For any given nonnegative random variable $X$ and $0 < \rho \leq 1$, I define the following object.

$$\pi_{\rho}(X) = \int_0^{\infty}(P\{X > t\})^{\rho}\, dt$$

The inequality that I want to prove is the following property of this object.

If $X,Y \geq 0$ and $0 < \rho \leq 1$, then $$\pi_{\rho}(X+Y) \leq \pi_{\rho}(X) + \pi_{\rho}(Y)$$

Here is my attempt.

$$X + Y > t \implies X > \frac{t}{2} \quad \text{or} \quad Y > \frac{t}{2}$$ Starting from this implication I reason as follows \begin{align}P\{X + Y > t\} \leq P\left\{X > \frac{t}{2}\right\} + P\left\{Y > \frac{t}{2}\right\}\end{align} \begin{align}(P\{X + Y > t\})^{\rho} &\leq \left(P\left\{X > \frac{t}{2}\right\} + P\left\{Y > \frac{t}{2}\right\}\right)^{\rho}\\ &\leq \left(P\left\{X > \frac{t}{2}\right\}\right)^{\rho} + \left(P\left\{Y > \frac{t}{2}\right\}\right)^{\rho} \quad (\triangle)\\ &\leq \left(P\left\{2X > t\right\}\right)^{\rho} + \left(P\left\{2Y > t\right\}\right)^{\rho}\end{align}

The inequality $(\triangle)$ follows from the observation that for $a,b \geq 0$ and $0\leq p \leq 1$, $$a^p + b^p \geq (a+b)^p$$

Using the fact that for $a > 0$, $$\pi_{\rho}(aX) = a\pi_{\rho}(X)$$ I arrive at

$$\pi_{\rho}(X+Y) \leq 2\pi_{\rho}(X) + 2\pi_{\rho}(Y)$$

I am not sure if what I have written is correct. In any case I didn't get the inequality I needed. Can someone please point out if one of the inequalities I used is too conservative or I should have taken a whole other path?



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