# 天文学入门|PHYS1160 Introduction to Astronomy代写 unsw

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Is there life beyond the Earth? How common might life be? Millions of dollars are spent by the Australian government and other countries each year on space exploration, so what do we actually learn from modern day telescopes and satellites? In this course, students will explore an introduction to the Universe and the study of astronomy.

Earth has an orbital period of 365 days and its mean distance from the Sun is $1.495 \times 10^{8} \mathrm{~km}$. The planet Pluto’s mean distance from the Sun is $5.896 \times 10^{9} \mathrm{~km}$. Using Kepler’s third law, calculate Pluto’s orbital period in Earth days.

\begin{aligned}
&T_{E}=365 \text { days }\
&r_{E}=1.495 \times 10^{8} \mathrm{~km}\
&r_{P}=5.896 \times 10^{9} \mathrm{~km}\
&T_{P}=\text { ? }\
&\left(\frac{T_{E}}{T_{P}}\right)^{2}=\left(\frac{r_{E}}{r_{P}}\right)^{3}\
&\left(\frac{365 \text { days }}{T_{P}}\right)^{2}=\left(\frac{1.495 \times 10^{8} \mathrm{~km}}{5.896 \times 10^{9} \mathrm{~km}}\right)^{3}\
&\left(\frac{365 \text { days }}{T_{P}}\right)^{2}=\left(2.54 \times 10^{-2}\right)^{3}\
&\left(\frac{1.32 \times 10^{5} \text { days }^{2}}{T_{P}{ }^{2}}\right)=1.63 \times 10^{-5}\
&T_{P}=\sqrt{\frac{1.32 \times 10^{5} \text { days }^{2}}{1.63 \times 10^{-5}}}\
&T_{P}=9.00 \times 10^{4} \text { days }
\end{aligned}

## PHYS1160 COURSE NOTES ：

A spherical planet orbits an F-v star with luminosity $1.5 \times 10^{34}$ erg $^{-1}$. The orbit is circular with a radius from the star of $2 \mathrm{AU}$. Assume that the planet is rapidly rotating, has an atmosphere and reflects $20 \%$ of the light that falls on it, but absorbs the other $80 \%$ and, in the assumed steady state, radiates it as a blackbody.
(a) Ignoring any greenhouse effect, what is the temperature of the planet?
Solution:
The solution to the planetary tempearture comes from assuming a state of “balanced power”. The energy received per second by the planet from its host star, $\dot{E}{\text {in }}$ is balanced by the energy it radiates per second as a blackbody, $\dot{E}{\text {out }}$. If this were not so, then the temperature of the planet would rise or fall until the balance was achieved. This would take something like the heat capacity of the atmosphere divided by $E_{\text {in }}$ or in the case of the earth a few weeks.

The energy in is the cross sectional area of the planet as viewed from the star, a disk with radius $R_{p}$ intersects the radiation that would have passed througha disk of area $\pi R_{p}^{2}$ (not $2 \pi R_{p}^{2}$, then one would have to integrate $\cos \theta$ for the incident radiation, which adds work). The flux passing through each square $\mathrm{cm}$ of the disk is $L_{} /\left(4 \pi d^{2}\right)$ where $d$ is the distance from the star to the planet. Additionally it was specified that only $80 \%$ of the flux gets through to the planets surface and contributes to its warming, so $$\dot{E}{\text {in }}=0.8 \frac{L{}}{4 \pi d^{2}}\left(\pi R_{p}^{2}\right)$$

# 物理学1A (航空)|PHYS1149 Physics 1A (Aviation)代写 unsw

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This is an introductory level course in physics for students from all disciplines. The course has both a laboratory and theoretical component. Topics covered include the description of motion; forces and momentum; the dynamics of particles; kinetic and potential energy; the conservation of energy; oscillations and simple harmonic motion

An aircraft is equipped with a wing of symmetrical airfoils. The lift curve slope of the total aircraft is estimated to be $\frac{\partial C_{L}}{\partial \alpha}=0.8 \cdot 2 \pi \frac{1}{\mathrm{rad}}$. The stall angle of attack (AOA) is $12^{\circ}$. Wing area is $16 \mathrm{~m}^{2}$. Use $g=g_{0}$ and $\rho=\rho_{0}$.
What is the aircraft’s mass during a flight on which a stall speed of $50 \mathrm{kt}$ was observed.

