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why there is no “error” term in survival analysis?

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why there is no “error” term in survival analysis?

Trying to understand formula for the Survival Function (survival analysis)Interaction term in survival analysisError term interpretation in the Cox PH modelHow to simulate survival times using true base line hazard functionMean Absolute Scaled ErrorApproximation error of the delta method: Berry Esseen type boundParameter estimates from extended cox model using the coxph() function from the survival package in RLikelihood-ratio test and hazard function in survival analysisMLE of Weibull in the context of Survival analysisTime ratio of AFT generalized gamma parametric regression in survival analysis

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1

$begingroup$

Where is the error terms behind the following model:

$$h_i(t) = h_0(t) exp left ( sum_k = 1^p beta_k z_ik right )$$

survival error

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edited 8 hours ago

Dimitris Rizopoulos

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asked 8 hours ago

masoudmasoud

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New contributor

masoud is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Apart from linear model based on normal distribution, I did not see any other model has error term.
  $endgroup$
  – user158565
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  the logistic model does not have a Gaussian distribution (normal distribution) but But it has the error term.
  $endgroup$
  – masoud
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Then you are wrong.
  $endgroup$
  – user158565
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  chapter 14 , the book Applied linear statistical model , author Kutner
  $endgroup$
  – masoud
  7 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  When I have time, I will write a long Answer to your Question to cover how to write statistical model.
  $endgroup$
  – user158565
  7 hours ago
  

  
  
  
  

add a comment | 

1

$begingroup$

Where is the error terms behind the following model:

$$h_i(t) = h_0(t) exp left ( sum_k = 1^p beta_k z_ik right )$$

survival error

share|cite|improve this question

edited 8 hours ago

Dimitris Rizopoulos

9,50011 gold badge66 silver badges2424 bronze badges

asked 8 hours ago

masoudmasoud

622 bronze badges

New contributor

masoud is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Apart from linear model based on normal distribution, I did not see any other model has error term.
  $endgroup$
  – user158565
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  the logistic model does not have a Gaussian distribution (normal distribution) but But it has the error term.
  $endgroup$
  – masoud
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Then you are wrong.
  $endgroup$
  – user158565
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  chapter 14 , the book Applied linear statistical model , author Kutner
  $endgroup$
  – masoud
  7 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  When I have time, I will write a long Answer to your Question to cover how to write statistical model.
  $endgroup$
  – user158565
  7 hours ago
  

  
  
  
  

add a comment | 

$begingroup$

Where is the error terms behind the following model:

$$h_i(t) = h_0(t) exp left ( sum_k = 1^p beta_k z_ik right )$$

survival error

share|cite|improve this question

edited 8 hours ago

Dimitris Rizopoulos

9,50011 gold badge66 silver badges2424 bronze badges

asked 8 hours ago

masoudmasoud

622 bronze badges

New contributor

masoud is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

share|cite|improve this question

edited 8 hours ago

Dimitris Rizopoulos

9,50011 gold badge66 silver badges2424 bronze badges

asked 8 hours ago

masoudmasoud

622 bronze badges

New contributor

masoud is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

share|cite|improve this question

edited 8 hours ago

Dimitris Rizopoulos

9,50011 gold badge66 silver badges2424 bronze badges

asked 8 hours ago

masoudmasoud

622 bronze badges

New contributor

masoud is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

edited 8 hours ago

Dimitris Rizopoulos

9,50011 gold badge66 silver badges2424 bronze badges

edited 8 hours ago

Dimitris Rizopoulos

9,50011 gold badge66 silver badges2424 bronze badges

asked 8 hours ago

masoudmasoud

622 bronze badges

New contributor

masoud is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked 8 hours ago

masoudmasoud

622 bronze badges

2 Answers
2

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2

$begingroup$

The distributional assumptions behind a relative risk model are hidden in the baseline hazard function $h_0(t)$. If you specify a form for this function, then you completely specify the distribution of your data.

For example, $h_0(t) = phi psi t^phi - 1$ corresponds to the Weibull distribution.

share|cite|improve this answer

answered 8 hours ago

Dimitris RizopoulosDimitris Rizopoulos

9,50011 gold badge66 silver badges2424 bronze badges

$endgroup$

add a comment | 

2

$begingroup$

There absolutely is an "error" in survival analysis.

