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In the context of a 'reliability function' (as my textbook calls it), I'm being asked to prove the following. Note that below I provide all the context which I believe should be sufficient to solve the problem, but I can't seem to prove one direction of the problem.

Suppose we have function $r(p)$ and that $r(p_0)=p_0$ for a certain $p_0\in(0,1)$. Prove that, for $p\in (0,1)$:

  1. $r(p)\geq p$ for $p\geq p_0$
  2. $r(p)\leq p$ for $p\leq p_0$

Now, one property I am able to use is that: For each $r(p)$ we know $r(p^\alpha) \geq \{r(p)\}^\alpha$ for $0\leq \alpha \leq 1$.

Now, here is where I have come so far: The case $p=p_0$, present in both question 1 and 2, is trivial. In this case, $r(p)=r(p_0)=p_0$ so both $r(p)\geq p$ and $r(p)\leq p$ hold.

For question 1, I have said the following: Assume $p>p_0$. Then, $p=p_0^\alpha$ for some $0\leq \alpha \leq 1$ (Since $p_0 < 1$, raising it to a power $\alpha\leq 1$ will raise it's value: $p_0^\alpha > p_0$). Now we can just fill in $r(p) = r(p_0^\alpha)\geq r(p_0)^\alpha = p_0^\alpha = p$, with the inequality through the property of $r(p)$, $r(p_0) = p_0$ through the property of $p_0$ and $p=p_0^\alpha$ through my assumption.

However, I do not see how to apply a similar logic to question 2, since $p<p_0$ requires an $\alpha>1$ to make $p=p_0^\alpha$ hold, and choosing a $0\leq \alpha \leq 1$ such that $p^\alpha = p_0$ also doesn't get me anywhere: this allows me to fill in $p_0 = r(p_0) = r(p^\alpha) \geq r(p)^\alpha$, but there I get stuck.

If anyone has any insights... I would love to hear them.

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You are almost there. Just take the last inequality to the power $\frac{1}{\alpha}$

$p=p_0^{\frac{1}{\alpha}} \geq r(p)$

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  • $\begingroup$ Ah, of course, thanks. I guess I was just staring myself blind at the problem. $\endgroup$ – Radical Nov 30 '16 at 16:41
  • $\begingroup$ No worries. yes, it happens. $\endgroup$ – Med Nov 30 '16 at 16:42

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