Survival analysis: Treatment "parameter" for Weibull 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?

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*}
