It is well-known that if $X[n]$ is a sequence of i.i.d. random variables, each with mean $\mu_X$, then as $n\rightarrow \infty$, we have $\mu_X[n]\triangleq \frac{1}{n}\sum_{k=1}^n{X[k]}\rightarrow \mu_X$ (a.s.).
This is known as the strong law of large numbers (SLLN).

Does this result still hold if $X[n]$ is a sequence of i.i.d. random vectors? Can this be proved with the scalar version of SLLN?

  • $\begingroup$ Yes, the result holds for random vectors as long as the vector $\mu_X$ exists. I believe the one-dimensional version is a special case of the random vector version. $\endgroup$ Mar 4, 2016 at 22:05
  • $\begingroup$ @Greenparker Thank you! Do you know of some reference? I wonder if the vector version of the SLLN can be proved as follows. Suppose $X[n]$ is a sequence of iid $L$-dimensional random vectors. Then $\mu_X[n]$ comprises $L$ sequences of iid random variables, i.e. $\mu_{X_j}[n], j=1\sim L$. Apparently, $\mu_{X_j}[n] \rightarrow \mu_{X_j} \mathrm{(a.s.), }$ for any $j\in[1, L]$, due to the scalar SLLN. The proof is now immediate, since the intersection of any finite number of almost sure events is itself an almost sure event. Are there some flaws in this argument? Thanks a lot! $\endgroup$
    – syeh_106
    Mar 5, 2016 at 2:26


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.