$$\begin{array}{ll} \text { Given: } \quad & \frac{\partial C_{L}}{\partial \alpha}=5.0265 \frac{1}{\mathrm{rad}} \quad \alpha_{\max }=12^{\circ}=0.20944 \mathrm{rad} \ & V=50 \mathrm{kt}=92.6 \mathrm{~km} / \mathrm{h}=25.72 \mathrm{~m} / \mathrm{s} \ & S_{W}=16 \mathrm{~m}^{2} \end{array}$$
$$C_{L, \max }=\frac{\partial C_{L}}{\partial \alpha} \cdot \alpha_{\max }=5.0265 \frac{1}{\mathrm{rad}} \cdot 0.20944 \mathrm{rad}=1.05275$$
$$L=\frac{1}{2} \rho V^{2} \cdot C_{L, \max } \cdot S_{W}=\frac{1}{2} \rho_{0} V^{2} \cdot C_{L, \max } \cdot S_{W}=W=m \cdot g=m \cdot g_{0}$$
$$m=\frac{\rho_{0} \cdot V^{2}}{2 g_{0}} \cdot C_{L, \max } \cdot S_{W}=\frac{1.225 \mathrm{~kg} \cdot \mathrm{s}^{2} \cdot 25.72^{2} \cdot \mathrm{m}^{2}}{\mathrm{~m}^{3} \cdot 2 \cdot 9.80665 \cdot \mathrm{m} \cdot \mathrm{s}^{2}} \cdot 1.05275 \cdot 16 \mathrm{~m}=696 \mathrm{~kg}$$
Answer: The aircraft’s mass is $696 \mathrm{~kg}$.

## PHYS1131 COURSE NOTES ：

The category of effect of a failure is judged to be hazardous. Following $A C J N o$. 1 . to $J A R$ $25.1309$
a) What is the largest permissible failure probability?
b) What is the mean time to failure MTTF?
Solution
$\begin{array}{ll}F(t) & \text { probability of failure, } \ \lambda & \text { failure rate } \ \text { MTTF } & \text { mean time to failure } \ \text { FH } & \text { flight hour }\end{array}$
a) hazardous: $F(t=1 \mathrm{FH}) \leq 10^{-7}$
b) For small probabilities of failure: $\lambda \approx F / t=10^{-7} \cdot \frac{1}{\mathrm{FH}}$
$$\mathrm{MTTF}=1 / \lambda=\frac{1}{10^{-7}} \mathrm{FH}=10000000 \mathrm{FH}$$
Answer: If a failure has a hazardous effect, the mean time to this failure may not be less than $10000000 \mathrm{FH}$

# 高级物理学1B|PHYS1231 Higher Physics 1B代写 unsw

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This is the second of the two introductory courses in Physics. It is a calculus based course. The course is examined at two levels, with Higher Physics 1B being the higher of the two levels. While the same content is covered as Physics 1B, Higher Physics 1B features more advanced assessment, including separate tutorial and laboratory

\begin{prob}

A car started moving from rest with a constant acceleration. At some moment of time, it covered the distance $x$ and reached the speed $v$. Find the acceleration and the time.

Solution. The formulas for the motion with constant acceleration read
$$v=a t, \quad x=\frac{1}{2} a t^{2},$$
where we have taken into account that the motion starts from rest (all initial values are zero). If $v$ and $x$ are given, this is a system of two equations with the unknowns $a$ and $t$. This system of equations can be solved in different ways.

For instance, one can express the time from the first equation, $t=v / a$, and substitute it to the second equation,
$$x=\frac{1}{2} a\left(\frac{v}{a}\right)^{2}=\frac{v^{2}}{2 a} .$$
From this single equation for $a$ one finds
$$a=\frac{v^{2}}{2 x}$$
Also, one can relate $x$ to $v$ as follows
$$x=\frac{1}{2} a t \times t=\frac{1}{2} v t .$$
After that one finds
$$t=\frac{2 x}{v}$$
and, further,
$$a=\frac{v}{t}=\frac{v}{2 x / v}=\frac{v^{2}}{2 x}$$

