CExamParaModels

The test is as follows:

c) The p-value is {$S(\chi^2)$}. (Note that the "values of P" given in the chart is {$F=1-S$}).

e) Or, look on the chart for values of {$P_1, P_2$} where {$P_1<F(\chi^2)<P_2$}. Then reject the null hypothesis at the {$1-P_1$} significance level; do not reject the null hypothesis at the {$1-P_2$} significance level.

e) Or, look on the chart for values of {$p_1, p_2$} where {$p_1<p<p_2$}. Then reject the null hypothesis at the {$1-p_1$} significance level; do not reject the null hypothesis at the {$1-p_2$} significance level.

This test statistic formula seems to be too complicated to be on the exam.

b) For the given significance level {$\alpha$}, the critical value is {$c=S^{-1}(\alpha).$} The rejection region is {$[c, \infty)$}.

d) If {$\chi^2\geq c$}, then reject the null hypothesis at the {$\alpha$} significance level; if {$\chi^2<c$}, then do not reject the null hypothesis at the {$\alpha$} significance level.

Hypothesis tests: this involves constructing of the following elements:

''Null hypothesis {$H_0$}: The data came from a population with the stated model.

Alternative hypothesis {$H_1$}: The data did not come from such a population.

Test statistic: a function of the observations, defining what the test is that we're using.

Rejection region, the boundary of which is the critical values, labeled {$c$}; this tells us that if the test statistic in in the region, the null hypothesis is rejected, otherwise we fail to reject it.

Type I error occurs when we reject a null hypothesis when it is true. Define {$\alpha$} to be the (significance) level, the probability of rejecting a null hypothesis when it is true.

Type II error occurs when we do not reject a null hypothesis when the alternative is true. The probability of this kind of error is denoted {$\beta$}.

b) For the given significance level {$\alpha$}, the critical value is {$c=S^{-1}(\alpha).$}

d) If {$c<\chi^2$}, then reject the null hypothesis at the {$\alpha$} significance level; if {$c>\chi^2$}, then do not reject the null hypothesis at the {$\alpha$} significance level.

e) Or, look on the chart for values of {$p_1, p_2$} where {$p_1<p<p_2}. Then reject the null hypothesis at the {$1-p_1$} significance level; do not reject the null hypothesis at the {$1-p_2$} significance level.

Select {$k-1$} arbitrary values, {$t=c_0 < c_1< \cdots < c_{k-1} < c_k=\infty$}. Define

Let {$\theta_0, \theta_1$} be the values of parameters that maximizes the likelihood function {$L$}, within the range of values allowed in the null and alternative hypotheses, respectively. Let {$L_0 = L(\theta_0), L_1 = L(\theta_1)$}. The test statistic is

d) If T<c, the null hypothesis is not rejected.

e) The p-value is {$S(T)$}.

d) Tell how much the model supports the alternative hypothesis by the 1%/10% rule with p-value stated above.

Let {$\theta_0, \theta_1$} be the values of parameters that maximizes the likelihood function {$L$}, within the range of values allowed in the null and alternative hypotheses, respectively. Let {$L_0 = L(\theta_0), L_1 = L(\theta_1)$|. The test statistic is

{$$T = 2 \ln (L_1/L_0).$$}

The critical value is determined as follows:

a) Start with a chi-square distribution with degrees of freedom = # of free parameters from the alternative hypothesis - # of free parameters from the null hypothesis; say the survival function is {$S$}.

b) For the given significance level {$\alpha$}, the critical value is {$c=S^{-1}(\alpha).$}

c) If T>c, the null hypothesis is rejected.

d) If T<c, the null phypothesis is not rejected.

e) The p-value is {$S(\T)$}.

f) Tell how much the model supports the alternative hypothesis by the 1%/10% rule with p-value stated above.

Here our hypotheses are:

H_0: The data came from a population with distribution A. (null hypothesis)

H_1: The data came from a population with distribution B, where A is a special case of B. (alternative hypothesis)

{$$D = \max_{t\leq x\leq u} |F_n(x) - F^*(x)|.$$}

The critical value is determined as follows:

a) Start with a chi-square distribution with degrees of freedom = {$k - 1 -$} # of estimated parameters; say the survival function is {$S$}.

b) For the given significance level {$\alpha$}, the critical value is {$S^{-1}(\alpha).$}

c) The p-value is {$S(\chi^2)$}.

d) Tell how good the fit is by the 1%/10% rule with p-value stated above.

Select {$n-1$} arbitrary values, {$t=c_0 < c_1< \cdots < c_{k-1} < c_k=\infty$}. Define

{$$\hat{p_j} = F^*(c_j)-F^*(c_{j-1}); \text{i.e., the probability that a (truncated) observations falls in } (c_{j-1}, c_j);$$} {$$p_{n,j} = F_n(c_j)-F_n(c_{j-1}); \text{i.e., the same probability with empirical distribution}.$$}

[The book is very unclear what do to once we find {$\chi^2$}.]

