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Actuary /
CExamParaModelsActuary.CExamParaModels HistoryHide minor edits - Show changes to output Changed line 21 from:
->If the data is censored at {$c$}, and {$n$} observations are censored, then the factor to use for those observations is {$ [F(c)]^n $}. to:
->If the data is censored at {$c$}, and {$n$} observations are censored, then the factor to use for those observations is {$ [1-F(c)]^n $}. Changed line 101 from:
-->The to:
-->The test is as follows: Changed line 104 from:
--->c) The p-value is {$S(\chi^2)$}. to:
--->c) The p-value is {$S(\chi^2)$}. (Note that the "values of P" given in the chart is {$F=1-S$}). Changed lines 106-107 from:
--->e) Or, look on the chart for values of {$ to:
--->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. Changed lines 106-107 from:
--->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. to:
--->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. Changed line 92 from:
This test statistic formula seems to be too complicated to be on the exam. to:
-->This test statistic formula seems to be too complicated to be on the exam. Changed line 103 from:
--->b) For the given significance level {$\alpha$}, the critical value is {$c=S^{-1}(\alpha).$} to:
--->b) For the given significance level {$\alpha$}, the critical value is {$c=S^{-1}(\alpha).$} The rejection region is {$[c, \infty)$}. Changed line 105 from:
--->d) If {$ to:
--->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. Changed lines 66-76 from:
-->Hypothesis tests: this involves constructing of the following ---> -->''Type I error'' occurs when we reject a null hypothesis when it is true. Define {$\alpha$} to be the ''significance level --> ->If ->If we calculate the test statistic and it ->We can compute a ''p-value'' such that -->If we conduct the test at a significance level {$geq p$}, then it'll lead to a rejection of the null hypothesis - to:
-->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 {$\beta$}. ->Sometimes we're given the rejection region, and we can calculate {$\alpha$}. More often, it seems, we're given {$\alpha$}, and the methods to calculate the test statistic and rejection region, and we determine whether to reject {$H_0$} based on whether the test statistic is in the rejection region. ->Given the same analysis on the same set of observations, if {$\alpha_1 > \alpha_2$}, then {$\alpha_1$} would result in a larger rejection region than {$\alpha_2$}, so we can have a situation where {$H_0$} is not rejected at the level of {$\alpha_2$}, but is rejected at the level of {$\alpha_1$}. ->We can also compute a ''p-value'' (''attained significance level'') -- the smallest level of significance {$\alpha$} for which {$H_0$} is rejected based on the test statistic. The way to calculate the p-value should be stated as part of the test method. Changed line 103 from:
--->b) For the given significance level {$\alpha$}, the critical value is {$S^{-1}(\alpha).$} to:
--->b) For the given significance level {$\alpha$}, the critical value is {$c=S^{-1}(\alpha).$} Changed lines 105-106 from:
--->d) to:
--->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. Changed line 21 from:
->If the data is censored at {$c$}, and {$n$} observations are censored, then the factor to use for those observations is {$[F(c)]^n$}. to:
->If the data is censored at {$c$}, and {$n$} observations are censored, then the factor to use for those observations is {$ [F(c)]^n $}. Changed line 94 from:
-->Select {$ to:
-->Select {$k-1$} arbitrary values, {$t=c_0 < c_1< \cdots < c_{k-1} < c_k=\infty$}. Define Changed line 5 from:
->Given a distribution {$X$} with parameter {$\theta$}, construct a ''likelihood function'' {$L(\theta)$} and find the value of {$\theta$} that maximizes {$L$}. Or maximizes the ''loglikelihood function'' {$l(\theta) = \log to:
