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）
As in Example 2.4, let $X$ have a standard exponential distribution, so that $F_X(x)=1-e^{-x}$, but this time let $y=g(x)=\frac{1}{x}$. Again find the SDF, PDF and HRF of $Y$.
We have $x=h(y)=\frac{1}{y}$, so $S_Y(y)=F_X[h(y)]=1-e^{-(1 / y)}$. Next we have $f_Y(y)=-\frac{d}{d y} S_Y(y)=y^{-2} \cdot e^{-(1 / y)}$. Finally, the HRF is
$$
\lambda_Y(y)=\frac{f_Y(y)}{S_Y(y)}=y^{-2} \cdot \frac{e^{-(1 / y)}}{1-e^{-(1 / y)}} .
$$
Again note that with $\lambda_X(x)=1$ and $\frac{d x}{d y}=-y^{-2}$, the last expression of Equation (2.71) can be used to find $\lambda_Y(y)$.
Show that Equations (2.74) and (2.75) are exact for the linear transformation $Y=g(X)=a X+b$.
We know that $E[Y]=a \cdot E[X]+b=a m+b=g(m)$, so (2.74) is exact. Furthermore, $\operatorname{Var}(Y)=a^2 \cdot \operatorname{Var}(X)$. Since $g^{\prime}(m)=a$, then $\left[g^{\prime}(m)\right]^2=a^2$, so $(2.75)$ is exact. [Note: It is easy to see that the approximations will be exact for linear transformations, since $g^{\prime \prime}(x)$ and all higher order derivatives are zero.]
A major use of these results will arise in Part II of the text, where we wish to find the variance of an estimator. Frequently that estimator will be a complex function of another random variable, whose mean and variance are known (at least approximately), so the method of statistical differentials can be used.
In some cases, we may encounter a random variable which is a function of two (or more) other random variables, and we seek its mean and variance. For example, if $Y=g\left(X_1, X_2\right)$, and if $E\left[X_1\right]=m_1, E\left[X_2\right]=m_2$, $\operatorname{Var}\left(X_1\right), \operatorname{Var}\left(X_2\right)$, and $\operatorname{Cov}\left(X_1, X_2\right)$ are all known, then we can proceed similarly to the univariate case.
Expand $g\left(X_1, X_2\right)$ in a bivariate Taylor series as
$$
\begin{aligned}
g\left(X_1, X_2\right)= & g\left(m_1, m_2\right)+\left(X_1-m_1\right) \cdot g_1^{\prime}\left(m_1, m_2\right)+\left(X_2-m_2\right) \cdot g_2^{\prime}\left(m_1, m_2\right) \
& +\frac{1}{2}\left(X_1-m_1\right)^2 \cdot g_1^{\prime \prime}\left(m_1, m_2\right)+\frac{1}{2}\left(X_2-m_2\right)^2 \cdot g_2^{\prime \prime}\left(m_1, m_2\right) \
& +\left(X_1-m_1\right)\left(X_2-m_2\right) \cdot g_{1,2}^{\prime \prime}\left(m_1, m_2\right)+\cdots,
\end{aligned}
$$
where $g_i^{\prime}\left(m_1, m_2\right)$ represents $\left.\frac{\partial}{\partial x_1} g\left(x_1, x_2\right)\right|{\substack{x_1=m_1, x_2=m_2}}$ for $i=1,2$. Similarly, $g_i^{\prime \prime}\left(m_1, m_2\right)$ represents $\left.\frac{\partial^2}{\partial x_i^2} g\left(x_1, x_2\right)\right|{\substack{x_1=m_1 \ x_2=m_2}}, i=1,2$, and $g_{1,2}^{\prime \prime}\left(m_1, m_2\right)$ represents $\left.\frac{\partial^2}{\partial x_1, x_2} g\left(x_1, x_2\right)\right|{\substack{x_1=m_1 \ x_2=m_2}}$. Taking the expected value of both sides of (2.76), we obtain $$ \begin{aligned} E\left[g\left(X_1, X_2\right)\right] & \approx g\left(m_1, m_2\right)+\frac{1}{2} g_1^{\prime \prime}\left(m_1, m_2\right) \cdot \operatorname{Var}\left(X_1\right) \ & +\frac{1}{2} g_2^{\prime \prime}\left(m_1, m_2\right) \cdot \operatorname{Var}\left(X_2\right)+g{1,2}^{\prime \prime}\left(m_1, m_2\right) \cdot \operatorname{Cov}\left(X_1, X_2\right)
\end{aligned}
$$
From Table 3.1, calculate (a) the probability that a life age 0 will die before age $3 ;$ (b) the probability that a life age 1 will survive to age 4.
(a) This is given directly by $F(3)=1-S(3)=.02840$.
(b) This conditional probability is given by ${ }_3 p_1=\frac{S(4)}{S(1)}=.99665$.
Show that the force of mortality is the limiting value of the probability of death over an interval, divided by the interval length (in years), as the interval length approaches zero.
SOLUTION Consider first a one-year interval, with $q_x=\frac{d_x}{\ell_x}$. Then consider a half-year interval with $\frac{1 / 2 q_x}{1 / 2}=\frac{\ell_x-\ell_{x+1 / 2}}{1 / 2 \cdot \ell_x}$. Now, in general, consider $\frac{\Delta x q_x}{\Delta x}=\frac{\ell_x-\ell_{x+\Delta x}}{\Delta x \cdot \ell_x}$, and show that $\lim {\Delta x \rightarrow 0} \frac{\Delta x q_x}{\Delta x}=\mu_x$. We have $$ \lim {\Delta x \rightarrow 0}\left[\frac{\ell_x-\ell_x+\Delta x}{\Delta x \cdot \ell_x}\right]=\frac{1}{\ell_x} \cdot \lim {\Delta x \rightarrow 0}\left[\frac{\ell_x-\ell{x+\Delta x}}{\Delta x}\right]=\frac{1}{\ell_x}\left[-\frac{d}{d x} \ell_x\right]=\mu_x,
$$
by Equation (3.8).
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！
