统计代写|ST227 Survival Models

MY-ASSIGNMENTEXPERT™可以为您提供 lse.ac.uk ST227 Survival Models 生存模型的代写代考和辅导服务！

这是伦敦政经学院 生存模型课程的代写成功案例。

ST227课程简介

This course is compulsory on the BSc in Actuarial Science. This course is available on the BSc in Business Mathematics and Statistics, BSc in Data Science and BSc in Mathematics, Statistics and Business. This course is available as an outside option to students on other programmes where regulations permit and to General Course students.

Students must have completed Mathematical Methods (MA100) and Elementary Statistical Theory (ST102).

“Students must have completed one of the following two combinations of courses: (a) ST102 and MA100, or (b) MA107 and ST109 and EC1C1. Equivalent combinations may be accepted at the lecturer’s discretion.”

Prerequisites 

An introduction to stochastic processes with emphasis on life history analysis and actuarial applications. Principles of modelling; model selection, calibration, and testing; Stochastic processes and their classification into different types by time space, state space, and distributional properties; construction of stochastic processes from finite-dimensional distributions, processes with independent increments, Poisson processes and renewal processes and their applications in general insurance and risk theory, Markov processes, Markov chains and their applications in life insurance and general insurance, extensions to more general intensity-driven processes, counting processes, semi-Markov processes, stationary distributions. Determining transition probabilities and other conditional probabilities and expected values; Integral expressions, Kolmogorov differential equations, numerical solutions, simulation techniques.

ST227 Survival Models HELP（EXAM HELP， ONLINE TUTOR）

SOLUTION Since, from (3.3), ${ }m d{x+n}=\ell_{x+n}-\ell_{x+n+m}$, then (3.24) becomes $_{n ! m} q_x=\frac{\ell_{x+n}-\ell_{x+n+m}}{\ell_x}={ }n p_x-{ }{n+m} p_x$. Since ${ }n p_x$ is the probability of surviving to age $x+n$, we can think of it as containing the probability of surviving to any age beyond $x+n$. If we remove from ${ }_n p_x$ the probability of surviving to $x+n+m$, which is ${ }{n+m} p_x$, we have the probability of surviving to $x+n$, but not to $x+n+m$, which is ${ }{n ! m} q{x \text {. }}$.

Next we wish to explore the conditional PDF for death at age $y$, given alive at age $x(y>x)$. From (2.39), we know this conditional PDF is $f(y \mid X>x)=\frac{f(y)}{S(x)}$. Now from (3.11) we have $f(y)=\frac{1}{\ell_0} \cdot \ell_y \mu_y$, and from (3.1) we have $S(x)=\frac{\ell_x}{\ell_0}$. Thus
$$
f(y \mid X>x)=\frac{\ell_y \mu_y}{\ell_x}={ }{y-x} p_x \mu_y $$ Letting $s=y-x$, so $y=x+s$, we have $$ f(s \mid X>x)={ }_s p_x \mu{x+s},
$$
where the random variable $S$ denotes the length of future lifetime of a person alive at age $x$. The conditional PDF given by (3.26) is a very useful function for developing other results.

If both numerator and denominator of (3.8a) are divided by $\ell_x$, we obtain
$$
\mu_{x+s}=\frac{-\frac{d}{d s} s p_x}{s p_x}
$$
which shows that
$$
\frac{d}{d s}{ }s p_x=-{ }_s p_x \mu{x+s}
$$
The expected future lifetime of a person alive at age $x$ is given by
$$
\stackrel{\circ}{e}x=E[S]=\int_0^{\infty} s \cdot{ }_s p_x \mu{x+s} d s=\int_0^{\infty} s p_x d s,
$$
since, in the integration by parts, $\int_s p_x \mu_{x+. s} d s=-{ }s p_x$ (from (3.28)), and $-s \cdot{ }_s p_x$ evaluates to zero at both limits. Furthermore, ${ }_s p_x=\frac{\ell{x+5}}{\ell_x}$, so (3.29) becomes
$$
\stackrel{\circ}{e}x=\frac{1}{\ell_x} \cdot \int_0^{\infty} \ell{x+s} d s=\frac{1}{\ell_x} \cdot \int_x^{\infty} \ell_y d y=\frac{T_x}{\ell_x},
$$
from (3.17).
By steps parallel to those leading from (3.18) to (3.21), we find
$$
E\left[S^2\right]=\frac{2 \cdot Y_x}{\ell_x}
$$
where
$$
Y_x=\int_x^{\infty} T_y d y
$$
Finally, from (3.31) and (3.30), we have
$$
\operatorname{Var}(S)=E\left[S^2\right]-{E[S]}^2=\frac{2 \cdot Y_x}{\ell_x}-\left(\frac{T_x}{\ell_x}\right)^2
$$

SOLUTION We recognize that the age-at-death random variable $X$ and the future-lifetime random variable $S$ are related by $X=x+S$, where $x$ is a constant. Thus, $\operatorname{Var}(X \mid X>x)=\operatorname{Var}(S \mid X>x)$. this variance is given by $\frac{2 \cdot Y_x}{\ell_x}-\left(\frac{T_x}{\hat{\ell}_x}\right)^2$. Thus we first find
$$
T_x=\int_x^{\infty} \ell_y d y=10,000 \int_x^{\infty}(y+1)^{-3} d y=5000(x+1)^{-2}
$$
and
$$
Y_x=\int_x^{\infty} T_y d y=5000 \int_x^{\infty}(y+1)^{-2} d y=5000(x+1)^{-1} .
$$
Then $\operatorname{Var}(X \mid X>x)=\frac{(x+1)^{-1}}{(x+1)^{-3}}-\left(\frac{(x+1)^{-2}}{2(x+1)^{-3}}\right)^2=\frac{3(x+1)^2}{4}$.

SOLUTION $d_0=\ell_0-\ell_1=2592$. $f_0=\frac{L_0-\ell_1}{d_0}=.13734$.
Similarly, ${ }_4 d_1=\ell_1-\ell_5=393$. ${ }_4 f_1=\frac{{ }_4 L_1-4 \cdot \ell_5}{4 \cdot{ }_4 d_1}=.41539$.

SOLUTION leads to $n \cdot{ }n m_x-n \cdot{ }_n m_x \cdot{ }_n q_x\left(1-{ }_n f_x\right)={ }_n q_x$, so that $\left(1-{ }_n f_x\right)=\frac{n \cdot{ }_n m_x-{ }_n q_x}{n \cdot{ }_n m_x \cdot{ }_n q_x}$. Substituting the given values, along with $n=5$, we obtain $\left(1-{ }_5 f{80}\right)=.53921$, whence ${ }5 f{80}=.46079$.

MY-ASSIGNMENTEXPERT™可以为您提供 LSE.AC.UK ST227 SURVIVAL MODELS 生存模型的代写代考和辅导服务！