You can define the "time to event" according to a probability model with some $$g(T) = b (X, t) + epsilon(X,t)$$

where $g$ would usually be something like a log transform. Of course requiring $epsilon$ to be normal, identically distributed, or even stationary is a rather strong assumption that just doesn't play out in real life. But if we allow $epsilon$ to be quite general, the Cox proportional hazard model is a special case of the above display. Is this an abuse of notation? Maybe. Note we are mpt guaranteed any of the desirable properties of independence between the parameters. But if we think carefully about what an error is, it's not that it doesn't exist, it's just not a helpful notation to facilitate scientific investigation.

This "fully parametric" approach can be very efficient when it's true. A fully parametric "Weibull" model is actually a lot like a linear regression model for survival data, where the scale parameter is a lot like an error variance (dispersion parameter)

You could predict survival time for a given subject, subtract that from observed survival time, and this "residual" can be flexibly modeled using semiparametric splines to describe the distribution and mean-variance relationship. More commonly, we use the difference of predicted and observed cumulative hazard (Schoenfeld) residuals and their theoretical basis to infer the appropriateness of the proportional hazards assumption.

share|cite|improve this answer

answered 8 hours ago

AdamOAdamO

37.7k22 gold badges6868 silver badges151151 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You're right, (like Cox-Snell residuals, martingale residuals),We predict in the statistics so there is an error term. can it be said that survival analysis itself is random implicit?
  $endgroup$
  – masoud
  7 hours ago
  

  
  
  
  

add a comment | 

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2

$begingroup$

The distributional assumptions behind a relative risk model are hidden in the baseline hazard function $h_0(t)$. If you specify a form for this function, then you completely specify the distribution of your data.

For example, $h_0(t) = phi psi t^phi - 1$ corresponds to the Weibull distribution.

share|cite|improve this answer

answered 8 hours ago

Dimitris RizopoulosDimitris Rizopoulos

9,50011 gold badge66 silver badges2424 bronze badges

$endgroup$

add a comment | 

2

$begingroup$

The distributional assumptions behind a relative risk model are hidden in the baseline hazard function $h_0(t)$. If you specify a form for this function, then you completely specify the distribution of your data.

For example, $h_0(t) = phi psi t^phi - 1$ corresponds to the Weibull distribution.

share|cite|improve this answer

answered 8 hours ago

Dimitris RizopoulosDimitris Rizopoulos

9,50011 gold badge66 silver badges2424 bronze badges

$endgroup$

add a comment | 

$begingroup$

The distributional assumptions behind a relative risk model are hidden in the baseline hazard function $h_0(t)$. If you specify a form for this function, then you completely specify the distribution of your data.

For example, $h_0(t) = phi psi t^phi - 1$ corresponds to the Weibull distribution.

share|cite|improve this answer

answered 8 hours ago

Dimitris RizopoulosDimitris Rizopoulos

9,50011 gold badge66 silver badges2424 bronze badges

$endgroup$

share|cite|improve this answer

answered 8 hours ago

Dimitris RizopoulosDimitris Rizopoulos

9,50011 gold badge66 silver badges2424 bronze badges

answered 8 hours ago

Dimitris RizopoulosDimitris Rizopoulos

9,50011 gold badge66 silver badges2424 bronze badges

answered 8 hours ago

Dimitris RizopoulosDimitris Rizopoulos

9,50011 gold badge66 silver badges2424 bronze badges

2

$begingroup$

There absolutely is an "error" in survival analysis.

You can define the "time to event" according to a probability model with some $$g(T) = b (X, t) + epsilon(X,t)$$

where $g$ would usually be something like a log transform. Of course requiring $epsilon$ to be normal, identically distributed, or even stationary is a rather strong assumption that just doesn't play out in real life. But if we allow $epsilon$ to be quite general, the Cox proportional hazard model is a special case of the above display. Is this an abuse of notation? Maybe. Note we are mpt guaranteed any of the desirable properties of independence between the parameters. But if we think carefully about what an error is, it's not that it doesn't exist, it's just not a helpful notation to facilitate scientific investigation.

