# Showing that the restricted mean survival time (RMST) of iid $\mathsf{exp}(\lambda)$ survival time random variables is decreasing in $\lambda$

I would like to show that $$\frac{1-\exp(-\lambda\tau)}{\lambda}$$ is a decreasing function of $$\lambda$$ where $$\lambda,\tau>0$$.

Motivation: Suppose we have iid continuous survival time random variables $$T_i\sim\mathsf{exp}(\lambda)$$. The restricted mean survival time (RMST), defined as the mean survival time of all subjects up to time $$\tau$$ is given by

$$\mathsf{RMST}_{\tau}(\lambda)=\int_0^{\tau} \exp(-\lambda t)dt=\frac{1-\exp(-\lambda\tau)}{\lambda}$$

We are interested in the posterior probability that $$\mathsf{RMST}_{\tau}(\lambda)>\theta$$ given our data. I would like to show that $$\mathsf{RMST}_{\tau}(\lambda)>\theta$$ for sufficiently small values of $$\lambda$$. Otherwise, if the function $$\mathsf{RMST}_{\tau}(\lambda)$$ were not monotone, it wouldn't make sense to simply evaluate the proportion of posterior samples $$\hat{\lambda}$$ such that $$\mathsf{RMST}_{\tau}\left(\hat{\lambda}\right)>\theta$$.

In order for the inequality $$\mathsf{RMST}_{\tau}(\lambda)>\theta$$ to hold for sufficiently small values of $$\lambda$$, the derivative

$$\frac{d}{d\lambda}\left(\frac{1-\exp(-\lambda\tau)}{\lambda}\right)=\frac{\exp(-\lambda\tau)\lambda\tau-1+\exp(-\lambda\tau)}{\lambda^2}$$

should be less than zero for all $$\lambda>0$$ and for some fixed $$\tau>0$$. However, it's not clear to me that this is the case. I believe it should be because $$\exp(-\lambda t)$$ is decreasing in $$\lambda$$ and we're integrating over positive values of $$t$$. How can I show this more rigorously? Are my concerns valid in that $$\mathsf{RMST}_{\tau}(\lambda)$$ should be monotone in order to make meaningful inference?

• What do you get when you multiply the derivative by the positive number $\exp(\lambda \tau)$?
– user225256
Commented May 24, 2022 at 14:40
• We get $\frac{\lambda\tau-\exp(\lambda\tau)+1}{\lambda^2}$. It's not clear to me that this is negative for all $\lambda>0$ either.
– Remy
Commented May 24, 2022 at 14:46
• I just came across a possible issue. There could potentially be values of $\theta$ and $\tau$ where $\mathsf{RMST}_{\tau}(\lambda)\leq\theta$ for all $\lambda>0$. I will post it as a separate question.
– Remy
Commented May 24, 2022 at 18:33

Since $$\lambda^2 >0$$ showing the derivative is negative reduces to showing that $$\exp(-\lambda\tau)\lambda\tau-1+\exp(-\lambda\tau) = \exp (-\lambda \tau )(1 +\lambda \tau) -1 \leq 0$$
The right hand side is equivalent to $$e^{-x}(1+x) \leq 1$$ or $$e^x \geq 1+x$$
The function $$g : x \mapsto e^x -x - 1$$ has derivative $$e^x - 1 \geq 0$$ for $$x \geq 0$$ hence $$g$$ is an increasing function with $$g(0) = 0$$ thus for $$x \geq 0$$, $$g(x) \geq 0 \Rightarrow e^x \geq 1+x$$
Then for $$x \geq 0$$, $$e^{-x}(1+x) \leq 1$$ and $$\exp (-\lambda \tau )(1 +\lambda \tau) -1 \leq 0$$ if $$\lambda \tau \geq 0$$