Confidence Interval of S(x) using Nelso Aalen

[wheez17]

I'm a little confused on calculating this...Do we first have to calculate the confidence interval for H(x) and then raise both points to e^(-x) or is there another way to do this?

[Abraham Weishaus]

Well, the alternative would be to use the delta method. Depends whether you feel survival time is normally or lognormally distributed.

[UFActuary]

Not hard, the last thing you do is convert the H(t) into S(t) using the old familiar exp[-H(t)]=s(t) for each of the values

[Ionic Order]

I'm a little confused on calculating this...Do we first have to calculate the confidence interval for H(x) and then raise both points to e^(-x) or is there another way to do this?

You don't. But you do have to find U (I'm assuming you're using ASM for this chapter)

[UFActuary]

You don't need 'u' if you are doing a linear confidence interval based on Nelson Ailen

You take squareroot of the variance of. Sum totals of s/r-squareds

Use the H(t) from your sums of s/r and then
H(t) +- (1.645 times sqrt of variance) for 90% linear confidence, 1.96 for 95%

AFTER you do all this Take the upper and lower limit of the confidence interval and
Exp(-H(t) it to turn it into an S(t) value instead

You need 'u' only if you are trying to transform the interval into the log form

For H(t). You would then take
e^(1.645*sqrt of variance/H(T) to get u

Upper limit is then H(T)*u
Lower limit is then H(T)/u

[mathmajor]

Yea, linear confidence is just like any other typical confidence interval. Find the variance of your estimator, make it a standard deviation, multiply that s.d. by the appropriate factor (based on confidence %) and add/subtract from the estimator. Voila.

Log estimators require using the exponentials and whatnot. Those needs to be firmly in your memory also.

[Ionic Order]

You take squareroot of the variance of. Sum totals of s/r-squareds

Use the H(t) from your sums of s/r and then
H(t) +- (1.645 times sqrt of variance) for 90% linear confidence, 1.96 for 95%

AFTER you do all this Take the upper and lower limit of the confidence interval and
Exp(-H(t) it to turn it into an S(t) value instead



Could you please explain why we build the linear confidence interval first then exponentiate the endpoints rather than the other way around?

[mathmajor]

If you have found a 90% CI on H(t), then you have also found a 90% CI on S(t). This is because they are a direct, 1 to 1 function of each other, and there may be other requirements. Maybe also because they are strictly increasing/decreasing in (0,1)...