## PHYS1131 COURSE NOTES ：

A missile launched from a cannon with the initial speed $v_{0}$ targets an object at the linear distance $d$ from the cannon and at the height $h$ with respect to the cannon. Investigate the possibility of hitting the object and the targeting angles.
Solution. The formula for the motion of the missile has the form (motion with constant acceleration)
$$z=v_{0 z} t-\frac{1}{2} g t^{2}, \quad x=v_{0 x} t .$$
The instance of these general formulas corresponding to hitting the target is
$$h=v_{0 z} t_{f}-\frac{1}{2} g t_{f}^{2}, \quad d=v_{0 x} t_{f} .$$
From the first equation one finds $t_{f}$ as in the preceding problem,
$$t_{f}=\frac{1}{g}\left(v_{0 z} \pm \sqrt{v_{0 z}^{2}-2 g h}\right) .$$

# 高级物理学1A|PHYS1131 Physics 1B代写 unsw

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This is the second of the two introductory courses in Physics. It is a calculus based course. The course is examined at two levels, with Physics 1A being the lower of the two levels.Electricity and Magnetism: electrostatics, Gauss’s law, electric potential, capacitance and dielectrics, magnetic fields and magnetism, Ampere’s and Biot-Savart law, Faraday’s law, induction and inductance. Physical Optics: light, interference, diffraction, gratings and spectra, polarization. Introductory quantum theory and the wave nature of matter. Introductory solid state and semiconductor physics: simple energy band picture.

\begin{prob}

(a) Consider an arbitrary electrostatic field configuration. A small test charge is placed at a null point (i.e., where $E=0$ ) of the configuration. Show that the equilibrium of the test charge is necessarily unstable.
(b) Verify this result for the simple configuration of two charges of the same magnitude and sign placed a certain distance apart.

(a) Let the equilibrium of the test charge be stable. If test charge is in equilibrium and it is displaced from its position in any direction, then it experiences a force towards a null point, where the electric field is zero. All the field lines near the null point are directed inwards and towards the null point. There is a net inward flux through a closed surface around the null point. According to Gauss’s law, the flux of electric field through a surface, which is not enclosing any charge, is zero. Hence, the equilibrium of the test charge can be stable.
(b) Two charges of same magnitude and same sign are placed at some distance. The mid-point of the joining line of the two charges is the null point. When a test charged is displaced along the line joining charges, it experiences a restoring force. If it is displaced normal to the joining line of charges, then the net force takes it away from the null point. Hence, the charge is not stable because stability of equilibrium requires restoring force in all directions.

## PHYS1131 COURSE NOTES ：

What is the force between two small charged sphere having charges of $2 \times$ $10^{-7} \mathrm{C}$ and $3 \times 10^{-7} \mathrm{C}$ placed $30 \mathrm{~cm}$ apart in air?
Solution:
Given:
Charge on the first sphere, $\mathrm{Q}{1}=2 \times 10^{-7} \mathrm{C}$ Charge on the first sphere, $Q{2}=3 \times 10^{-7} \mathrm{C}$
Distance between the spheres is, $r=30 \mathrm{~cm}, \mathrm{r}=0.3 \mathrm{~m}$
The electrostatic force between two charges is given by the formula, $\mathrm{F}=\frac{\mathrm{kQ}{1} \mathrm{Q}{2}}{\mathrm{r}^{2}}$
Where $\mathrm{k}=\frac{1}{4 \pi \varepsilon}=9 \times 10^{9}$
$$\mathrm{F}=\frac{9 \times 10^{9} \times 2 \times 10^{-7} \times 3 \times 10^{-7}}{(0.3)^{2}}=6 \times 10^{-3} \mathrm{~N}$$
Hence, the force between the spheres is $6 \times 10^{-3} \mathrm{~N}$. Charges are of the same polarity so force between them will be repulsive in nature.

# 高级物理学1A|PHYS1131 Higher Physics 1A代写 unsw

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This course provides an introduction to Physics. It is a calculus based
course. The course is examined at two levels, with Higher Physics 1A
being the higher of the two levels. While the same content is covered as
Physics 1A, Higher Physics 1A features more advanced assessment.

\begin{prob}

As an oil well is drilled, each new section of drill pipe supports its own weight and that
of the pipe and drill bit beneath it. Calculate the stretch in a new 6.00 m length of steel
pipe that supports 3.00 km of pipe having a mass of 20.0 kg/m and a 100-kg drill bit.
The pipe is equivalent in strength to a solid cylinder 5.00 cm in diameter.

