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What is the „intuitive” reason behind the following statements...

Let $\left(X_{n}\right)_{n}$ be a sequence of random variables. Let us assume $\mathbf{Q}\ll\mathbf{P}$, i.e. the $\mathbf{Q}$ probability measure is absolutely continuous with respect to the $\mathbf{P}$ probability measure.

  1. If $\left(X_{n}\right)_{n}$ is almost surely convergent under $\mathbf{P}$, then ${(X_n)}_n$ is almost surely convergent under $\mathbf{Q}$ as well.

  2. If $\left(X_{n}\right)_{n}$ is convergent in the stochastic convergence under $\mathbf{P}$, then ${(X_n)}_n$ is also convergent in the stochastic convergence under $\mathbf{Q}$ as well.

I think the first statement makes sense to me: If I randomly choose an $\omega$ according to $\mathbf{P}$, then it is 100% sure (i.e. it has $1$ probability), that ${(X_n)}_n$ will be convergent. Since the $\mathbf{Q}$-zero probabilities are the same as the $\mathbf{P}$-zero probabilities, then the $\mathbf{Q}$-one probabilities are the same as the $\mathbf{P}$-one probabilities. Therefore, if I randomly choose an $\omega$ according to $\mathbf{Q}$, then ${(X_n)}_n$ will also be convergent under $\mathbf{Q}$ almost surely.

However, it is not as obvious to me in case of stochastic convegence as in the case of almost sure convergence. Can you give me some „intuitive” explanation why it is indeed true? Also, is the $X$ limit the same under different measures? (I think it is, but I need confirmation...)

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  • $\begingroup$ What do you mean by stochastic convergence? Do you mean convergence in Probability? Or do you mean convergence in $L^{p}$ norm? Or do you mean convergence in distribution. $\endgroup$ Apr 17 at 10:22
  • $\begingroup$ I mean by that $\lim_{n\rightarrow\infty} \mathbf{P} (|X_{n}-X|>\varepsilon)=0$ for all $\varepsilon>0$. $\endgroup$
    – Kapes Mate
    Apr 17 at 13:52

1 Answer 1

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Clarification: When $\mathbf{Q}\ll\mathbf{P}$, every $\mathbf{P}$ null set is a $\mathbf{Q}$ nullset, but not necessarily vice versa.

I'm not sure how intuitive you will find these, but here are two approaches to the second statement.

1. If $X_n\to X$ in probability/$\mathbf{P}$ then there is a subsequence $\Bbb{K}\subset\Bbb{N}$ with $\lim_{n\to\infty, \, n\in\Bbb K}X_n=X$ a.s./$\mathbf{P}$. Conversely, if every subsequence $\Bbb K\subset\Bbb N$ has a further subsequence $\Bbb J\subset\Bbb K$ along which $X_n\to X$ a.s./$\mathbf{P}$, then $X_n\to X$ in probability/$\mathbf{P}$ .

Now suppose $X_n\to X$ in probability/$\mathbf{P}$. Let $\Bbb K\subset\Bbb N$. Then $(X_n)_{n\in\Bbb K}$ converges in probability/$\mathbf{P}$ to $X$, and so there is a subsequence $\Bbb J\subset\Bbb K$ with $\lim_{n\to\infty,\,n\in\Bbb J}X_n=X$ a.s./$\mathbf{P}$. By statement 1., $\lim_{n\to\infty,\,n\in\Bbb J}X_n=X$ a.s./$\mathbf{Q}$ as well. From the converse stated above (but now applied to $\mathbf{Q}$) we deduce that $X_n\to X$ in probability/$\mathbf{Q}$.

2. Use the $\epsilon$$\delta$ characterization of $\ll$. Namely, $\mathbf{Q}\ll\mathbf{P}$ if and only if for each $\epsilon>0$ there exists $\delta>0$ such that for each event $A$, if $\mathbf{P}(A)<\delta$ then $\mathbf{Q}(A)<\epsilon$. Using this characterization it's not hard to check that if $\mathbf{Q}\ll\mathbf{P}$ and if $(A_n)$ is a sequence of events with $\lim_n\mathbf{P}(A_n)=0$, then $\lim_n\mathbf{Q}(A_n)=0$.

Now take $A_n:=\{|X_n-X|>\eta\}$ for a fixed $\eta>0$.

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  • $\begingroup$ These look fine, specially the first one. Huge thanks. $\endgroup$
    – Kapes Mate
    Apr 17 at 16:25

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