{$$\chi^2 =\sum_{j=1}^k \frac{n [\hat{p}_j -p_{n,j}]^2}{\hat{p}_j }.$$}

{$$\chi^2 =\sum_{j=1}^k \frac{[E_j-O_j]^2}{E_j },$$}

{$$\hat{p_j} = F*(c_j)-F^*(c_{j-1}); \text{i.e., the probability that a (truncated) observations falls in } (c_{j-1}, c_j);$$} {$$\{p_{n,j}} = F_n(c_j)-F_n(c_{j-1}); \text{i.e., the same probability with empirical distribution}.$$}

{$$\chi^2 \sum_{j=1}^k \frac{[E_j-O_j]^2}{E_j },$$}

{$$\xi^2 \sum_{j=1}^k \frac{[E_j-O_j]^2}{E_j },$$}

{$$\Xi^2 \sum_{j=1}^k \frac{[E_j-O_j]^2}{E_j },$$}

{$$\{p_{n,j}} = F_n(c_j)-F_n(c_{j-1}); \text{i.e., the same probability with empirical distribution.$$}

test statistic is

{$$\Chi^2 \sum_{j=1}^k \frac{n [\hat{p}_j -p_{n,j}]^2}{\hat{p}_j }.$$}

Easier to use the following: Let {$E_j = n\hat{p_j}$} = # of Expected observations in the interval, {$O_j = n\hat{p_{n,j}}$} = # of actual observations in the interval. Then

{$$\Chi^2 \sum_{j=1}^k \frac{[E_j-O_j]^2}{E_j },$$}

The critical values are

(These values appear to be always given in the problem.)

For individual data, this integral can be evaluated as a complicated sum.

The critical values are

{$$ 1.933/\sqrt{n} \text{ for }\alpha = 0.1, \qquad 2.492/\sqrt{n} \text{ for }\alpha = 0.05, \qquad 3.857/\sqrt{n} \text{ for }\alpha = 0.01.$$} This test statistic formula seems to be too complicated to be on the exam.

Select {$n-1$} arbitrary values, {$t=c_0 < c_1< \cdots < c_k=\infty$}. Define

{$$\hat{p_j} = F*(c_j)-F^*(c_{j-1}); \text{i.e., the probability that a (truncated) observations falls in } (c_{j-1}, c_j;$$} {$$\hat{p_j} = F*(c_j)-F^*(c_{j-1}); \text{i.e., the probability that a (truncated) observations falls in } (c_{j-1}, c_j;$$}

{$$ 1.22/\sqrt{n} \text{ for }\alpha = 0.1, \qquad 1.36/\sqrt{n} \text{ for }\alpha = 0.05, \qquad 1.63/\sqrt{n} \text{ for }\alpha = 0.01.$$} (These values appear to be always given in the problem.)

If {$t$} is the truncation point and {$u$} is the censoring point, test statistic is

{$$A^2 =n \int_t^u \frac{[F_n(x)-F^*(x)]^2}{F^*(x)(1-F^*(x))} f^*(x)\,dx.$$}

For individual data, this integral equals

{$$$$} The critical values are {$$ 1.22/\sqrt{n} \text{ for }\alpha = 0.1, \qquad 1.36/\sqrt{n} \text{ for }\alpha = 0.05, \qquad 1.63/\sqrt{n} \text{ for }\alpha = 0.01.$$}

If {$t$} is the truncation point and {$u$} is the censoring point, test statistic is

{$$D = \max_{t\leq x\leq u} \abs{F_n(x) - F^*(x)}.$$} The critical values are {$$ 1.22/\sqrt{n} \text{ for }\alpha = 0.1, 1.36/\sqrt{n} \text{ for }\alpha = 0..05, 1.63/\sqrt{n} \text{ for }\alpha = 0.01.$$}

Point(?): Compare {$F^*$} with {$F_n$} (the empirical esitmate).

