Stochastic Dominance between random variables

Question

Let $\xi,\eta,\nu$ be three independent random variables with supports contained in a bounded interval on $\mathbb{R}$. Suppose $\xi$ and $\eta$ are identical in the sense that $\mathbb{P}_\xi \equiv \mathbb{P}_\eta$ as push-forward measures on $\mathbb{R}$ (I think this is the definition of identical random variables? Please correct me if I'm mistaken). Further, suppose that $\xi,\eta$ first order stochastically dominate $\nu$. The definition I have here is that, for all $x\in\mathbb{R}$, I have $\mathbb{P}(\eta\geq x)\geq \mathbb{P}(\nu\geq x)$, and $\mathbb{P}(\eta\geq x^*)> \mathbb{P}(\nu\geq x^*)$ for some $x^*\in\mathbb{R}$. Similarly those statements hold for $\xi$.

I want to show the following: $\mathbb{P}(\eta\geq\xi) > \mathbb{P}(\nu\geq\xi)$.

I first tried to show the discrete case. Applying the law of total probability, I have $$\mathbb{P}(\nu\geq\xi) = \sum_{x\in X} \mathbb{P}(\nu\geq\xi\ |\ \xi=x)\mathbb{P}(\xi = x) =\sum_{x\in X} \mathbb{P}(\nu\geq x)\mathbb{P}(\xi = x)$$ where $X$ is the set of values upon which $\xi$ places positive probability. On the other hand, the LHS of the inequality is $$\mathbb{P}(\eta\geq\xi) = \sum_{x\in X} \mathbb{P}(\eta\geq\xi\ |\ \xi=x)\mathbb{P}(\xi = x) =\sum_{x\in X} \mathbb{P}(\eta\geq x)\mathbb{P}(\xi = x).$$ Note that both of these sums are convergent since $\mathbb{P}(\nu\geq \xi),\mathbb{P}(\eta\geq\xi)\in [0,1]$. Then, one can use the fact that, for all $x\in\mathbb{R}$, we have $\mathbb{P}(\eta\geq x)\geq \mathbb{P}(\eta\geq x)$, and $\mathbb{P}(\eta\geq x^*)> \mathbb{P}(\eta\geq x^*)$ for some $x^*\in\mathbb{R}$ to show the desired inequality.

I now try to prove for the continuous case. The idea is to apply a "variation" of the law of total probability. I found this formula on MSE and applied it here. Using this, I rewrite $$\mathbb{P}(\nu\geq\xi) = \int_\mathbb{R}\mathbb{P}(\nu\geq x)f_\xi(x)\ \mathrm{d}x$$ where $f_\xi$ is the density function of $\xi$. Similarly, I have $$\mathbb{P}(\eta\geq \xi) = \int_\mathbb{R}\mathbb{P}(\eta\geq x)f_\xi(x)\ \mathrm{d}x.$$ Note that $\mathbb{P}(\nu\geq x) = 1-F_\nu(x)$ where $F_\nu$ is the CDF of $\nu$, so $\mathbb{P}(\nu\geq x)$ is absolutely continuous with respect to $x$. The similar statement holds for $\mathbb{P}(\eta\geq x)$. I then have $$\mathbb{P}(\eta\geq \xi) - \mathbb{P}(\nu\geq \xi) = \int_\mathbb{R} \big[\mathbb{P}(\eta\geq x)-\mathbb{P}(\nu\geq x)\big]f_\xi(x)\ \mathrm{d}x.$$ We know that $\exists x^*$ such that $\big[\mathbb{P}(\eta\geq x^*)-\mathbb{P}(\nu\geq x^*)\big]>0$, but I'm struggling to show that $f_\xi(x^*)>0$ for such $x^*$ as well. If that can be done, then the continuity of the integrand will naturally complete the proof. Additionally, I'm also worried about the use of the "variation" of law of total probability here.

Lastly, I want to show this for the general case. I attempted to do so by writing $$\mathbb{P}(\nu\geq\xi) = \int_\mathbb{R}\mathbb{P}(\nu\geq x)\ \mathrm{d}\mathbb{P}_\xi(x)$$ and similarly for $\eta$: $$\mathbb{P}(\eta\geq\xi) = \int_\mathbb{R}\mathbb{P}(\eta\geq x)\ \mathrm{d}\mathbb{P}_\xi(x).$$ My idea is that I can use the Riemann-Stieltjes integral, which would mirror the proof of the continuous case. But I'm having trouble pinning down the integrability of $\mathbb{P}(\nu\geq\xi)$ and $\mathbb{P}(\eta\geq x)$ as functions of $x$.

Answer 1

There are some typos in line 6.

We say that $X$ and $Y$ are identically distributed if the push forward measures are equal. We cannot call them identical in this case; identical would mean $X(\omega)=Y(\omega)$ for all sample points $\omega$.

The result you are trying to prove follows immediately from Fubini's Theorem:

$P(\eta \geq \xi) =\int P(\eta \geq x)dP_{\xi} (x) \geq \int P(\nu \geq x)dP_{\xi} (x)=P(\nu \geq \xi)$ and strict inequality follows from the additional hypothesis that $P(\eta >x^{*}) >P(\nu >x^{*})$.