I have troubles applying the Law of Large Numbers (abbreviated LLN) when the random variable is transformed. Let all $X_i$ be independent and identically distributed and $X_i\in L^2(P)$.

For example I've read that when $$S_n=X_1^2+\cdots + X_n^2$$ Then the LLN says that $$\sum\limits_{n=1}^\infty\frac{S_n}{n}=\mathbb{E}(X^2)$$

Does that mean when $f\in L^1$ $$S_n=f(X_1)+\cdots + f(X_n)$$ Then the LLN says that $$\sum\limits_{n=1}^\infty\frac{S_n}{n}=\mathbb{E}(f(X))$$

Second example. If I have $f(S_n)$ like for instance $$\frac{1}{\sqrt{1-(n^{-1}S_n)^2}}=\frac{1}{\sqrt{1-(n^{-1}(X_1+\cdots+X_n))^2}}$$

Does that mean that the LLN say that $$\frac{1}{\sqrt{1-(n^{-1}S_n)^2}}\xrightarrow{\text{a.s.}} \frac{1}{\sqrt{1-(\mathbb{E}(X))^2}}$$

Thank you for explanation.

  • $\begingroup$ Never use an acronym such as LLN without a full text explanation, for example in this way : "the Law of Large Numbers (subsequently abbreviated as LLN)...". $\endgroup$ – Jean Marie Sep 6 '16 at 10:42
  • $\begingroup$ @JeanMarie I changed it. $\endgroup$ – Matriz Sep 6 '16 at 10:44
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    $\begingroup$ I don't really understand your difficulty. What i can tel is that, whenever you have $f$ a deterministic function, $X_1,X_2,...$ a sequence of iid rv such that $f(X_1)\in L^2(P)$, then the following convergence holds: $(1/n)\sum f(X_i)\to E(f(X_1))$. $\endgroup$ – anonymus Sep 6 '16 at 10:50
  • $\begingroup$ where do you apply $f(S_n)$? you just have $f(X_i)$ as far as I can tell... $\endgroup$ – user190080 Sep 6 '16 at 10:54
  • $\begingroup$ @user190080 The first example is different from the second example. I have in the second one $f(x)=\frac{1}{\sqrt{1-(n^{-1}x)^2}}$ and then $f(S_n)$. $\endgroup$ – Matriz Sep 6 '16 at 10:59

The LLN says that, if $Y_1,\ldots,Y_n,\ldots$ are independent and identically distributed random variables in $L^1$, then $$\frac{1}{n}\sum_{k=1}^nY_k\stackrel{\text{a.s.}}{\rightarrow}\mu,$$ where $\mu=E[Y_i]$ for all $i$.

If you call $S_n=Y_1+\ldots +Y_n$, then $$\frac{S_n}{n}\stackrel{\text{a.s.}}{\rightarrow}\mu$$ (your notation with the $\sum_{n=1}^{\infty}$ is not correct).

If $Y_i=f(X_i)$, you need $f(X_i)\in L^1$ to apply the LLN and conclude that $$\frac{1}{n}\sum_{k=1}^nf(X_k)\stackrel{\text{a.s.}}{\rightarrow}E[f(X)].$$

In your final question, $n^{-1}S_n\rightarrow E[X]$ by the LLN. Assuming that $n^{-1}S_n\in (-1,1)$ and $E[X]\in (-1,1)$, as the function $x\mapsto 1/\sqrt{1-x^2}$ is continuous on $(-1,1)$, you can apply the continuous mapping theorem: $$ \frac{1}{\sqrt{1-(n^{-1}S_n)^2}}\stackrel{\text{a.s.}}{\rightarrow} \frac{1}{\sqrt{1-E[X]^2}}.$$

  • $\begingroup$ Thank you very much, this was very helpful! $\endgroup$ – Matriz Sep 6 '16 at 11:20

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