Use the equation $\Delta L=\frac{1}{Y} \frac{F}{A} L_{0}$, where $L_{0}=6.00 \mathrm{~m}, Y=1.6 \times 10^{10} \mathrm{~N} / \mathrm{m}^{2}$. To calculate the mass supported by the pipe, we need to add the mass of the new pipe to the mass of the $3.00 \mathrm{~km}$ piece of pipe and the mass of the drill bit:
\begin{aligned} &m=m_{\mathrm{p}}+m_{3 \mathrm{~km}}+m_{\text {bit }} \ &=(6.00 \mathrm{~m})(20.0 \mathrm{~kg} / \mathrm{m})+\left(3.00 \times 10^{3} \mathrm{~m}\right)(20.0 \mathrm{~kg} / \mathrm{m})+100 \mathrm{~kg}=6.022 \times 10^{4} \mathrm{~kg} \end{aligned}
So that the force on the pipe is:
$$F=w=m g=\left(6.022 \times 10^{4} \mathrm{~kg}\right)\left(9.80 \mathrm{~m} / \mathrm{s}^{2}\right)=5.902 \times 10^{5} \mathrm{~N}$$
Finally the cross sectional area is given by: $A=\pi r^{2}=\pi\left(\frac{0.0500 \mathrm{~m}}{2}\right)^{2}=1.963 \times 10^{-3} \mathrm{~m}^{2}$

## PHYS1131 COURSE NOTES ：

Derive the equation for the vertical acceleration of a rocket

The force needed to give a small mass $\Delta m$ an acceleration $a_{\Delta m}$ is $F=\Delta m a_{\Delta m}$. To accelerate this mass in the small time interval $\Delta t$ at a speed $v_{\mathrm{e}}$ requires $v_{\mathrm{e}}=a_{\Delta m} \Delta t$, so $F=v_{e} \frac{\Delta m}{\Delta t}$. By Newton’s third law, this force is equal in magnitude to the thrust force acting on the rocket, so $F_{\text {thnust }}=v_{e} \frac{\Delta m}{\Delta t}$, where all quantities are positive. Applying Newton’s second law to the rocket gives $F_{\text {thrus }}-m g=m a \Rightarrow a=\frac{v_{e}}{m} \frac{\Delta m}{\Delta t}-g$, where $m$ is the mass of the rocket and unburnt fuel.

# 物理学1A|PHYS1121 Physics 1A代写 unsw

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This course provides an introduction to Physics. It is a calculus based
course. The course is examined at two levels, with Physics 1A being the
lower of the two levels.
Mechanics: particle kinematics in one dimension, motion in two and
three dimensions, particle dynamics, work and energy, momentum and
collisions.
Thermal physics: temperature, kinetic theory and the ideal gas, heat and
the first law of thermodynamics.
Waves: oscillations, wave motion, sound waves.

\begin{prob}

(a) Calculate the momentum of a 2000-kg elephant charging a hunter at a speed of
7.50 m/s. (b) Compare the elephant’s momentum with the momentum of a 0.0400-kg
tranquilizer dart fired at a speed of 600 m/s. (c) What is the momentum of the 90.0-
kg hunter running at 7.40 m/s after missing the elephant?

(a) $p_{\mathrm{e}}=m_{\mathrm{e}} v_{e}=2000 \mathrm{~kg} \times 7.50 \mathrm{~m} / \mathrm{s}=1.50 \times 10^{4} \mathrm{~kg} \cdot \mathrm{m} / \mathrm{s}$
(b) $p_{\mathrm{b}}=m_{\mathrm{b}} v_{\mathrm{b}}=0.0400 \mathrm{~kg} \times 600 \mathrm{~m} / \mathrm{s}=24.0 \mathrm{~kg} \cdot \mathrm{m} / \mathrm{s}$, so
$$\frac{p_{\mathrm{c}}}{p_{\mathrm{b}}}=\frac{1.50 \times 10^{4} \mathrm{~kg} \cdot \mathrm{m} / \mathrm{s}}{24.0 \mathrm{~kg} \cdot \mathrm{m} / \mathrm{s}}=625$$
The momentum of the elephant is much larger because the mass of the elephant is much larger.
(c) $p_{\mathrm{b}}=m_{\mathrm{h}} v_{\mathrm{h}}=90.0 \mathrm{~kg} \times 7.40 \mathrm{~m} / \mathrm{s}=6.66 \times 10^{2} \mathrm{~kg} \cdot \mathrm{m} / \mathrm{s}$
Again, the momentum is smaller than that of the elephant because the mass of the hunter is much smaller.

