# Tag Info

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Note that the p values from cox.zph() shouldn't be thought of in the same way that you use p values in standard tests of null hypotheses. The burden is on you to demonstrate that PH holds; it's not enough to show that you can't rule out the PH null hypothesis because p > 0.05. A p value on the order of 0.1, as you have for some of your cox.zph() results, is ...

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If an adjustment for covariates is needed to illustrate the relationship of your continuous variable of interest to outcome fairly, then you should consider displaying 2 predicted survival curves that differ only in terms of that variable of interest. For the predicted curves, choose representative values of the other covariates (same set of covariate ...

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BW Turnbull - ‎1976 paper: The empirical distribution function with arbitrarily grouped, censored and truncated data

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This is called interval-censored data - the true (unobserved) value you are trying to estimate for each individual lies within an (observed) interval. There are a number of ways to deal with this. Knowing nothing else, you could take a multiple imputation approach in which the imputation model is simply a discrete uniform variable covering your interval. In ...

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The difference here is between when the study started administratively (presumably Study Start) and when each Subject entered the study (presumably on Study Start plus the days of Delay_Entry). Then the Actual_Follow_Up_Days represents the follow-up time for each Subject after that Subject's entry into the study. The Actual_Follow_Up_Days is almost ...

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Survival analysis is used to model the time it takes for a certain event to occur from a defined starting point. So say that you had a group of people in good status and you were interested in how long it took them to develop an intermediate or bad status, survival analysis might be a useful approach. But if your "purpose is to determine the probability ...

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The primary complication in the analysis of survival data is censoring, i.e., the fact that you do not observe the event of interest for some subjects in your study. Many statistical procedures do not work with censored data, such as the empirical cumulative distribution function. To obtain valid estimates you will need to use statistical techniques that ...

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Simple Linear Regression Model $$Y_i=B_0+B_1 X_i+ε_i$$ Where $Y_i$ is the value of the response variable in the ith trial $ε_i$ is a random error term with mean $E[ε_i]=0$ and variance $σ^2 [ε_i ]=σ^2$ $$E[Y_i ]=B_0+B_1 X_i$$ Consider the simple linear regression model Y_i=...

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This Answer is limited to frequentist statistics and statistical model without random effect. In fact, the statistical modeling is to find the conditional distribution of response variable conditional on fixed values of the covariates, i.e., distribution of $Y|X=x$. When writing the statistical model, abiding following 3 steps will keep you from ...

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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 ...

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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.

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Let $(X_1,Y_1)$ be a bivariate continuous random vector, independently and identically distributed with the vector $(X_2, Y_2)$. $X_1$ may be dependent with $Y_1$, and $X_2$ may be dependent with $Y_2$. Kendall's tau is defined as $$\tau_{X,Y} = \Pr[(X_1-X_2)(Y_1-Y_2) > 0] - \Pr[(X_1-X_2)(Y_1-Y_2) < 0]$$ Now, we have $$\Pr[(X_1-X_2)(Y_1-Y_2) < 0] ... 0 In Cox proportional hazard model, there is unspecified baseline hazard function \lambda_0(t). It is not constant and is the function of time. For subject id=11, the first table says the subject went through from t_1 = 62 to t_2=243. So the subject should has the hazard \lambda_0(t)exp(X\beta), for  t=62 \text{ to } 243. From table 2, the subject ... 1 This looks like yet another misapplication of statistical significance testing. This statement: Many interventional studies have found that the intervention reduced the primary end-point (a composite which includes cardiovascular mortality) without affecting total mortality. is incorrect, at least in the example shown. It is true that overall ... 1 You have made quite a bit of progress; all that remains is to show that the numerator of your derivative h' is negative. Expanding the products of integrals into double-integrals you get:$$ \begin{aligned} \text{NUM} &\equiv \left(\int_0^\infty \lambda e^{-\lambda t}f(\lambda) \ d\lambda\right)^2 -\left(\int_0^\infty e^{-\lambda t}f(...

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