Show that $\exp(-\exp(\log(-\log(\hat{S}(t)))+A)=\hat{S}(t)^{\exp(A)}$

Question

Let $L(t)=\log(-\log S(t))$ where $S(t)$ is a survival function. Then a confidence interval for $S(t)=\exp(-\exp(L(t))$ is given by $$[\exp(-\exp(\hat{L}(t)+A));\exp(-\exp(\hat{L}(t)-A));$$ and taking $\hat{L}(t)=\log(-\log (\hat{S}(t)))$ the confidence interval is $$[\hat{S}(t)^{e^{A}};\hat{S}(t)^{e^{-A}}]$$

where $A=\sigma(\hat{L}(t)$)

This result can be found here http://bcb.dfci.harvard.edu/DOCS/Notes/TS16/train16.pdf slide number 13.

I already understood the expression for the variance of the estimator, but the problem is the resulting confidence interval.

$$\exp(-\exp(\hat{L}(t)+A))=\exp(-\exp(\log(-\log(\hat{S}(t)))+A)$$

$$=\exp(\log(\hat{S}(t))\times \exp(A))=\hat{S}(t)^{\exp(A)}$$

I do not know if it is solved that way or if I forced the result. Can anyone help?

Answer 1

That looks okay to me.

It's not clear to me what "forced the result" means - if it's a valid manipulation of the expression, it's a valid manipulation.

You'd do the same kind of thing for the lower bound.