## PHYS1121 COURSE NOTES ：

A cruise ship with a mass of 1.00 10 kg 7 × strikes a pier at a speed of 0.750 m/s. It
comes to rest 6.00 m later, damaging the ship, the pier, and the tugboat captain’s
finances. Calculate the average force exerted on the pier using the concept of impulse.
(Hint: First calculate the time it took to bring the ship to rest.)

Given: $m=1.00 \times 10^{7} \mathrm{~kg}, v_{0}=0.75 \mathrm{~m} / \mathrm{s}, v=0 \mathrm{~m} / \mathrm{s}, \Delta x=6.00 \mathrm{~m}$. Find: net force on the pier. First, we need a way to express the time, $\Delta t$, in terms of known quantities.
Using the equations $\bar{v}=\frac{\Delta x}{\Delta t}$ and $\bar{v}=\frac{v_{0}+v}{2}$ gives:
$$\Delta x=\bar{v} \Delta t=\frac{1}{2}\left(v+v_{0}\right) \Delta t \text { so that } \Delta t=\frac{2 \Delta x}{v+v_{0}}=\frac{2(6.00 \mathrm{~m})}{(0+0.750) \mathrm{m} / \mathrm{s}}=16.0 \mathrm{~s} .$$
net $F=\frac{\Delta p}{\Delta t}=\frac{m\left(v-v_{0}\right)}{\Delta t}=\frac{\left(1.00 \times 10^{7} \mathrm{~kg}\right)(0-0750) \mathrm{m} / \mathrm{s}}{16.0 \mathrm{~s}}=-4.69 \times 10^{5} \mathrm{~N}$.
By Newton’s third law, the net force on the pier is $4.69 \times 10^{5} \mathrm{~N}$, in the original direction of the ship.

# 统计学专题|MATH5805 Special Topics in Statistics代写 unsw代写

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This course will cover topics in the general area of Monte Carlo methods and their application
domains. The topics include Markov chain Monte Carlo and Sequential Monte Carlo methods,
Quantum and Diffusion Monte Carlo techniques, as well as branching and interacting particle
methodologies. The lectures cover discrete and continuous time stochastic models, starting from
traditional sampling techniques (perfect simulation, Metropolis-Hasting, and Gibbs-Glauber
models) to more refined methodologies such as gradient flows diffusions on constraint state space
and Riemannian manifolds, ending with the more recent and rapidly developing Branching and
mean field type Interacting Particle Systems techniques.

The mean and the standard deviation of the score at the exam of Statistics for the past 2 cohorts of students was:

• cohort 2011-2012 (Italian; $n=171):$ mean $=25.5$, std $=3.959$
• cohort 2012-2013 (English; $\mathrm{n}=16$ ): mean $=26.4$, std $=4.11$
Compute the overall average and compare the variability.

The overall mean is a weighted average:
$$\text { Mean }=[(25.2 \cdot 171)+(26.4 \cdot 16)] /(171+16)=4782.9 / 187=25.58$$
The variability is compared by looking at the coefficient of variation:
CV for cohort 2011-2012: $25.5 / 3.959=16 \%$
CV for cohort 2011-2012: $26.6 / 4.3=16 \%$