{$$f^*(x)= \frac{f(x)}{1-F(t)}, \text{ for } x\geq t, \text{ and } 0 \text{ for } x<t;$$}

5. Determine the acceptability of a fitted model and/or compare models

{$$F^*(x)= \frac{F(x)-F(t)}{1-F(t)}, \text{ for } x\geq t, \text{ and } 0 \text{ for } x<t;$$}

{$$ \frac{\partial}{\partial \theta_r}\frac{\partial}{\partial \theta_s} l(\vec\theta) \approx \frac{ (l(\vec{\theta}+\frac{1}{2}h_r\vec{e_r} + \frac{1}{2}h_s \vec{e_s}) -l(\vec{\theta}+\frac{1}{2}h_r\vec{e_r} - \frac{1}{2}h_s \vec{e_s}) -l(\vec{\theta}-\frac{1}{2}h_r\vec{e_r} + \frac{1}{2}h_s \vec{e_s}) +l(\vec{\theta}-\frac{1}{2}h_r\vec{e_r}- \frac{1}{2}h_s \vec{e_s}) }{h_rh_s}$$}

b) Instead of taking derivatives in b) above, use the following approximation on 2nd derivatives:

{$$ \frac{\partial}{\partial \theta_r}\frac{\partial}{\partial \theta_s} l(\vec\theta) \approx \frac{ (l(\vec{theta}+\frac{1}{2}h_r\vec{e_r} + \frac{1}{2}h_s \vec{e_s} -l(\vec{theta}+\frac{1}{2}h_r\vec{e_r} - \frac{1}{2}h_s \vec{e_s} -l(\vec{theta}-\frac{1}{2}h_r\vec{e_r} + \frac{1}{2}h_s \vec{e_s} +l(\vec{theta}-\frac{1}{2}h_r\vec{e_r}- \frac{1}{2}h_s \vec{e_s} }{h_rh_s}$$}

{$$ \frac{\partial}{\partial \theta_r}\frac{\partial}{\partial \theta_s} l(\vec\theta) \approx \frac{(l(theta}{h_rh_s}$$}

{$$I(\theta)_{rs} = -E\left[ \frac{\partial}{\partial \theta_r}\frac{\partial}{\partial \theta_s} l(\theta)\right] = E\left[ \frac{\partial}{\partial \theta_r} l(\theta)\frac{\partial}{\partial \theta_s} l(\theta)\right].$$}

c) Find the expected values of the 2nd derivatives (by basically substituting the know expected value of the distribution we started with)

d) Put the negatives of the the result from c) into matrix form. The result is an information matrix in terms of the parameters.

e) Estimate the parameters by setting the 1st derivatives to 0.

f) Estimate the values of the information matrix using the estimated parameters from e).

g) Estimate a confidence interval using the fact that variance = reciprocal of the diagonal values in the information matrix in f).

{$$$$}

a) Start with a distribution (book example uses lognormal; see here for an example using Bernoulli). Find a formula for the likelihood function {$L$} and the loglikelyhood function {$l$}, in terms of the parameters, the number of sample points {$n$}, and each observed value {$x_1, \ldots, x_n$}.

b) Take the 2nd partial derivatives with respect to each parameter.

c) Find the expected values of the 2nd derivatives

{$$I(\theta) = -E\left[ \frac{\partial^2}{\partial \theta^2} l(\theta)\right] = E\left[ \left(\frac{\partial}{\partial \theta} l(\theta)\right)^2\right]$$}

The multivariable version of this is: {$$I$$} is an information matrix, where the {$(r,s)$}th element is {$$I(\theta)_{rs} = E\left[ \frac{\partial}{\partial \theta_r} l(\theta)\frac{\partial}{\partial \theta_s} l(\theta)\right].$$} So that the variances of the individual RVs (estimators) are on the diagonal, and the off diagonal entries are the covariances.

{$$I(\theta) = -E\left[ \frac{\partial^2}{\partial^2 \theta} l(\theta)\right] = E\left[ \left(\frac{\partial}{\partial \theta} l(\theta)\right)^2\right]$$}

{$I$} is called (Fisher's) information.

Use the unshifted approach: Ignore all observations under {$t$}, and keep original observations {$x_j$} above {$t$}, but instead of {$f(x_j)$}, use {$f(x_j)/(1-F(t))$}, i.e., probability of {$x_j$} given that {$x_j \geq t$}.

{$f(x_j)/(1-F(t))$}, i.e., probability of {$x_j$} given that {$x_j \geq t$}.

{$$L(\theta) = \prod_{j=1}^{k} [F(c_j) - F(c_{j-1})]^{n_j} \text{ (All cdf values given }\theta).$$}

1. Estimate the parameters of failure time and loss distributions

{$$L(\theta) = \prod_{j=1}^{n} Prob(X\in A_j | \theta).$$}

{$$L(\theta) = \prod_{j=1}^{k} [F(c_j) - F(c_{j-1})]^{n_j} (\text{ All cdf values given }\theta).$$}

I

Construction and Selection of Parametric Models (25-30%)

1. Estimate the parameters of failure time and loss distributions using:

2. Estimate the parameters of failure time and loss distributions with censored and/or truncated data using maximum likelihood.

3. Estimate the variance of estimators and the confidence intervals for the parameters and functions of parameters of failure time and loss distributions.

4. Apply the following concepts in estimating failure time and loss distributions:

5. Determine the acceptability of a fitted model and/or compare models using: