In Auburngrads/teachingApps: Apps for Teaching Statistics, R Programming, and Shiny App Development

Derivation of Greenwood's Formula

• Greenwood's formula is

• The text states that using the Delta Method with $\theta=q_j$ and $g(\theta)=S(t_i)$ results in $$\displaystyle \hat{S}(t_{i})\approx S(t_i)+ \sum_{j=1}^{i}\frac{\partial S}{\partial q_{j}}\vert_{q_{j}}(\hat{q}{j}-q{j})$$

• However, there's not much detail showing how this result was achieved

• The following example walks through the derivation of Equation 3.8

First, We Must Identify The Parameter $\theta$

• Our goal is to find a expression for $\widehat{Var}[\hat{F}(t_i)]$

  • From Equation 3.6, we see that $\hat{S}(t_i)$ is a function of $\hat{p_j}$

  • Since $\hat{F}(t_{i})=1-\hat{S}(t_{i})$, we know that $Var[\hat{F}(t_{i})]=Var[\hat{S}(t_{i})]$

  • Similarly, $\hat{q_j}=1-\hat{p_j}$ and $Var[\hat{q_j}]=Var[\hat{p_j}]$

• Therefore, $\theta = p_{j} \;\text{or}\;q_{j}$

Next, Find The Variance of The Parameter $Var[\theta]$

• Since $q_{j}=1-p_{j}$, we know that $Var[q_{j}]=Var[p_{j}]$

• Recall that $\hat{p}=\frac{x}{n}$ where $x$ is the number of observed "successes"

• Also, note that $\hat{p}$ is an unbiased estimator for $p$

$$E[\hat{p}]=E\left[\frac{x}{n}\right]=\frac{E[x]}{n}=\frac{np}{n}=p$$

• Since $\hat{p}$ is an unbiased estimator for $p$,

$$ \begin{aligned} Var[\hat{p}]&=Var[p]\\ &=Var\left[\frac{x}{n}\right]=\frac{1}{n^{2}}Var[x]=\frac{Var[x]}{n^{2}}\\\ &=\frac{np(1-p)}{n^{2}}=\frac{p(1-p)}{n}\ \end{aligned} $$

• Therefore, $Var[p]=\frac{p(1-p)}{n}=\frac{1-q(q)}{n}$

Now, What is The Function of the Parameter$g(\theta)?$

• From Equation 3.6

$$ \begin{aligned} S(t_{i})&=\prod_{j=1}^{i}[1-p_{j}], i=1,...,m \ &=(1-p_{1})(1-p_{2})...(1-p_{i})\ &=(q_{1})(q_{2})...(q_{i}) \end{aligned} $$

• Therefore, $g[p]=\prod_{j=1}^{i}[1-p_{j}]=\prod_{j=1}^{i}[q_{j}], i=1,...,m$

Finally, What is $\frac{\partial g(\theta_{i})}{\partial \theta_{i}}, \;\; i=1,...m$?

• In this case, it's easier to compute the derivatives $\frac{\partial S(t_{i})}{\partial q_{i}}, \;\; i=1,...m$

• We know that $\forall i \in 1,2,...m$

$$ \begin{aligned} \frac{\partial S(t_{i})}{\partial q_{i}}&=\frac{\partial \left( q_{1}q_{2}...q_{i-1}q_{i}\right)}{\partial q_{i}}\\\ &=q_{1}q_{2}...q_{i-1}\\\ &=\frac{S(t_{i})}{q_{i}}\ \end{aligned} $$

• Therefore, $\frac{\partial g(\theta_{i})}{\partial \theta_{i}}, \;\; i=1,...m=\frac{S(t_{i})}{q_{i}}$

Putting Everything Together

$$ \begin{aligned} Var[S(t_{i})]&=\sum_{j=1}^{i}\left[\frac{\partial S(t_{i})}{\partial q_{j}}\right]^{2}Var(q_{j})\\ &=\sum_{j=1}^{i}\left[\frac{S(t_{i})}{q_{j}}\right]^{2}\frac{p_{j}(1-p_{j})}{n_{j}}\\ &=S(t_{i})^{2}\sum_{j=1}^{i}\frac{p_{j}(1-p_{j})}{n_{j}(1-p_{j})^{2}}\\ &=S(t_{i})^{2}\sum_{j=1}^{i}\frac{p_{j}}{n_{j}(1-p_{j})} \end{aligned} $$

3.6.2 - Greenwood's formula

Substituting the estimated values into (3.8) gives what is known as Greenwood's formula

$$\displaystyle \widehat{Var}\left[\hat{F}(t_{i})\right]=\widehat{Var}\left[\hat{S}(t_i)\right]=\left[\hat{S}(t_{i})\right]^{2}\sum_{j=1}^{i}\frac{\hat{p}{j}}{n{j}(1-\hat{p_{j}})}$$

Greenwood's formula can then be used to estimate the standard error of $\hat{F}(t_{i})$ as

$$\hat{se}{\hat{F}}=\sqrt{\widehat{Var}[\hat{F}(t{i})]}$$

Auburngrads/teachingApps documentation built on June 17, 2020, 4:57 a.m.