## MATH5805 COURSE NOTES ：

Exercise
There are two urns containing coloured balls. The first urn contains 50 red balls and 50 blue balls. The second urn contains 30 red balls and 70 blue balls. One of the two urns is randomly chosen (both urns have equal probability of being chosen) and then a ball is drawn at random from that urn.
(a) What is the probability to draw a red ball?
(b) If a red ball is drawn, what is the probability that it comes from the first urn?
Solution\begin{aligned}
&\mathrm{P}(\mathrm{U} 1)=\mathrm{P}(\mathrm{U} 2)=1 / 2 \
&\mathrm{P}(\text { Red } \mid \mathrm{U} 1)=1 / 2 \
&\mathrm{P}(\text { Red } \mid \mathrm{U} 2)=3 / 10
\end{aligned}
$$a) P( Red )=P( Red|U1 ) \cdot P(U 1)+P( Red| U 2) \cdot P(U 2)=0.5 \cdot 0.5+0.3 \cdot 0.5=0.4 b) P(U 1 \mid Red )=P(U 1 \& Red ) / P( Red )=P( Red \mid U 1) \cdot P(U 1) / P( Red )=0.5 \cdot 0.5 / 0.4=0.625 # 概率和随机过程|MATH3801 Probability and Stochastic Processes代写 unsw代写 0 This course is an introduction to the theory of stochastic processes. Informally, a stochastic process is a random quantity that evolves over time, like a gambler’s net fortune and the price fluctuations of a stock on any stock exchange, for instance. The main aims of this course are: 1) to provide a thorough account of basic probability theory: 2) to introduce the ideas and tools of the theory of stochastic processes; and 3) to discuss in depth important classes of stochastic processes, including Markov Chains (both in discrete and continuous time), Poisson processes, the Brownian motion and Martingales. The course will also cover other important but less routine topies, like Markov decision processes and some elements of queucing theory. 这是一份unsw新南威尔斯大学MATH3801的成功案例 问题 1. Suppose the claim is wrong and that$$
\mathrm{P}\left[A_{\infty}=\infty, \sup {n}\left|X{n}\right|<\infty\right]>0 .
$$Then,$$
\mathrm{P}\left[T(c)=\infty ; A_{\infty}=\infty\right]>0
$$where T(c) is the stopping time$$
T(c)=\inf \left{n|| X_{n} \mid>c\right} .
$$证明 . Now$$
\mathrm{E}\left[X_{T(c) \wedge n}^{2}-A_{T(c) \wedge n}\right]=0
$$and X^{T(c)} is bounded by c+K. Thus$$
\mathrm{E}\left[A_{T(c) \wedge n}\right] \leq(c+K)^{2}
$$for all n. This is a contradiction to \mathrm{P}\left[A_{\infty}=\infty, \sup {n}\left|X{n}\right|<\infty\right]>0. ## MATH3801 COURSE NOTES ： it is seen that the rate into the box around Node 0 is \mu p_{1}; the rate out of the box around Node 0 is \lambda p_{0}; thus, “rate in” = “rate out” yields$$
\mu p_{1}=\lambda p_{0}
$$The rate into the box around Node 1 is \lambda p_{0}+\mu p_{2}; the rate out of the box around Node 1 is (\mu+\lambda) p_{1}; thus$$
\lambda p_{0}+\mu p_{2}=(\mu+\lambda) p_{1}
$$Continuing in a similar fashion and rearranging, we obtain the system$$
\begin{aligned}
p_{1} &=\frac{\lambda}{\mu} p_{0} \text { and } \
p_{n+1} &=\frac{\lambda+\mu}{\mu} p_{n}-\frac{\lambda}{\mu} p_{n-1} \text { for } n=1,2, \cdots .
\end{aligned}
$$# 随机分析入门|MATH5975 Introduction to Stochastic Analysis代写 unsw代写 0 In this course, you will learn the basic concepts and techniques of Stochastic Analysis, such as: Brownian motion, martingales, Itˆo stochastic integral, Itˆo’s formula, stochastic differential equations, equivalent change of a probability measure, integral representation of martingales with respect to a Brownian filtration, relations to second order partial differential equations, the Feynman-Kac formula, and jump processes. 这是一份unsw新南威尔斯大学MATH5975的成功案例 问题 1. Let \left(X_{n}\right){n \in \mathbb{N}} be a submartingale w.r.t. the filtration \left(\mathcal{F}{n}\right){n \in \mathbb{N}} such that$$ \forall n \in \mathbb{N}, \exists k \geq n: \mathbb{E}\left(X{k}\right) \leq \mathbb{E}\left(X_{n}\right)
$$Show that \left(X_{n}\right)_{n \in \mathbb{N}} is a martingale! 证明 . We consider the Doob composition X_{n}=M_{n}+A_{n} with M_{n} a martingale and A_{n} \geq 0 an increasing predictable process. It follows$$
