Suppose $\{X_n\}$ is a sequence of square-integrable i.i.d. random variables under the measure $\mathbb{P}$. Under the strong law of large numbers we have that \begin{align*} \mathbb{P}\left(\lim_{n\to\infty}\frac{1}{n}\sum_{j=1}^nX_j=\mathbb{E}_{\mathbb{P}}[X_1]\right)=1. \end{align*} If $\mathbb{Q}$ is an equivalent measure to $\mathbb{P}$ with $\frac{d\mathbb{Q}}{d\mathbb{P}}\in L^2(\mathbb{P})$ then each $X_i$ is still integrable under $\mathbb{Q}$ and the strong law of large numbers applied to $\mathbb{Q}$ will give \begin{align*} \mathbb{Q}\left(\lim_{n\to\infty}\frac{1}{n}\sum_{j=1}^nX_j=\mathbb{E}_{\mathbb{Q}}[X_1]\right)=1. \end{align*} As the measures are equivalent, this also means that \begin{align*} \mathbb{P}\left(\lim_{n\to\infty}\frac{1}{n}\sum_{j=1}^nX_j=\mathbb{E}_{\mathbb{Q}}[X_1]\right)=1 \end{align*} which implies that $\mathbb{E}_{\mathbb{P}}[X_1]=\mathbb{E}_{\mathbb{Q}}[X_1]$. However this is generally not the case.

Can anyone find where the argument above breaks down?


It occurred to me that the $\{X_n\}$ may not longer be independent under $\mathbb{Q}$ and so the strong law of large numbers doesn't necessarily hold for $\mathbb{Q}$.

  • $\begingroup$ yes, that is the reason. $\endgroup$
    – user159517
    Oct 25 '20 at 9:37
  • $\begingroup$ I originally thought this was assumed in your question. So would another way to state the conclusion is "If $P$ and $Q$ are equivalent measures such that the $X_i$ are i.i.d. under both measures, then $E_P[X_1] = E_Q[X_1]$ necessarily."? $\endgroup$
    – angryavian
    Oct 25 '20 at 21:20
  • $\begingroup$ @angryavian Yes, as the strong law can then also be applied to $\mathbb{Q}$. $\endgroup$
    – user375366
    Oct 25 '20 at 21:37

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