->Given a distribution {$X$} with parameter {$\theta$}, construct a ''likelihood function'' {$L(\theta)$} and find the value of {$\theta$} that maximizes {$L$}. Or maximizes the ''loglikelihood function'', {$l(\theta) = \log{L(\theta)}$}. More specifically, given a series of events {$A_1, \ldots, A_n$}, where each {$A_n$} is an observed value (or interval), then Changed line 111 from:
-->Let {$\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)$ to:
-->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 Changed lines 117-118 from:
--->d) If T<c, the null --->e) The p-value is {$S( to:
--->d) If T<c, the null hypothesis is not rejected. --->e) The p-value is {$S(T)$}. Changed lines 105-106 from:
--->d) Tell how to:
--->d) Tell how much the model supports the alternative hypothesis by the 1%/10% rule with p-value stated above. Changed lines 111-113 from:
to:
-->Let {$\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. Added lines 108-113:
-->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) Added lines 77-78:
->If {$p$} is above 10%, then the data has no evidence to support the alternative hypothesis. ->If {$p$} is below 1%, then the data gives strong support for the alternative hypothesis. Changed line 82 from:
{$$D = \max_{t\leq x\leq u} to:
{$$D = \max_{t\leq x\leq u} |F_n(x) - F^*(x)|.$$} Changed lines 101-102 from:
--> to:
-->The critical value is determined as follows: --->a) Start with a chi-square distribution with degrees of freedom = {$k - 1 -$} # of estimated parameters; say the survival function is {$S$}. --->b) For the given significance level {$\alpha$}, the critical value is {$S^{-1}(\alpha).$} --->c) The p-value is {$S(\chi^2)$}. --->d) Tell how good the fit is by the 1%/10% rule with p-value stated above. Changed lines 92-94 from:
-->Select {$n-1$} arbitrary values, {$t=c_0 < c_1< \cdots < c_k { {$$ to:
-->Select {$n-1$} arbitrary values, {$t=c_0 < c_1< \cdots < c_{k-1} < c_k=\infty$}. Define {$$\hat{p_j} = F^*(c_j)-F^*(c_{j-1}); \text{i.e., the probability that a (truncated) observations falls in } (c_{j-1}, c_j);$$} {$$p_{n,j} = F_n(c_j)-F_n(c_{j-1}); \text{i.e., the same probability with empirical distribution}.$$} Changed lines 99-100 from:
to:
-->[The book is very unclear what do to once we find {$\chi^2$}.] Changed lines 74-76 from:
->We can compute a to:
->We can compute a ''p-value'' such that -->If we conduct the test at a significance level {$geq p$}, then it'll lead to a rejection of the null hypothesis. -->If we conduct the test at a significance level {$< p$}, then it'll lead to a failure to reject the null hypothesis. Added line 74:
->We can compute a ran Changed line 94 from:
{$$\ to:
{$$\chi^2 =\sum_{j=1}^k \frac{n [\hat{p}_j -p_{n,j}]^2}{\hat{p}_j }.$$} Changed lines 96-98 from:
{$$\chi^2 \sum_{j=1}^k \frac{[E_j-O_j]^2}{E_j },$$} to:
{$$\chi^2 =\sum_{j=1}^k \frac{[E_j-O_j]^2}{E_j },$$} Changed lines 90-91 from:
{$$\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.$$} to:
{$$\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}.$$} Changed lines 95-97 from:
{$$\ to:
{$$\chi^2 \sum_{j=1}^k \frac{[E_j-O_j]^2}{E_j },$$} Changed lines 95-97 from:
{$$\ to:
{$$\xi^2 \sum_{j=1}^k \frac{[E_j-O_j]^2}{E_j },$$} Changed lines 95-97 from:
{$$\ to:
{$$\Xi^2 \sum_{j=1}^k \frac{[E_j-O_j]^2}{E_j },$$} Changed lines 91-97 from:
{$$\ to:
{$$\{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 },$$} Changed line 78 from:
The critical values are to:
-->The critical values are Changed line 80 from:
(These values appear to be always given in the problem.) to:
-->(These values appear to be always given in the problem.) Changed lines 84-87 from:
-->For individual data, this integral {$$$$} The critical values are {$$ 1 to:
-->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. Added lines 89-91:
-->Select {$n-1$} arbitrary values, {$t=c_0 < c_1< \cdots < c_k=\infty$}. Define {$$\hat{p_j} = F*(c_j)-F^*(c_{j-1}); \text{i.e., the probability that a (truncated) observations falls in } (c_{j-1}, c_j;$$} {$$\hat{p_j} = F*(c_j)-F^*(c_{j-1}); \text{i.e., the probability that a (truncated) observations falls in } (c_{j-1}, c_j;$$} Changed line 64 from:
->Point to:
->Point: Compare {$F^*$} with {$F_n$} (the empirical esitmate). There are several ways to do this: Changed lines 79-80 from:
{$$ 1.22/\sqrt{n} \text{ for }\alpha = 0.1, 1.36/\sqrt{n} \text{ for }\alpha = 0. to:
{$$ 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.$$} (These values appear to be always given in the problem.) Added lines 82-87:
-->If {$t$} is the truncation point and {$u$} is the censoring point, test statistic is {$$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.$$} Added lines 72-73:
->If we calculate the test statistic and it's bigger than the critical value, we can reject the null hypothesis. ->If we calculate the test statistic and it's smaller than the critical value, we do not reject the null hypothesis and view it as plausible. Added lines 76-79:
-->If {$t$} is the truncation point and {$u$} is the censoring point, test statistic is {$$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.$$} Changed lines 64-71 from:
Point(?): Compare {$F^*$} with {$F_n$} (the empirical esitmate). to:
->Point(?): Compare {$F^*$} with {$F_n$} (the empirical esitmate). There are several ways to do this: -->Graphically: graph a few things (either {$F^*$} with {$F_n$} directly, or some other values) and visually tell how close the functions are. -->Hypothesis tests: this involves constructing of the following ''hypotheses'': --->H_0: The data came from a population with the stated model. (null hypothesis) --->H_1: The data did not come from such a population. (alternative hypothesis) -->along with the ''test statistic'' (a function of the observations) and a ''rejection region'', the boundary of which is the ''critical values'' to tell you if the null hypothesis is accepted or rejected. -->''Type I error'' occurs when we reject a null hypothesis when it is true. Define {$\alpha$} to be the ''significance level'', the probability of rejecting a null hypothesis when it is true. This is usually set in advance. -->''Type II error'' occurs when we do not reject a null hypothesis when the alternative is true. Added line 64:
Point(?): Compare {$F^*$} with {$F_n$} (the empirical esitmate). Added line 63:
{$$f^*(x)= \frac{f(x)}{1-F(t)}, \text{ for } x\geq t, \text{ and } 0 \text{ for } x<t;$$} Changed lines 60-62 from:
5. Determine the acceptability of a fitted model and/or compare models to:
5. Determine the acceptability of a fitted model and/or compare models ->If the data has been modeled with a parametric model, with a resulting distribution function {$F$} and density function {$f$}, and the data is truncated at {$t$}, then the modified functions are: {$$F^*(x)= \frac{F(x)-F(t)}{1-F(t)}, \text{ for } x\geq t, \text{ and } 0 \text{ for } x<t;$$} Changed lines 46-52 from:
to:
->There is also a method called ''delta method'' that is used to approximate expected values and variances of a function of estimators. Here is the one-dimensional statement of the theorem: -->Suppose {$\hat{\theta}$} is an estimator of {$\theta$} that has an asymptotic normal distribution with mean {$\theta$} and variance {$\sigma^2/n$}. Then {$g(\hat{\theta})$} has an asymptotic normal distribution with mean {$g(\theta)$} and asymptotic variance {$g'(\theta)^2 \sigma^2/n$}. ->Steps for the calculation: -->a)Mind the mle {$\hat{\theta}$} of the parameter {$\theta$}. -->a)Figure out the quantity we're estimating {$g(\theta)$} This could be, say, a probability. -->b)Then the new mle is {$g(\hat{\theta})$}. -->c)Find the mean and variance of the first estimator {$\hat{\theta}$}; then find the mean and variance of the 2nd estimator according to the theorem. Changed lines 45-46 from:
{$$ \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}$$} to:
{$$ \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}$$} Changed lines 44-46 from:
-->b) Instead of taking derivatives in b) above, use the {$$ \frac to:
-->b) Instead of taking derivatives in b) above, use the following approximation on 2nd derivatives: {$$ \frac{\partial}{\partial \theta_r}\frac{\partial}{\partial \theta_s} l(\vec\theta) \approx \frac{ (l(\vec{theta}+\frac{1}{2}h_r\vec{e_r} + \frac{1}{2}h_s \vec{e_s} -l(\vec{theta}+\frac{1}{2}h_r\vec{e_r} - \frac{1}{2}h_s \vec{e_s} -l(\vec{theta}-\frac{1}{2}h_r\vec{e_r} + \frac{1}{2}h_s \vec{e_s} +l(\vec{theta}-\frac{1}{2}h_r\vec{e_r}- \frac{1}{2}h_s \vec{e_s} }{h_rh_s}$$} Changed lines 45-46 from:
{$$$$} to:
{$$ \frac{\partial}{\partial \theta_r}\frac{\partial}{\partial \theta_s} l(\vec\theta) \approx \frac{(l(theta}{h_rh_s}$$} Changed line 32 from:
{$$I(\theta)_{rs} = E\left[ \frac{\partial}{\partial \theta_r} l(\theta)\frac{\partial}{\partial \theta_s} l(\theta)\right].$$} to:
{$$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].$$} Changed lines 37-38 from:
-->c) Find the expected values of the 2nd derivatives to:
-->c) Find the expected values of the 2nd derivatives (by basically substituting the know expected value of the distribution we started with) -->d) Put the negatives of the the result from c) into matrix form. The result is an information matrix in terms of the parameters. -->e) Estimate the parameters by setting the 1st derivatives to 0. -->f) Estimate the values of the information matrix using the estimated parameters from e). -->g) Estimate a confidence interval using the fact that variance = reciprocal of the diagonal values in the information matrix in f). ->Variations: -->a) Instead of taking expected values in c) above, plug in observed values. -->b) Instead of taking derivatives in b) above, use the ''delta approximation'' {$$$$} Changed lines 35-36 from:
--> -->Take to:
-->a) Start with a distribution (book example uses lognormal; see [[http://en.wikipedia.org/wiki/Fisher_information#Single-parameter_Bernoulli_experiment|here]] for an example using Bernoulli). Find a formula for the likelihood function {$L$} and the loglikelyhood function {$l$}, in terms of the parameters, the number of sample points {$n$}, and each observed value {$x_1, \ldots, x_n$}. -->b) Take the 2nd partial derivatives with respect to each parameter. -->c) Find the expected values of the 2nd derivatives Changed lines 30-33 from:
The multivariable version of this is: {$$I$$} is an ''information matrix'', where the {$(r,s)$}th element is to:
->{$I$} is called ''(Fisher's) information''. ->The multivariable version of this is: {$$I$$} is an ''information matrix'', where the {$(r,s)$}th element is Changed lines 33-34 from:
So that the variances of the individual RVs (estimators) are on the diagonal, and the off diagonal entries are the covariances. to:
->So that the variances of the individual RVs (estimators) are on the diagonal, and the off diagonal entries are the covariances. ->Steps for the calculation: --> Find a formula for the likelihood function {$L$} and the loglikelyhood function {$l$}, in terms of the parameters, the number of sample points {$n$}, and each observed value {$x_1, \ldots, x_n$}. -->Take the 2nd partial derivatives with respect to each parameter. Changed lines 28-32 from:
->We can estimate the variance of the mle (maximum likelihood estimator) by the following (due to a long and complicated theorem): to:
->We can estimate the variance of the mle (maximum likelihood estimator) by the following (due to a long and complicated theorem): If the underlying density function {$f$} is smooth enough, then {$Var(\hat{\theta}) \approx \frac{1}{I(\theta)}$}, where {$$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]$$} Changed lines 33-34 from:
to:
The multivariable version of this is: {$$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].$$} So that the variances of the individual RVs (estimators) are on the diagonal, and the off diagonal entries are the covariances. Changed lines 28-29 from:
to:
->We can estimate the variance of the mle (maximum likelihood estimator) by the following (due to a long and complicated theorem): {$Var(\hat{\theta}) \approx \frac{1}{I(\theta)}$}, where {$$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''. Changed lines 24-26 from:
