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View Full Version : Confidence Interval of S(x) using Nelso Aalen

wheez17
01-05-2012, 09:20 PM
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
01-05-2012, 09:54 PM
Well, the alternative would be to use the delta method. Depends whether you feel survival time is normally or lognormally distributed.

UFActuary
01-08-2012, 12:46 AM
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
01-08-2012, 04:54 PM
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
01-08-2012, 09:54 PM
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
01-09-2012, 10:46 AM
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
02-05-2012, 02:11 PM
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
02-06-2012, 10:06 AM
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)...