How to derive expected value $E(\delta_i y_i)$?

Question

Weibull density function is defined as following:

$$f(x)=\frac{\alpha}{\theta}x^{\alpha-1}e^{-\frac{x^{\alpha}}{\theta}}\quad, x>0$$

Let $T_i$ denote the survival time and $C_i$ denote the censoring time for subject $i$. $T_i$ and $C_i$ are independent.

$T_i\sim \text{Weibull}(\alpha,K)$, and $C_i \sim \text{Weibull}(\alpha,\lambda)$

Let Y denote observed follow-up time, $Y_i=\min(T_i,C_i)$. Let $\delta_i$ be the indicator variable $\delta_i=(T_i\leq C_i)$.

How to derive the expected value of $E(\delta Y)$, which is the observed survival time?

I derived $$Y\sim \text{Weibull}\left(\alpha,{(\frac{1}{\lambda^\alpha}+\frac{1}{K^\alpha})^{-1/\alpha}}\right)$$ and $$P(\delta=1)=\frac{1}{1+\frac{K^\alpha}{\lambda^\alpha}}$$

But not sure what to next to get $E(\delta Y)$.

Any help will be appreciated!

P.S:

I guessed the following:

$Y=\min(T,C)$, and $Y\sim \text{Weibull}\left(\alpha,{(\frac{1}{\lambda^\alpha}+\frac{1}{K^\alpha})^{-1/\alpha}}\right)$, then $Y$ and $\delta$ should be independent. then it is easier to derive the expected value as the following: $$E(Y\delta)=E(Y)E(\delta)$$ It appears counter intuitive but the simulation confirms this:

\begin{verbatim}
nsim<-1000000
s.time<-rweibull(nsim,shape=2,scale=0.3)
c.time<-rweibull(nsim,shape=2,scale=0.6)
y<-pmin(s.time,c.time)
d<-(s.time<=c.time)
mean(y*d)
K<-0.3
lambda<-0.6
alpha<-2
mean.1<-(1/(lambda^alpha)+1/(K^alpha))^(-1/alpha)*gamma(1+1/alpha)
cen.prob<-1/(1+(K/lambda)^alpha)
mean.1*cen.prob
\end{verbatim}

Answer 1

This is most likely not the easiest way to do it, but it might be of help anyway.

There is a general result saying:

Let $U$ is a non-negative random variable with density $f$ and survival function $\bar{F}$, and $V$ is a non-negative random variable with density $g$ and survival function $\bar{G}$. Assume that $U$ and $V$ are independent and let $X=\min(U,V)$ and $Y=\mathbf{1}_{\{U\leq V\}}$. Then for any measurable and bounded function $T:\mathbb{R}\times\{0,1\}\to\mathbb{R}$ we have $$ {\rm E}[T(X,Y)]=\int_0^\infty T(x,0)\bar{F}(x)g(x)\,\mathrm dx+\int_0^\infty T(x,1)\bar{G}(x)f(x)\,\mathrm dx.\tag{1} $$

Now let $U\sim\mathrm{Weibull}(\alpha,K)$ and $V\sim\mathrm{Weibull}(\alpha,\lambda)$ be independent, and let $T(x,y)=x\cdot y$. Then $(1)$ says that $$ \begin{align} {\rm E}[\min(U,V)\mathbf{1}_{\{U\leq V\}}]&=\int_0^\infty x\cdot\bar{G}(x)f(x)\,\mathrm dx\\ &=\int_0^\infty x\exp\left(-\frac{x^\alpha}{\lambda}\right)\frac{\alpha}{K}x^{\alpha-1}\exp\left(-\frac{x^\alpha}{K}\right)\,\mathrm dx \end{align} $$ which is the same integral as that of Did's answer (which is more elementary anyway).

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Proof of the result: $$ \begin{align} {\rm E}[T(X,Y)]&={\rm E}[T(X,Y)\mathbf{1}_{\{U\leq V\}}]+{\rm E}[T(X,Y)\mathbf{1}_{\{U> V\}}]\\ &=\underbrace{{\rm E}[T(U,1)\mathbf{1}_{\{U\leq V\}}]}_{*}+\underbrace{{\rm E}[T(V,0)\mathbf{1}_{\{U>V\}}]}_{**}\\ \end{align} $$ Now $$ \begin{align} *&=\int_{\mathbb{R}^2}T(u,1)\mathbf{1}_{\{u\leq v\}}f_{(U,V)}(u,v)\,\mathrm dv\,\mathrm du\\ &=\int_{\mathbb{R}}\int_{\mathbb{R}}T(u,1)\mathbf{1}_{\{u\leq v\}}f(u)g(v)\,\mathrm dv\,\mathrm du\\ &=\int_0^\infty\left\{\int_0^\infty \mathbf{1}_{\{u\leq v\}}g(v) \,\mathrm dv\right\}T(u,1)f(u)\,\mathrm du\\ &=\int_0^\infty T(u,1)\bar{G}(u)f(u)\,\mathrm du, \end{align} $$ where $f_{(U,V)}$ denotes the joint density. The other expression $**$ is treated similarly.

Answer 2

Since $Y\delta=T\delta$ and $P(C\geqslant t)=\exp(-t^\alpha/\theta)$ for every $t\geqslant0$, one is looking for $$ E(Y\delta)=E(T;T\leqslant C)=E(T\exp(-T^\alpha/\theta))=\int_0^\infty t\mathrm e^{-t^\alpha/\theta}(\alpha/\theta)t^{\alpha-1}\mathrm e^{-t^\alpha/\theta}\mathrm dt. $$ The change of variable $x=2t^\alpha/\theta$ yields $$ E(Y\delta)=\int_0^\infty \mathrm e^{-x}(\theta x/2)^{1/\alpha}\mathrm dx/2=2^{-1-1/\alpha}\theta^{1/\alpha}\Gamma(1+1/\alpha).. $$ On the other hand, for every $y\geqslant0$, $$ [Y\geqslant y,\delta=1]=[C\geqslant T\geqslant y],\qquad[Y\geqslant y,\delta=0]=[T\geqslant C\geqslant y]. $$ Since $(T,C)$ and $(C,T)$ are identically distributed, these two events are equiprobable hence $Y$ and $\delta$ are independent. In particular, $$ E(Y\delta)=\tfrac12E(Y)=E(Y)E(\delta). $$