-->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$}. to:
-->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$}. Changed lines 11-15 from:
->If the data is truncated at {$t$}, we can either -->Use the shift approach --> to:
Added line 14:
Added line 17:
Changed lines 21-26 from:
to:
->If the data is censored at {$c$}, and {$n$} observations are censored, then the factor to use for those observations is {$[F(c)]^n$}. ->If the data is truncated at {$t$}, we can either -->Use the shift approach: Ignore all observations under {$t$}, and shift all observations above {$t$} by subtracting {$t$} off of every one of them. -->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$}. Changed lines 11-12 from:
to:
->If the data is censored at {$c$}, and {$n$} observations are censored, then the factor to use for those observations is {$[F(c)]^n$}. ->If the data is truncated at {$t$}, we can either -->Use the shift approach --> Changed line 16 from:
->If there are {$p$} parameters, and {$g_1, \ldots, g_p$} are arbitrary values between 0 and 100, form and solve a system of {$p$} equations by equation the {$g_1, \ldots, g_p$}th percentile of the data and of the distribution; more accurately, use ''smoothed'' empirical estimate of a {$g$}th percentile: Look at the {$n+1$} intervals formed by {$(-\infty, x_1), (x_1, x_2), \ldots, (x_n, \infty)$}. Determine the interval {$(x_{j}, x_{j+1})$} where the {$g$}th percentile lands in by {$j = \text{greatest integer }<= g to:
->If there are {$p$} parameters, and {$g_1, \ldots, g_p$} are arbitrary values between 0 and 100, form and solve a system of {$p$} equations by equation the {$g_1, \ldots, g_p$}th percentile of the data and of the distribution; more accurately, use ''smoothed'' empirical estimate of a {$g$}th percentile: Look at the {$n+1$} intervals formed by {$(-\infty, x_1), (x_1, x_2), \ldots, (x_n, \infty)$}. Determine the interval {$(x_{j}, x_{j+1})$} where the {$g$}th percentile lands in by {$j = \text{greatest integer }<= g/100 \cdot (n+1) $}. Let {$h$} be the leftover decimal part. Then the {$g$}th percentile is a weighted average between {$x_j$} and {$x_{j+1}$} by {$h$}. Changed line 12 from:
to:
Changed line 16 from:
->If there are {$p$} parameters, and {$g_1, \ldots, g_p$} are arbitrary values between 0 and 100, form and solve a system of {$p$} equations by equation the {$g_1, \ldots, g_p$}th percentile of the data and of the distribution; more accurately, use ''smoothed'' empirical estimate of a {$g$}th percentile: Look at the {$n+1$} intervals formed by {$(-\infty, x_1), (x_1, x_2), \ldots, (x_n, \infty)$}. Determine the interval {$(x_{j}, x_{j+1)$} where the {$g$}th percentile lands in by {$j = greatest integer <= g% \cdot (n+1) $}. Let {$h$} be the leftover decimal part. Then the {$g$}th percentile is a weighted average between {$x_j$} and {$x_{j+1}$} by {$h$}. to:
->If there are {$p$} parameters, and {$g_1, \ldots, g_p$} are arbitrary values between 0 and 100, form and solve a system of {$p$} equations by equation the {$g_1, \ldots, g_p$}th percentile of the data and of the distribution; more accurately, use ''smoothed'' empirical estimate of a {$g$}th percentile: Look at the {$n+1$} intervals formed by {$(-\infty, x_1), (x_1, x_2), \ldots, (x_n, \infty)$}. Determine the interval {$(x_{j}, x_{j+1})$} where the {$g$}th percentile lands in by {$j = \text{greatest integer }<= g% \cdot (n+1) $}. Let {$h$} be the leftover decimal part. Then the {$g$}th percentile is a weighted average between {$x_j$} and {$x_{j+1}$} by {$h$}. Changed line 16 from:
->If there are {$p$} parameters, and {$g_1, \ldots, g_p$} are arbitrary values between 0 and 100, form and solve a system of {$p$} equations by equation the {$g_1, \ldots, g_p$}th percentile of the data and of the distribution; more accurately, use ''smoothed'' empirical estimate of a {$g$}th percentile: Look at the {$n+1$} intervals formed by {(-\infty, x_1), (x_1, x_2), \ldots, (x_n, \infty)$}. Determine the interval {$(x_{j}, x_{j+1)$} where the {$g$}th percentile lands in by {$j = greatest integer <= g% \cdot (n+1) $}. Let {$h$} be the leftover decimal part. Then the {$g$}th percentile is a weighted average between {$x_j$} and {$x_{j+1}$} by {$h$}. to:
