Variance of Kaplan-Meier estimator Here and here and on the Wikipedia page it is stated that for estimating the variance of Kaplan Meier estimator $S(t)$ using delta method one can use the fact that: $$Var(log\hat{S}(t)) \approx \frac{1}{\hat{S}(t)^2}Var(\hat{S}(t))$$
My question is why $\frac{1}{\hat{S}(t)^2}$ is used instead of $\frac{1}{E(\hat{S}(t))^2}$, while the latter is the one that should be used in delta method? What is the justification?

For $t\in [y_j,y_{j+1})$, I denote the survival function by
$$S(t)=\prod_{j: y_j \le t} p_j$$
Notations have their usual meaning.
Suppose $g(x)=e^x$, so that by delta method,
$$\operatorname{Var}(\hat S(t))=\operatorname{Var}(g(\ln\hat S(t)))\approx [g'(E(\ln \hat S(t)))]^2 \operatorname{Var} (\ln \hat S(t))$$
Again by delta method,
$$E(\ln \hat S(t))=\sum_{j:y_j\le t}E(\ln \hat p_j)\approx \sum_{j: y_j \le t}\ln E(\hat p_j)=\sum_{j: y_j \le t}\ln p_j$$
Since $(g'(x))^2=e^{2x}$, note that
$$[g'(E(\ln \hat S(t)))]^2 \approx \exp\left\{2\sum_{j:y_j\le t} \ln p_j\right\}=\exp\left\{\ln \left(\prod_{j:y_j\le t} p_j \right)^2\right\}=\left(\prod_{j:y_j\le t} p_j \right)^2=(S(t))^2$$
Hence,
$$\operatorname{Var}(\hat S(t)) \approx (S(t))^2 \operatorname{Var} (\ln \hat S(t))$$
And
$$\widehat{\operatorname{Var}}(\hat S(t)) \approx (\hat S(t))^2 \widehat{\operatorname{Var}} (\ln \hat S(t))$$