This "fully parametric" approach can be very efficient when it's true. A fully parametric "Weibull" model is actually a lot like a linear regression model for survival data, where the scale parameter is a lot like an error variance (dispersion parameter)

You could predict survival time for a given subject, subtract that from observed survival time, and this "residual" can be flexibly modeled using semiparametric splines to describe the distribution and mean-variance relationship. More commonly, we use the difference of predicted and observed cumulative hazard (Schoenfeld) residuals and their theoretical basis to infer the appropriateness of the proportional hazards assumption.

share|cite|improve this answer

answered 8 hours ago

AdamOAdamO

37.7k22 gold badges6868 silver badges151151 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You're right, (like Cox-Snell residuals, martingale residuals),We predict in the statistics so there is an error term. can it be said that survival analysis itself is random implicit?
  $endgroup$
  – masoud
  7 hours ago
  

  
  
  
  

add a comment | 

2

$begingroup$

There absolutely is an "error" in survival analysis.

You can define the "time to event" according to a probability model with some $$g(T) = b (X, t) + epsilon(X,t)$$

where $g$ would usually be something like a log transform. Of course requiring $epsilon$ to be normal, identically distributed, or even stationary is a rather strong assumption that just doesn't play out in real life. But if we allow $epsilon$ to be quite general, the Cox proportional hazard model is a special case of the above display. Is this an abuse of notation? Maybe. Note we are mpt guaranteed any of the desirable properties of independence between the parameters. But if we think carefully about what an error is, it's not that it doesn't exist, it's just not a helpful notation to facilitate scientific investigation.

This "fully parametric" approach can be very efficient when it's true. A fully parametric "Weibull" model is actually a lot like a linear regression model for survival data, where the scale parameter is a lot like an error variance (dispersion parameter)

You could predict survival time for a given subject, subtract that from observed survival time, and this "residual" can be flexibly modeled using semiparametric splines to describe the distribution and mean-variance relationship. More commonly, we use the difference of predicted and observed cumulative hazard (Schoenfeld) residuals and their theoretical basis to infer the appropriateness of the proportional hazards assumption.

share|cite|improve this answer

answered 8 hours ago

AdamOAdamO

37.7k22 gold badges6868 silver badges151151 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You're right, (like Cox-Snell residuals, martingale residuals),We predict in the statistics so there is an error term. can it be said that survival analysis itself is random implicit?
  $endgroup$
  – masoud
  7 hours ago
  

  
  
  
  

add a comment | 

$begingroup$

There absolutely is an "error" in survival analysis.

You can define the "time to event" according to a probability model with some $$g(T) = b (X, t) + epsilon(X,t)$$

where $g$ would usually be something like a log transform. Of course requiring $epsilon$ to be normal, identically distributed, or even stationary is a rather strong assumption that just doesn't play out in real life. But if we allow $epsilon$ to be quite general, the Cox proportional hazard model is a special case of the above display. Is this an abuse of notation? Maybe. Note we are mpt guaranteed any of the desirable properties of independence between the parameters. But if we think carefully about what an error is, it's not that it doesn't exist, it's just not a helpful notation to facilitate scientific investigation.

This "fully parametric" approach can be very efficient when it's true. A fully parametric "Weibull" model is actually a lot like a linear regression model for survival data, where the scale parameter is a lot like an error variance (dispersion parameter)

You could predict survival time for a given subject, subtract that from observed survival time, and this "residual" can be flexibly modeled using semiparametric splines to describe the distribution and mean-variance relationship. More commonly, we use the difference of predicted and observed cumulative hazard (Schoenfeld) residuals and their theoretical basis to infer the appropriateness of the proportional hazards assumption.

share|cite|improve this answer

answered 8 hours ago

AdamOAdamO

37.7k22 gold badges6868 silver badges151151 bronze badges

$endgroup$

share|cite|improve this answer

answered 8 hours ago

AdamOAdamO

37.7k22 gold badges6868 silver badges151151 bronze badges

answered 8 hours ago

AdamOAdamO

37.7k22 gold badges6868 silver badges151151 bronze badges

answered 8 hours ago

AdamOAdamO

37.7k22 gold badges6868 silver badges151151 bronze badges