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If $X$ and $Y$ have same distributions and $X \leq Y$ almost surely, does $X=Y$ almost surely? Here is a specific problem I met.

Assume $f(\omega)$ be a real-valued random variable which is defined on a metric dynamical system $(\mathcal{\Omega},\mathcal{F},\theta_{t},\mathbb{P})$ and $f(\theta_{t}\omega)\leq f(\omega)$ $\mathbb{P}$-a.s. Can we claim that $f(\theta_{t}\omega)=f(\omega)$ $\mathbb{P}$-a.s. ?

Recall that a metric dynamical system $(\mathcal{\Omega},\mathcal{F},\theta_{t},\mathbb{P})$ means that $(\mathcal{\Omega},\mathcal{F},\mathbb{P})$ is a probability space. $\theta_t\circ\theta_s=\theta_{t+s}$, $(t,\omega)\mapsto\theta_{t}\omega$ is measurable. And $\mathbb{P}$ is measure-preserving, that is, $\mathbb{P}(B)=\mathbb{P}(\theta_{t}B)$ for all $B\in\mathcal{F}$.

Here is my option. Clearly, $f(\theta_{t}\omega)$ and $f(\omega)$ has same distribution since $\mathbb{P}$ is measure-preserving. Thus we have $\mathbb{P}\{\omega:f(\theta_{t}\omega)\in[0,f(\theta_{t}\omega)]\}=\mathbb{P}\{\omega:f(\omega)\in[0,f(\theta_{t}\omega)]\}$.

Since $\mathbb{P}\{\omega:f(\theta_{t}\omega)\in[0,f(\theta_{t}\omega)]\}=1$, we have $\mathbb{P}\{\omega:f(\omega)\in[0,f(\theta_{t}\omega)]\}=1$, that is $f(\omega)\leq f(\theta_{t}\omega)$ $\mathbb{P}$-a.s.

Similarly, $\mathbb{P}\{\omega:f(\omega)\in[0,f(\omega)]\}=\mathbb{P}\{\omega:f(\theta_t\omega)\in[0,f(\omega)]\}$. And $\mathbb{P}\{\omega:f(\omega)\in[0,f(\omega)]\}=1$. Hence $\mathbb{P}\{\omega:f(\theta_t\omega)\in[0,f(\omega)]\}=1$, that is $f(\theta_{t}\omega)\leq f(\omega)$ $\mathbb{P}$-a.s.

However, we don't use the assumption $f(\theta_{t}\omega)\leq f(\omega)$ almost surely, we directly follow the conclusion $f(\theta_{t}\omega)=f(\omega)$ almost surely from the validity of the same distribution of $f(\theta_{t}\omega)$ and $f(\omega)$. It is not true in general. Could you tell me where I deduced wrong? Thanks!

JvT
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R-CH2OH
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    If $f$ is integrable, the answer is yes since $\int f\circ\theta_t=\int f$, for then $\int f-f\circ \theta_=0$ and since $f\geq f\circ\theta_t$ a.s., then $f=f\circ\theta$ a.s. – Mittens Dec 09 '23 at 15:39

2 Answers2

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Yes, $\arctan X\le \arctan Y$ (because $\arctan$ is increasing) and $E\arctan X=E \arctan Y$ (because $\arctan X$ and $\arctan Y$ have the same distribution). Hence, $\arctan X=\arctan Y$ a.s. which implies $X=Y$ as. [Use of $\arctan$ is to make expectations exist].

George Giapitzakis
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Kavi Rama Murthy
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Here is a slightly different proof. Let $z \in \mathbb{R}$. Then

\begin{align*} \mathbf{P}(Y \leq z < X) &= \mathbf{P}(Y \leq z) - \mathbf{P}(X \leq z, Y \leq z) \\ &= \mathbf{P}(Y \leq z) - \mathbf{P}(X \leq z) \tag{$\because$ $Y \leq X$ a.s.} \\ &= 0. \tag{$\because$ $Y \sim X$} \end{align*}

Since this is true for any $z \in \mathbb{R}$, for any countable dense subset $\mathcal{D}$ of $\mathbb{R}$ (such as $\mathbb{Q}$),

$$ \mathbf{P}(Y < X) = \mathbf{P}\biggl(\bigcup_{z \in \mathcal{D}} \{ Y \leq z < X \} \biggr) \leq \sum_{z \in \mathcal{D}} \mathbf{P}(Y \leq z < X) = 0. $$

Therefore $\mathbf{P}(Y \geq X) = 1$, and this in turn implies the desired conclusion.

Addendum. Note that the key idea of this proof is essentially the same as in @geetha290krm's answer, especially by noting that

$$ \mathbf{P}(Y \leq z < X) = \mathbf{E}[\mathbf{1}_{(z, \infty)}(X) - \mathbf{1}_{(z, \infty)}(Y)]. $$

Replacing $\mathbf{1}_{(z, \infty)}$ by $\arctan$ leads to @geetha290krm's solution.

Sangchul Lee
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