Como comparar a sobrevida mediana entre os grupos?

12

Estou estudando a sobrevida mediana usando Kaplan-Meier em diferentes estados para um tipo de câncer. Existem grandes diferenças entre os estados. Como posso comparar a sobrevida mediana entre todos os estados e determinar quais são significativamente diferentes da sobrevida mediana média em todo o país?

Misha
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Você poderia dar alguma indicação sobre tamanhos de amostra, período de tempo,% de sobrevivência etc. para que possamos ter uma idéia melhor do design do seu estudo?
chl
existem valores censurados nos dados - além dos valores maiores?
Ronaf
De fato, existem valores censurados nos dados e a população total é de aproximadamente 1500, a sobrevida global média é de 18 meses (intervalo de 300 a 600 dias) ... o período é o período 2000-2007.
M26:

Respostas:

6

Uma coisa a ter em mente com a curva de sobrevivência de Kaplan-Meier é que ela é basicamente descritiva e não inferencial . É apenas uma função dos dados, com um modelo incrivelmente flexível que está por trás deles. Isso é uma força porque significa que praticamente não há suposições que possam ser quebradas, mas uma fraqueza porque é difícil generalizá-la e que se encaixa em "ruído" e "sinal". Se você deseja fazer uma inferência, basicamente precisa introduzir algo desconhecido que deseja saber.

Agora, uma maneira de comparar os tempos médios de sobrevivência é fazer as seguintes suposições:

  1. Tenho uma estimativa do tempo médio de sobrevivência ti para cada um dos estados, dada pela curva de Kaplan-Meier.i
  2. Espero que o verdadeiro tempo médio de sobrevivência, para ser igual a essa estimativa. E ( T i | t i ) = t iTiE(Ti|ti)=ti
  3. Estou 100% certo de que o verdadeiro tempo médio de sobrevivência é positivo. Pr(Ti>0)=1

Agora, a maneira "mais conservadora" de usar essas suposições é o princípio da entropia máxima, para que você obtenha:

p(Ti|ti)=Kexp(λTi)

Onde e λ são escolhidos de forma que o PDF seja normalizado e o valor esperado seja t i . Agora temos:KλtEu

= K [ - e x p ( - λ T i )

1=0p(Ti|ti)dTi=K0exp(λTi)dTi
e agora temos E ( T i ) = 1
=K[exp(λTi)λ]Ti=0Ti==KλK=λ
E(Ti)=1λλ=ti1

And so you have a set of probability distributions for each state.

p(Ti|ti)=1tiexp(Titi)(i=1,,N)

Which give a joint probability distribution of:

p(T1,T2,,TN|t1,t2,,tN)=i=1N1tiexp(Titi)

H0:T1=T2==TN=t¯t¯=1Ni=1Nti is the mean median survivial time. The severe alternative hypothesis to test against is the "every state is a unique and beautiful snowflake" hypothesis HA:T1=t1,,TN=tN because this is the most likely alternative, and thus represents the information lost in moving to the simpler hypothesis (a "minimax" test). The measure of the evidence against the simpler hypothesis is given by the odds ratio:

O(HA|H0)=p(T1=t1,T2=t2,,TN=tN|t1,t2,,tN)p(T1=t¯,T2=t¯,,TN=t¯|t1,t2,,tN)
=[i=1N1ti]exp(i=1Ntiti)[i=1N1ti]exp(i=1Nt¯ti)=exp(N[t¯tharm1])

Where

tharm=[1Ni=1Nti1]1t¯

is the harmonic mean. Note that the odds will always favour the perfect fit, but not by much if the median survival times are reasonably close. Further, this gives you a direct way to state the evidence of this particular hypothesis test:

assumptions 1-3 give maximum odds of O(HA|H0):1 against equal median survival times across all states

Combine this with a decision rule, loss function, utility function, etc. which says how advantageous it is to accept the simpler hypothesis, and you've got your conclusion!

There is no limit to the amount of hypothesis you can test for, and give similar odds for. Just change H0 to specify a different set of possible "true values". You could do "significance testing" by choosing the hypothesis as:

HS,i:Ti=ti,Tj=T=t¯(i)=1N1jitj

So this hypothesis is verbally "state i has different median survival rate, but all other states are the same". And then re-do the odds ratio calculation I did above. Although you should be careful about what the alternative hypothesis is. For any one of these below is "reasonable" in the sense that they might be questions you are interested in answering (and they will generally have different answers)

  • my HA defined above - how much worse is HS,i compared to the perfect fit?
  • my H0 defined above - how much better is HS,i compared to the average fit?
  • a different HS,k - how much is state k "more different" compared to state i?

