# 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 $)=0.5 \cdot 0.5 / 0.4=0.625$