统计代写|ST227 Survival Models

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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 $E[T]=\int_0^{\infty} t \cdot f(t) d t=\int_0^{\infty} t \cdot \lambda e^{-\lambda t} d t$. Integration by parts produces $\int_0^{\infty} e^{-\lambda t} d t$, whence $E[T]=-\left.\frac{1}{\lambda} e^{-\lambda t}\right|_0 ^{\infty}=\frac{1}{\lambda}$. We also have
$$
E\left[T^2\right]=\int_0^{\infty} t^2 \cdot \lambda e^{-\lambda t} d t=2 \int_0^{\infty} t \cdot e^{-\lambda t} d t=\frac{2}{\lambda} \int_0^{\infty} e^{-\lambda t} d t=\frac{2}{\lambda^2} .
$$
Then
$$
\operatorname{Var}(T)=\frac{2}{\lambda^2}-\left{\frac{1}{\lambda}\right}^2=\frac{1}{\lambda^2} \text {. }
$$
The exponential distribution, with its property of a constant hazard rate, is frequently used in reliability engineering as a survival model for inanimate objects such as machine parts (see Chapter 12). Like the uniform distribution, however, it is not appropriate as a model for human survival over a broad range, but is used extensively over short intervals, such as one year, due to its mathematical simplicity. This will be explored in Section 3.5.2.

Since we do not contemplate using the uniform or exponential as a model for human survival, we use $T$, rather than $X$ for our failure time random variable. For the next three distributions, we use $X$ to suggest that they are more useful as models of human survival.

[SOLUTION We seek the probability of death prior to age $x+20$, given alive at age $x+10$. We denote this probability by ${ }{10} q{[x]+10}$. It is equal to $1-\operatorname{Pr}$ (Survival to $x+20$, given survival to $x+10)=1-10 p[x]+10$. Now if the conditional probability ${ }{10} p{[x]+10}$ is multiplied by the probability of obtaining the condition, which is $S(10 ; x)$, the result is the unconditional probability of survival to $x+20$, namely $S(20 ; x)$. Thus
$$
{ }{10} q{[x]+10}=1-10 p_{[x]+10}=1-\frac{S(20 ; x)}{S(10 ; x)} .
$$
Consider next the PDF for death at age $y$, given alive at age $x$, for $y>x$. If this conditional density is multiplied by the probability of obtaining the condition, which is $S(x)$, then the unconditional density, which is $f(y)$, results. Thus the conditional density is
$$
\frac{f(y)}{S(x)}
$$
We will derive this conditional density more formally in the next subsection. Finally, consider the conditional HRF (or force of mortality) for death at age $y$, given alive at age $x(y>x)$. Recall that the HRF is itself always conditional on survival to the age at which it applies. (There is no such thing as an unconditional HRF.) Thus, since $y>x$ and $\mu_y$ itself is conditional on survival to $y$, then the statement “given survival to $x$ ” is redundant. Therefore, this “conditional” HRF to which we allude is clearly the same as $\mu_y$ itself. This intuitive result will be shown more formally in the following subsection.

SOLUTION If $X$ is exponential, the HRF is constant, with $\lambda(y)=\lambda$ for all $y$. Then, from Equation (2.57), $m_x=\frac{\lambda \cdot \int_x^{x+1} S(y) d y}{\int_x^{x+1} S(v) d y}=\lambda$. Furthermore, we also have $p_x=\frac{S(x+1)}{S(x)}=\frac{e^{-\lambda(x+1)}}{e^{-\lambda(x)}}=e^{-\lambda}$. Thus $\lambda=-\ln p_x$, and since $m_x=\lambda$, then $m_x=-\ln p_x$.

variable $Y$.
SOLUTION Note that $y=g(x)$ is strictly increasing, and $x=h(y)=y^2$. Then $F_Y(y)=F_X\left(y^2\right)$. Since $X$ is exponential, then $F_X(x)=1-e^{-x}$, so we have $F_Y(y)=1-e^{-y^2}$, and $S_Y(y)=e^{-y^2}$. Next, $f_Y(y)=\frac{d}{d y} F_Y(y)=2 y \cdot e^{-y^2}$. Finally, $\lambda_Y(y)=\frac{f_Y(y)}{1-F_Y(y)}=\frac{2 y \cdot e^{-y^2}}{e^{-y^2}}=2 y$. Alternatively, since we have $\lambda_X(x)=\lambda_X[h(y)]=1$, then, from (2.66), $\lambda_Y(y)$ is simply $\frac{d x}{d y}=2 y$. Note that $Y$ has a Weibull distribution with $k=2$ and $n=1$.

If $y=g(x)$ is a strictly decreasing function, as shown in Figure 2.2, then our reasoning and our results change a bit.

Here we can see that if $X$ is less than $x$, then $Y$ will be greater than the value of $y$ which corresponds to the given value of $x$; or, conversely, if $Y>y$, then $Xy)=\operatorname{Pr}(X<h(y))=\operatorname{Pr}(X \leq h(y))
$$

or
$$
S_Y(y) \equiv F_X[h(y)]=1-S_X[h(y)]
$$
Then
$$
F_Y(y) \quad 1-S_Y(y)=1-F_X[h(y)]=S_X[h(y)]
$$
and
$$
f_Y(y)=\frac{d}{d y} F_Y(y)=-\frac{d}{d y} F_X[h(y)]=-f_X[h(y)] \cdot \frac{d x}{d y}
$$
by the chain rule. Since $x=h(y)$ is a decreasing function, then $\frac{d x}{d y}$ is negative and $f_Y(y)$ is positive, as required. Finally,
$$
\lambda_Y(y)=\frac{f_Y(y)}{S_Y(y)}=\frac{-f_X[h(y)] \cdot \frac{d x}{d y}}{1-S_X[h(y)]}=-\lambda_X[h(y)] \cdot \frac{S_X[h(y)]}{1-S_X[h(y)]} \cdot \frac{d x}{d y} .
$$
Note that the expression for $\lambda_Y(y)$ in terms of $\lambda_X(x)$ is not as convenient as it is for the case where $y=g(x)$ is a strictly increasing function.

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