Likelihood ratio test

[Ruud Van Nistelrooij]

how to determine the Chi-square degree of freedom???

[atomic]

You're smart, read the text or study manual.

Seriously though--help comes to those who help themselves. What I mean by this is that the quality of the feedback you get on a question depends on the amount of detail you provide. If I asked you, "How do you calculate integrals," what kind of response would that generate? The scope is too broad. On the other hand, if I asked you, "How do you calculate



then at least the response will be specific. Your question is too vague because an adequate response would basically involve quoting the text or study manual--essentially, to teach you the entire topic.

[Ruud Van Nistelrooij]

Originally Posted by atomic

You're smart, read the text or study manual.

Seriously though--help comes to those who help themselves. What I mean by this is that the quality of the feedback you get on a question depends on the amount of detail you provide. If I asked you, "How do you calculate integrals," what kind of response would that generate? The scope is too broad. On the other hand, if I asked you, "How do you calculate



then at least the response will be specific. Your question is too vague because an adequate response would basically involve quoting the text or study manual--essentially, to teach you the entire topic.

thank you for your advise

just like question 14 of exam C speing 2007, this question about likelihood ratio test. the test statistic = -2log^

i do not undestand the following statement:
"the test statistic −2 log Λ will be asymptotically χ2 distributed with degrees of freedom equal to the difference in dimensionality of Θ and Θ0." reference: http://en.wikipedia.org/wiki/Likelihood-ratio_test

[Ruud Van Nistelrooij]

Originally Posted by Ruud Van Nistelrooij

thank you for your advise

just like question 14 of exam C speing 2007, this question about likelihood ratio test. the test statistic = -2log^

i do not undestand the following statement:
"the test statistic −2 log Λ will be asymptotically χ2 distributed with degrees of freedom equal to the difference in dimensionality of Θ and Θ0." reference: http://en.wikipedia.org/wiki/Likelihood-ratio_test

btw, a question irrelevant to SOA exam

is Lakewood city in New Jersey close to New York city?

[NoName]

Originally Posted by Ruud Van Nistelrooij

btw, a question irrelevant to SOA exam

is Lakewood city in New Jersey close to New York city?

http://maps.google.com/maps?f=d&hl=e...UTF8&z=10&om=1

[Jim Daniel]

To atomic:

Good explanation, but ... : The LR test is valid provided that the model with fewer parameters is a special case of the model with more parameters. It is _not_ valid for comparing an Exponential and a 2-parameter Pareto, for instance.

Jim Daniel

[atomic]

Originally Posted by Ruud Van Nistelrooij

thank you for your advise

just like question 14 of exam C speing 2007, this question about likelihood ratio test. the test statistic = -2log^

i do not undestand the following statement:
"the test statistic −2 log Λ will be asymptotically χ2 distributed with degrees of freedom equal to the difference in dimensionality of Θ and Θ0." reference: http://en.wikipedia.org/wiki/Likelihood-ratio_test

The Wikipedia article examines the topic in greater detail than is needed or tested on the exam. However, I will address the article and apply the concepts as they are used in Exam 4/C. Hereafter, we will discuss the likelihood ratio test as it is tested for the exam.

We use the LR test when we want to select a parametric model to fit empirical data. However, the number of parameters in the model is not predetermined; for example, we can use a 1-parameter model (e.g., exponential or single-parameter Pareto), a 2-parameter model (gamma, Pareto), or an N-parameter model. We might expect that as the number of parameters is increased, we are more likely to obtain a model whose MLE fit of the data is better--at the cost of the complexity of the model.

To quantify this notion, we use the LR test. The idea is extraordinarily simple. For various models, we compute the loglikelihood of the MLE fit to the empirical data. Then twice the pairwise difference between the loglikelihoods is chi-square distributed, with degree of freedom equal to the difference in the number of parameters in each model. So if we want to compare a 2-parameter fit to a 4-parameter fit, the LR test gives us a chi-square statistic with df = 4 - 2 = 2. It's that easy.

For practical considerations on the exam, the value of the MLE fit is often given, since it is usually tedious to compute.

[Ruud Van Nistelrooij]

Originally Posted by atomic

The Wikipedia article examines the topic in greater detail than is needed or tested on the exam. However, I will address the article and apply the concepts as they are used in Exam 4/C. Hereafter, we will discuss the likelihood ratio test as it is tested for the exam.

We use the LR test when we want to select a parametric model to fit empirical data. However, the number of parameters in the model is not predetermined; for example, we can use a 1-parameter model (e.g., exponential or single-parameter Pareto), a 2-parameter model (gamma, Pareto), or an N-parameter model. We might expect that as the number of parameters is increased, we are more likely to obtain a model whose MLE fit of the data is better--at the cost of the complexity of the model.

To quantify this notion, we use the LR test. The idea is extraordinarily simple. For various models, we compute the loglikelihood of the MLE fit to the empirical data. Then twice the pairwise difference between the loglikelihoods is chi-square distributed, with degree of freedom equal to the difference in the number of parameters in each model. So if we want to compare a 2-parameter fit to a 4-parameter fit, the LR test gives us a chi-square statistic with df = 4 - 2 = 2. It's that easy.

For practical considerations on the exam, the value of the MLE fit is often given, since it is usually tedious to compute.

thank you for your explanation
in question 14 of spring 07, the distribution is pareto with parameters alpha =2. theta unknown
we test if theta = 3.1 by LR test
why the degree of freedom is 1?

[atomic]

Originally Posted by Ruud Van Nistelrooij

thank you for your explanation
in question 14 of spring 07, the distribution is pareto with parameters alpha =2. theta unknown
we test if theta = 3.1 by LR test
why the degree of freedom is 1?

Because you are comparing a 1-parameter (unknown theta) model with a 0-parameter (known theta) model.

To be more precise, the models you are comparing are both 2-parameter Paretos. But the hypothesis is testing between one model where both parameters are known and fixed, versus another model whose theta parameter is variable. So there is exactly 1 degree of freedom, since only one quantity is allowed to change.

[JohnLocke]

You're asking a lotta questions about 4/C. If you get a good manual (like ASM), it should have all the answers you seek.