037 Our integral over the mass reduces to an area integral in the case of a laminar 1 17 dA7gt dz dy 0 0 We know that I 1y 0 since the mass is all at 2 07 also for the case of a lamina7 I I Iyy and the symmetry of this triangle means that I my 07 1 17x 2 a I d d 72 m aO zO yr 12 1 17 Iggy7a dz dyzy777l 0 0 a 24 Writing out the inertia tensor I 71 2 0 0 Now7 we want to nd the principal axes and corresponding moments of inertial To nd the moments we solve the secular equation detI 7 A1 0 and then use the moments to nd the associated principal axis detI71 27A2747 47A 47A3742 47A37A17A 0 7A134 Plugging these values of A into the secular equation gives us the equations de ning the associated principal vectorsi 17m just going to write them down 1761 110 3 Has 1710 4 764 001 1040 First let s write down Eulerls equations for zero external torque E1 A1031 7 A2 7 A3w2w3 0 EgAgw39g 7 A3 7 A wgwl 0 EgAgw39g 7 A1 7 A2w1w2 0 Now if we sum over those equations with an additional factor of Maui as sug gested 0 ZAiWiEi P3011051 A W2OJ2 gwswsl i w1w2w3 w1w2 0J3 W2 0J3 W1 w3w1 4 2 E Aiwiwi 139 Note this did not demonstrate that L is constant but that its magnitude is constant The procedure with the kinetic energy is much the same except we multiply Ei by Lulu 0 EwiEi ZAiwiw39i d aT Remember when considering vectors A B 0 does not imply one of the vectors are zero but that they are perpendicular 1044 We begin again with Eulerls equations however in this case there is an external torque in the 2 direction and A1 A2 E A Nil 7 A 7 A3w2wg 0 Aug 7 A3 7 Aw3w1 0 Agwg F The 2 equation is uncoupled from the other directions we can immediately solve it by integrating and including the initial condition F wg t 7t wgo A3 Now we insert this equation into the ml and mg equationsi As in the book we will use the transformation to a complex variable containing both terms 77 wl iwgi In terms of this variable the rst two equations become A 7A z 3 w3tn This is easily solved however note that M3 depends on t We must integrate the factor on the RHS to get the exponent for the exponential solution A 7A F 7775 MOBIL7 3A 04075 it2gtl 2A3 So converting from the complex 77 back to W1w2 we can write our w vector A 7A T h wt M10608 3A ltw30t t2gtJ x 37A 75 Lt quot A w30 2A3 y 375 2 A3 W30 A w103m 115 In this problem we consider the twocart threespring problem with the right most spring removed This gives us the following K 2k 7k K lt7k k gt So following our normal procedure we consider the secular equation detk 7 wQM 2k 7 m2k 7 m2 7 k2 k2 73k7mu2mw22 0 3 i J3 k k k 2 if m 162 27 62 27 w 2 m lt gt mlt gt m ow we insert our two normal mode frequencies into the matrix and solve for the associated vector normal mode mgi 2 m 14 7k K7w2Mlt 3k 714 2 5 7 1 w 3 17 L 2 This is a mode in which the second mass oscillates out of phase with the rst mass and at a smaller amplitude The second mode w33 z 2 m hu k 7k 7 2 7 2 K wM7lt 7k 712 Ek a1 In this mode the masses oscillate in phase and the second mass has a larger amplitude of oscillation than the rst 119 The equations of motion for this case are as follow k k k k 11 2711I2I2 2I2Ii m m m m Now we want to consider the normal coordinates 51 12 11 12 and 51 12 11 7 12 Using these de nitions we can write down the equations of motion for 51 2 l l k k k k 51 11I2 27114r712 27124r 11 2 2 m m m m 1lt k k gt k 77117712 751 2 m m m 1 1 k k k k 52 11 i 12 i 2711 712 2712 11 2 2 m m m m 1 k k k 77 lt73711 7 3712 73752 2 m m m As we can see in the normal coordinates the equations are uncoupled and recognizable as simple harmonic motion SHO We can immediately write down the solution for the normal coordinates and from that construct the solution for 1112 11t 51t 52t A1003 gt 61gt A2003 gt 62 12t 5175 7 5275 A1003 gt 51gt 7 A2003 gt 62 The unknowns above A1A2 6162 would all be xed by the initial condi tions


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