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I got following question:

We let $T^*_1,...,T^*_n$ be independent survival times for n patients and we let $X_i\in\{0,1\}$ indicate if the i-th patient is treated $(X_i=1)$ or not $(X_i=0)$. We are interested in the estimating of the treatment effect by following parameter: $$P=\frac{median(T_1^*|X_1=1)}{median(T_1^*|X_1=0)}$$

We assume that $T_i^*$ follows a Weibull distribution with shape parameter $\gamma>0$ and scale parameter $\alpha_{x_i}$ so hazard rate is: $$\lambda(t)=\alpha_{x_i} \gamma t^{\gamma-1}$$

Then I have to show that $$p=(\frac{\alpha_0}{\alpha_1})^{\frac{1}{\gamma}}$$

It look like a very simple exercise but i have looked a long time in my book after a formula for these conditional medians and looking for how I maybe can use the given hazard rate. Can anyone help me with some hints?

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1 Answer 1

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The median of a continous random variable $X$ is the value $\alpha$ such that $$ \mathbb P(X \geq \alpha) = \mathbb P(X \leq \alpha) = \frac{1}{2}. $$

In the case of a Weibull random variable $T$ with hazard rate

$$ \lambda(t) = \alpha \gamma t^{\gamma -1} $$

we have

\begin{align*} \mathbb P(T \geq t) &=\exp \left ( - \int_0^t \lambda(u)du \right) \\ &= \exp\left(\alpha \gamma \int_0^t u^{\gamma-1}du \right) \\ &=\exp \left(-\alpha t^\gamma \right). \end{align*}

The median is then the value $t$ for which $\mathbb P(T \geq t) = \frac{1}{2}$, \begin{align*} \mathbb P(T \geq t) = \frac{1}{2} &\iff \exp \left(-\alpha t^\gamma \right) = \frac{1}{2} \\ &\iff -\alpha t^\gamma = -\log(2) \\ &\iff t^\gamma = \frac{\log(2)}{\alpha} \\ &\iff t = \left( \frac{\log(2)}{\alpha} \right )^{\frac{1}{\gamma}}. \end{align*}

For $i \in \{0,1 \}$ the median $m_i$ of the $i$th group is given by $$ m_i = \left( \frac{\log(2)}{\alpha_i} \right )^{\frac{1}{\gamma}} $$

Taking the ratio of $m_1$ and $m_0$ we get \begin{align*} \frac{m_1}{m_0} &= \frac{\left( \frac{\log(2)}{\alpha_1} \right )^{\frac{1}{\gamma}}}{\left( \frac{\log(2)}{\alpha_0} \right )^{\frac{1}{\gamma}}} \\ &= \left( \frac{\log(2)}{\alpha_1} \right )^{\frac{1}{\gamma}} \times \left( \frac{\alpha_0}{\log(2)} \right )^{\frac{1}{\gamma}} \\ &= \left(\frac{\alpha_0}{\alpha_1} \right)^{\frac{1}{\gamma}} \end{align*}

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  • $\begingroup$ Thank you. That makes sense $\endgroup$
    – Lifeni
    Commented Oct 25, 2021 at 13:33

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