Now one thing which has been over-looked here is correlations between states - this structure assumes that knowing the median survival rate in one state tells you nothing about the median survival rate in another state. While this may seem "bad" it is not to difficult to improve on, and the above calculations are good initial results which are easy to calculate.

Adding connections between states will change the probability models, and you will effectively see some "pooling" of the median survival times. One way to incorporate correlations into the analysis is to separate the true survival times into two components, a "common part" or "trend" and an "individual part":

Ti=T+Ui

And then constrain the individual part Ui to have average zero over all units and unknown variance σ to be integrated out using a prior describing what knowledge you have of the individual variability, prior to observing the data (or jeffreys prior if you know nothing, and half cauchy if jeffreys causes problems).

probabilityislogic
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(+1) Very interesting. Your post also made me insert a comment in my answer.
GaBorgulya
Perhaps I have missed it, but where is M1 defined?
cardinal
@cardinal, my apologies - its a typo. will be removed
probabilityislogic
no apology necessary. Just wasn't sure if I had skipped over it while reading or was simply missing something obvious.
cardinal
4

Thought I just add to this topic that you might be interested in quantile regression with censoring. Bottai & Zhang 2010 proposed a "Laplace Regression" that can do just this task, you can find a PDF on this here. There is a package for Stata for this, it has yet not been translated to R although the quantreg package in R has a function for censored quantile regression, crq, that could be an option.

I think the approach is very interesting and might be much more intuitive to patients that hazards ratios. Knowing for instance that 50 % on the drug survive 2 more months than ones that don't take the drug and the side effects force you to stay 1-2 months at the hospital might make the choice of treatment much easier.

Max Gordon
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I don't know "Laplace Regression", but regarding your 2nd paragraph I wonder if I'm understanding it correctly. Usually in survival analysis (thinking in terms of accelerated failure time), we would say something like 'the 50th percentile for the drug group comes 2 months later than the 50th % for the control group'. Is that what you mean, or does the output of LR afford a different interpretation?
gung - Reinstate Monica
@gung: I think you're right in your interpretation - changed the text, better? I haven't used the regression models myself although I've encountered them recently in a course. Tt's an interesting alternative to regular Cox-models that I've used a lot. Although I probably need to spend more time digesting the idea I feel that it's probably easier for me to explain to my patients since I frequently use KM curves when explaining to my patients. HR demands that you really understand the difference between relative and absolute risks - a concept that can take some time to explain...
Max Gordon
Thank you @Misha for the link. The author has a reply here: onlinelibrary.wiley.com/doi/10.1002/bimj.201100103/abstract
Max Gordon
3

First I would visualize the data: calculate confidence intervals and standard errors for the median survivals in each state and show CIs on a forest plot, medians and their SEs using a funnel plot.

The “mean median survival all across the country” is a quantity that is estimated from the data and thus has uncertainty so you can not take it as a sharp reference value during significance testing. An other difficulty with the mean-of-all approach is that when you compare a state median to it you are comparing the median to a quantity that already includes that quantity as a component. So it is easier to compare each state to all other states combined. This can be done by performing a log rank test (or its alternatives) for each state.
(Edit after reading the answer of probabilityislogic: the log rank test does compare survival in two (or more) groups, but it is not strictly the median that it is comparing. If you are sure it is the median that you want to compare, you may rely on his equations or use resampling here, too)

You labelled your question [multiple comparisons], so I assume you also want to adjust (increase) your p values in a way that if you see at least one adjusted p value less than 5% you could conclude that “median survival across states is not equal” at the 5% significance level. You may use generic and overly conservative methods like Bonferroni, but the optimal correction scheme will take the correlations of the p values into consideration. I assume that you don't want to build any a priori knowledge into the correction scheme, so I will discuss a scheme where the adjustment is multiplying each p value by the same C constant.

As I don't know how to derive the formula to obtain the optimal C multiplyer, I would use resampling. Under the null hypothesis that the survival characteristics are the same across all states, so you can permutate the state labels of the cancer cases and recalculate medians. After obtaining many resampled vectors of state p values I would numerically find the C multiplyer below which less than 95% of the vectors include no significant p values and above which more then 95%. While the range looks wide I would repeatedly increase the number of resamples by an order of magnitude.

GaBorgulya
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Good advice about visualising the data. (+1)
probabilityislogic
@probabilityislogic Thanks! I also welcome criticism, particularly if constructive.
GaBorgulya
the only criticism I have is the use of p-values, but this is more a "chip on my shoulder" than anything in your answer - seems like if you are going to use p-values, then what you recommend is good. I just don't think using p-values is good. see here for my exchange with @eduardo in the comments about p-values.
probabilityislogic