\mathbb{E}\left[X_{n}\right]=\underbrace{\mathbb{E}\left[M_{n}\right]}{=\mathbb{E}\left[M{n+1}\right]}+\underbrace{\mathbb{E}\left[A_{n}\right]}{\geq \mathbb{E}\left[A{n-1}\right]} \geq \mathbb{E}\left[X_{n-1}\right]
$$Hence, the sequence of \left(\mathbb{E}\left[X_{n}\right]\right){n \in \mathbb{N}} is an increasing sequence of real numbers. On the other hand, by$$ \forall n \in \mathbb{N}, \exists k \geq n: \mathbb{E}\left(X{k}\right) \leq \mathbb{E}\left(X_{n}\right)
$$we can a extract a decreasing a subsequence \left(\mathbb{E}\left[X_{n_{k}}\right]\right){k \in \mathbb{N}}. Hence, the whole sequence \left(\mathbb{E}\left[X{n}\right]\right){n \in \mathbb{N}} must be constant. By (1), this is only possible if \mathbb{E}\left[A{n}\right]=0 for all n. Since A_{n} \geq 0, it follows A_{n}=0. Hence, X_{n}=M_{n} is a martingale. ## MATH5975 COURSE NOTES ： Exercise ：Let \tau_{1}(\omega) and \tau_{2}(\omega) be stopping times with respect to the filtration \mathbb{F}=\left(F_{1}: t \in T\right) taking values in T. Here T could be either \mathbb{R}^{+}or \mathbb{N}. Use the definition of stopping time to show that \sigma(\omega)=\min \left(\tau_{1}(\omega), \tau_{2}(\omega)\right) is a F-stopping time. Solution ：We have$$
{\sigma(\omega) \leq t}=\left{\omega: \tau_{1}(\omega) \leq t \text { or } \tau_{2}(\omega) \leq t\right}=\left{\tau_{1}(\omega) \leq t\right} \cup\left{\tau_{2}(\omega) \leq t\right} \in \mathcal{F}_{t},
$$so \sigma is a stopping time. # 高等线性模型|MATH2931 Higher Linear Models代写 unsw代写 0 Statistics is about using probability models to make decisions from data in the face of uncertainty. This course gives an introduction to the process of building statistical models using an important class of models (linear models). In a linear model we try to predict or explain variation in a response variable in terms of related quantities (predictors). The relationship between the expected response and predictors is assumed to be linear. Topics covered in the course include how to estimate parameters in linear models, how to compare models, how to select a good model or models when prediction of the response is the goal. Concepts are illustrated with applications from the natural and social sciences. 这是一份unsw新南威尔斯大学MATH2931的成功案例 问题 1. The regression coefficients \hat{\beta}{1}, \hat{\beta}{2}, \ldots, \hat{\boldsymbol{\beta}}{k} in \hat{\boldsymbol{\beta}}{1} can be standardized so as to show the effect of standardized x values (sometimes called z scores). We illustrate this for k=2. The model in centered form$$
\hat{y}{i}=\bar{y}+\hat{\beta}{1}\left(x_{i 1}-\bar{x}{1}\right)+\hat{\beta}{2}\left(x_{i 2}-\bar{x}_{2}\right)
$$证明 . This can be expressed in terms of standardized variables as$$
\frac{\hat{y}{i}-\bar{y}}{s{y}}=\frac{s_{1}}{s_{y}} \hat{\beta}{1}\left(\frac{x{i 1}-\bar{x}{1}}{s{1}}\right)+\frac{s_{2}}{s_{y}} \hat{\beta}{2}\left(\frac{x{i 2}-\bar{x}{2}}{s{2}}\right),
$$where s_{j}=\sqrt{s_{j j}} is the standard deviation of x_{j}. We thus define the standardized coefficients as$$
\hat{\beta}{j}^{*}=\frac{s{j}}{s_{y}} \hat{\beta}_{j}
$$## MATH2931 COURSE NOTES ： The hypothesis H_{0}: \alpha_{1}=\alpha_{2}=\alpha_{3} can be expressed as H_{0}: \alpha_{1}-\alpha_{2}=0 and \alpha_{1}-\alpha_{3}=0. Thus H_{0} is testable if \alpha_{1}-\alpha_{2} and \alpha_{1}-\alpha_{3} are estimable. To check \alpha_{1}-\alpha_{2} for estimability, we write it as$$
\alpha_{1}-\alpha_{2}=(0,1,-1,0,0,0) \boldsymbol{\beta}=\boldsymbol{\lambda}{1}^{\prime} \boldsymbol{\beta} $$and then note that \boldsymbol{\lambda}{1}^{\prime} can be obtained from \mathbf{X} as$$
(1,0,-1,0,0,0) \mathbf{X}=(0,1,-1,0,0,0)
$$and from \mathbf{X}^{\prime} \mathbf{X} as$$
\left(0, \frac{1}{2},-\frac{1}{2}, 0,0,0\right) \mathbf{X}^{\prime} \mathbf{X}=(0,1,-1,0,0,0)