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For sequence ${X_{n}}$ and ${Y_{n}}$, for each $n$, ${X_{n}}$ and ${Y_{n}} $share same distribution and they are independent. (think of ${Y_{n}}$ constructed in that way :independent.) $0 \le X_{n}$, $Y_{n}\le 1$ for all n.

(${(1-X_{n+1})}$ / ${X_{n+1}}) =(({1-X_{n}}$) / ${X_{n}}$) * ((${1-Y_{n}}$) /${Y_{n}}$)

(${(1-Y_{n+1})}$/ ${Y_{n+1}}) =(({1-X_{n}}$ /${X_{n}}$) *((${1-Y_{n}}$) /${Y_{n}}$)

Then, how can we calculate $ \lim\limits_{n \to \infty}\operatorname{Var}(X_{n})$?

Bernard
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LANIOP
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1 Answers1

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Start by rewriting equations

$$X_{n}(1-X_{n+1})Y_{n} = (1-X_{n})X_{n+1}(1-Y_{n})$$ and $$X_{n}Y_{n}(1-Y_{n+1}) = (1-X_{n})Y_{n+1}(1-Y_{n}).$$

Also recall by assumption that $$X_{n} \overset{d}{\to} X_{\infty}$$ and $$Y_{n} \overset{d}{\to} Y_{\infty}$$ and so $$\mathbb{E}[Y_{n}(Y_{n+1}-Y_{n})] \to 0.$$

Then take expectations and pass to limits (by bounded convergence theorem):

$$(\mathbb{E}[X_{\infty}]-\mathbb{E}[X_{\infty}^{2}])\mathbb{E}[Y_{\infty}] = (\mathbb{E}[X_{\infty}]-\mathbb{E}[X_{\infty}^{2}])(1-\mathbb{E}[Y_{\infty}]),$$

and deduce that either $X_{\infty} \overset{p}{=} 0$, $X_{\infty} \overset{p}{=} 1$ or $\mathbb{E}[Y_{\infty}] = \frac{1}{2}$. Do the same with the other equation, follow up with a similar calculation for $1/X_{\infty}$ and see what you can deduce.

Christopher K
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  • Thank you for the elaboration. But where does the assumption(Convergence in distribution) come from? And where does E[Yn(Yn+1−Yn)]→0. also come from?? – LANIOP Nov 27 '21 at 23:09