1

Let $X$ and $Y$ be nonnegative integer valued random variables (with same mean). It is customary to define convex order stochastic dominance for such variables (denoted $X<Y$) if $\sum_{j\geq n}\mathbb{P}(X>j)\leq \sum_{j\geq n}\mathbb{P}(Y>j)$ for all $n\in\mathbb{N}_0$.

The general definition is, however, stronger: $\int_t^\infty\mathbb{P}(X>x)dx\leq \int_t^\infty\mathbb{P}(Y>x)dx$ for all $t\in\mathbb{R}_0^+$. My question is: how is the first condition equivalent to the stronger one when the support is restricted to $\mathbb{N}_0$? More specifically, how does it imply the stronger one?

I made an attempt, but when considering non integral $t$, I do not seem to find a way to exploit the first condition in order to get also $\mathbb{P}(X>\lfloor t\rfloor)\leq \mathbb{P}(Y>\lfloor t\rfloor)$, which is the bit I am not able to compare. This would imply the result, since one can decompose the integral, say, $\int_t^\infty\mathbb{P}(X>x)dx = (1- \left\lbrace t \right\rbrace )\mathbb{P}(X>\lfloor t\rfloor)+\sum_{j\geq\lceil t\rceil}\mathbb{P}(X>j)$. Perhaps I am missing the main point and there is something more sophisticated going on. Anybody familiar with this reduction? It seems fairly common in reliability theory.

xyz
  • 1,366

1 Answers1

1

It follows from piecewise linearity. Let $$F(t) = \int_t^\infty P(X > x)-P(Y > x)dx$$ where $t > 0$. The first condition basically states for all nonnegative integers $n$, $F(n) \leq 0$. However, note that for all $t \in (n,n+1),$ $$F'(t) = P(X > n)-P(Y > n).$$ I.e., $F'$ is constant in all intervals $(n,n+1)$, and continuous, so $F$ is a piecewise linear function. Now note that from the first condition, $F(n) = F(n+1) \leq 0$, so necessarily any convex combination of them is $\leq 0$ as well. I.e., for $n < t < n+1$, $$F(t) = F(n) + (t-n)(F(n+1)-F(n))= (n+1-t)F(n) + (t-n)F(n+1) \leq 0.$$ This proves that the first condition implies the second.

1mdlrjcmed
  • 1,191