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Andrew NC
Andrew NC
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Let $\{T_n\}_n$ be a sequence of random variables, and let $X$ be another random variable.

Each $\mathbb{E}(T_n|X)$ is a random variable, therefore the statement "$\mathbb{E}(T_n|X)\rightarrow 0$ almost surely" makes sense in the sense of almost sure convergence of random variables.

What makes this difficult to unpack is that almost sure convergence is phrased in terms of individual $\omega$'s in the probability space, but the definition of $\mathbb{E}(T_n|X)$ is via the existence of the Radon-Nikodym derivative, and so is non-constructive...

The question is whether you can have an equivalent definition of this statement that does not use the Radon-Nikodym derivative as a black box. For example, is it equivalent to the statement: "for every measurable set $U$ in the co-domain of $X$, the integralmean of $T_n$ over $X^{-1}(U)$ converges to $0$"?

Let $\{T_n\}_n$ be a sequence of random variables, and let $X$ be another random variable.

Each $\mathbb{E}(T_n|X)$ is a random variable, therefore the statement "$\mathbb{E}(T_n|X)\rightarrow 0$ almost surely" makes sense in the sense of almost sure convergence of random variables.

What makes this difficult to unpack is that almost sure convergence is phrased in terms of individual $\omega$'s in the probability space, but the definition of $\mathbb{E}(T_n|X)$ is via the existence of the Radon-Nikodym derivative, and so is non-constructive...

The question is whether you can have an equivalent definition of this statement that does not use the Radon-Nikodym derivative as a black box. For example, is it equivalent to the statement: "for every measurable set $U$ in the co-domain of $X$, the integral of $T_n$ over $X^{-1}(U)$ converges to $0$"?

Let $\{T_n\}_n$ be a sequence of random variables, and let $X$ be another random variable.

Each $\mathbb{E}(T_n|X)$ is a random variable, therefore the statement "$\mathbb{E}(T_n|X)\rightarrow 0$ almost surely" makes sense in the sense of almost sure convergence of random variables.

What makes this difficult to unpack is that almost sure convergence is phrased in terms of individual $\omega$'s in the probability space, but the definition of $\mathbb{E}(T_n|X)$ is via the existence of the Radon-Nikodym derivative, and so is non-constructive...

The question is whether you can have an equivalent definition of this statement that does not use the Radon-Nikodym derivative as a black box. For example, is it equivalent to the statement: "for every measurable set $U$ in the co-domain of $X$, the mean of $T_n$ over $X^{-1}(U)$ converges to $0$"?

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Andrew NC
Andrew NC
  • 2.1k
  • 15
  • 20

Is "$\mathbb{E}(T_n|X)->0$\rightarrow 0 $ a.s." equivalent to a statement that does not involve the Radon–Nikodym derivative as a black box?

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Andrew NC
Andrew NC
  • 2.1k
  • 15
  • 20

Is "$\mathbb{E}(T_n|X)->0$ a.s." equivalent to a statement that does not involve the Radon–Nikodym derivative as a black box?

Let $\{T_n\}_n$ be a sequence of random variables, and let $X$ be another random variable.

Each $\mathbb{E}(T_n|X)$ is a random variable, therefore the statement "$\mathbb{E}(T_n|X)\rightarrow 0$ almost surely" makes sense in the sense of almost sure convergence of random variables.

What makes this difficult to unpack is that almost sure convergence is phrased in terms of individual $\omega$'s in the probability space, but the definition of $\mathbb{E}(T_n|X)$ is via the existence of the Radon-Nikodym derivative, and so is non-constructive...

The question is whether you can have an equivalent definition of this statement that does not use the Radon-Nikodym derivative as a black box. For example, is it equivalent to the statement: "for every measurable set $U$ in the co-domain of $X$, the integral of $T_n$ over $X^{-1}(U)$ converges to $0$"?