# Strong law of large numbers under equivalent measures

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?

• Is this the notion of equivalence you are using? Oct 25 '20 at 4:29
• @angryavian yes that's the one Oct 25 '20 at 4:43

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 $$\mathbb{Q}$$.
• 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."? Oct 25 '20 at 21:20
• @angryavian Yes, as the strong law can then also be applied to $\mathbb{Q}$. Oct 25 '20 at 21:37