inequality in statistics $P (ω : P (ω) ≤ \alpha) ≤ \alpha.$

Question

We say that a random variable $E (ω)$ is an e-value if

$E [E] ≤ 1,$

and we say that a random variable $P (ω)$ is a p-value if for any $\alpha \in [0, 1]$:

$P (ω : P (ω) ≤ \alpha) ≤ \alpha.$

Prove that $P = \min \{1, 1/E\}$ is a p-value. The range of the e-value E is non-negative.

My attempt:

I am trying to prove $P (ω : \min \{1, 1/E(w)\} ≤ \alpha) ≤ \alpha.$

I can write $P (ω : \min \{1, 1/E(w)\} ≤ \alpha) = 1 - P (ω : \min \{1, 1/E(w)\} > \alpha)$

$= 1-P(1>\alpha, 1/E(w) > \alpha)$

In the first case, $\alpha =1$, then any probability is $\leq 1$, so the statement holds. In the second case, $0 \leq \alpha <1$, we then have

$= 1-P(1>\alpha, 1/E(w) > \alpha)= 1-P( 1/E(w) > \alpha)$

Then I am stuck. I don't know how to use e-value condition.

Answer 1

$1\geq E [E] \geq E[EI_{E \geq \frac 1 {\alpha}}] \geq \frac 1 {\alpha} P(E \geq \frac 1 {\alpha})$ so $P(E \geq \frac 1 {\alpha}) \leq \alpha$. Finally, if $\alpha <1$ then $P(\min \{1,\frac 1 {E} \}\leq \alpha )) = P(E \geq \frac 1 {\alpha}) \leq \alpha$.