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Let, $X_1, X_2, \ldots, X_n$ be i.i.d. RVs with mean $0$ variance $1$ and finite fourth order raw moment. Find the limiting distribution of

$Z_{n}=\frac{\sqrt{n}(X_{1}X_{2}+X_{3}X_{4}+\cdots+X_{2n-1}X_{2n})}{X^2_{1}+X^2_{2}+\cdots+X^2_{2n}}$.

Ans: $E(X_{i}X_{j})=0$

$V(X_{i}X_{j})=1$, for all $i\neq j$. Then by CLT,

$U_{n}=\frac{X_{1}X_{2}+X_{3}X_{4}+\cdots+X_{2n-1}X_{2n}-0}{\sqrt{n}} \rightarrow N(0,1)$.

Again,let, $V_{n}=\frac{X^2_{1}+X^2_{2}+\cdots+X^2_{2n}}{n}$

Now, $E(V_{n})=2$, $V(V_{n})=\frac{4}{n}\rightarrow 0$ as $n\rightarrow \infty$.

Hence, $V_{n}$ converges to $2$ in probability. Therefore, by Slutsky's theorem, $Z_{n}\rightarrow N(0,\frac{1}{4})$.

Is it correct or not??

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2 Answers 2

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Yes, that is correct. Slutsky's Lemma it is.

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    $\begingroup$ Delete voters (and people not voting to delete) -- there's a meta question on this sort of "is this correct" question (and the obvious problem with a "yes" answer) here: stats.meta.stackexchange.com/questions/2685/… .... you may like to read the answers, consider them and perhaps vote for what you see as the best policy (or suggest a better one). $\endgroup$
    – Glen_b
    Commented Aug 20, 2018 at 10:06
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You have approached the question in a correct manner and your answer is correct. However, I want to rephrase with a little more explanation so to have more confidence.

Slutky's theorem: If $X_n \xrightarrow{p} a$ and $Y_n \xrightarrow{p} b$, then $\dfrac{X_n}{Y_n} \xrightarrow{d} \dfrac{a}{b}$.

Coming back to your question, you have correctly expressed $Z_n$ as $\dfrac{U_n}{V_n}$, where $U_n \xrightarrow{p} N(0, 1)$ and $V_n \xrightarrow{p} 2$. Using Slutky's theorem, $\dfrac{U_n}{V_n} \xrightarrow{d} N(0, 1/4)$

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