->If there are {$p$} parameters, and {$g_1, \ldots, g_p$} are arbitrary values between 0 and 100, form and solve a system of {$p$} equations by equation the {$g_1, \ldots, g_p$}th percentile of the data and of the distribution; more accurately, use ''smoothed'' empirical estimate of a {$g$}th percentile: Look at the {$n+1$} intervals formed by {$(-\infty, x_1), (x_1, x_2), \ldots, (x_n, \infty)$}. Determine the interval {$(x_{j}, x_{j+1)$} where the {$g$}th percentile lands in by {$j = greatest integer <= g% \cdot (n+1) $}. Let {$h$} be the leftover decimal part. Then the {$g$}th percentile is a weighted average between {$x_j$} and {$x_{j+1}$} by {$h$}. Changed lines 10-11 from:
{$$L(\theta) = \prod_{j=1}^{k} [F(c_j) - F(c_{j-1})]^{n_j} to:
{$$L(\theta) = \prod_{j=1}^{k} [F(c_j) - F(c_{j-1})]^{n_j} \text{ (All cdf values given }\theta).$$} Changed lines 2-3 from:
1. Estimate the parameters of failure time and loss distributions to:
1. Estimate the parameters of failure time and loss distributions ->Assuming {$n$} independent observations throughout, label them {$x_1, \ldots, x_n$}. The general problem is: Given a distribution, find the parameters of the distribution. Added lines 5-12:
->Given a distribution {$X$} with parameter {$\theta$}, construct a ''likelihood function'' {$L(\theta)$} and find the value of {$\theta$} that maximizes {$L$}. Or maximizes the ''loglikelihood function'' {$l(\theta) = \log L(\theta)$}. More specifically, given a series of events {$A_1, \ldots, A_n$}, where each {$A_n$} is an observed value (or interval), then {$$L(\theta) = \prod_{j=1}^{n} Prob(X\in A_j | \theta).$$} ->When {$A_j$} is a single point observation {$x_j$}, we can estimate {$Prob(X=x_j)$} by the density function {$f(x_j)$}. When {$A_j$} is an interval we can use cdf {$F$}. ->If there is more than one parameter, we can maximize the likelihood function by taking partial derivatives. ->If the data is grouped into {$(0, c_1], (c_1, c_2], \ldots, (c_k,\infty)$}, and there are {$n_j$} observations in each {$(c_{j-1}, c_j]$}, then {$$L(\theta) = \prod_{j=1}^{k} [F(c_j) - F(c_{j-1})]^{n_j} (\text{ All cdf values given }\theta).$$} I Added line 14:
->If there are {$p$} parameters, form and solve a system of {$p$} equations by equating the 1st to {$p$}th moment of the data, and of the distribution. Added line 16:
->If there are {$p$} parameters, and {$g_1, \ldots, g_p$} are arbitrary values between 0 and 100, form and solve a system of {$p$} equations by equation the {$g_1, \ldots, g_p$}th percentile of the data and of the distribution; more accurately, use ''smoothed'' empirical estimate of a {$g$}th percentile: Look at the {$n+1$} intervals formed by {(-\infty, x_1), (x_1, x_2), \ldots, (x_n, \infty)$}. Determine the interval {$(x_{j}, x_{j+1)$} where the {$g$}th percentile lands in by {$j = greatest integer <= g% \cdot (n+1) $}. Let {$h$} be the leftover decimal part. Then the {$g$}th percentile is a weighted average between {$x_j$} and {$x_{j+1}$} by {$h$}. Added lines 1-25:
!!Construction and Selection of Parametric Models (25-30%) 1. Estimate the parameters of failure time and loss distributions using: ->a) Maximum likelihood ->b) Method of moments ->c) Percentile matching ->d) Bayesian procedures 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: ->a) Unbiasedness ->b) Asymptotic unbiasedness ->c) Consistency ->d) Mean squared error ->e) Uniform minimum variance estimator 5. Determine the acceptability of a fitted model and/or compare models using: ->a) Graphical procedures ->b) Kolmogorov-Smirnov test ->c) Anderson-Darling test ->d) Chi-square goodness-of-fit test ->e) Likelihood ratio test ->f) Schwarz Bayesian Criterion |