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Let $\{U_n\}$ be a sequence of i.i.d Uniform$(0,1)$ random variables and $\{a_k\}$ be a sequence of i.i.d random variables such that, $P(a_k = \pm1) = \frac{1}{2}$ and are independent of the $\{U_n\}$. Define the quantities,

$$X_n = \sum\limits_{k=1}^nU_ka_k, \;\;\;\; Y_n = \sum\limits_{k=1}^n(U_k^2 - 1/2)a_k$$

Find the limit, $\lim\limits_{n\rightarrow\infty}P\left(|X_n| > Y_n\sqrt{\frac{20}{7}}\right)$.

We can easily calculate that $$E[U_ka_k] = 0, \; E[U_k^2] = \frac{1}{3}$$ and similarly, $$E[(U_k^2-1/2)a_k] = 0, \; E[(U_k^2-1/2)^2] = \frac{7}{60}$$ So by the Central Limit Theorem, we get,

$$ \frac{1}{\sqrt{n}}X_n \xrightarrow{D} N(0,1/3), \;\;\;\; \frac{1}{\sqrt{n}}Y_n \xrightarrow{D} N(0,7/60)$$

Ideally, I can use something like the Continuous Mapping Theorem to deduce the limiting distribution of $\dfrac{|X_n|}{Y_n}$, but the question makes no additional assumptions on the joint distribution of $(X_n, Y_n)$. Is there another approach I can take from here?

EDIT: I went ahead and calculated the covariance of $U_ka_k$ and $(U_k^2-1/2)a_k$,

$$ E[U_ka_k(U_k^2-1/2)a_k] = E[U_k(U_k^2-1/2)] =\int_0^1(u^3-1/2u)du = 0$$

since $a_k^2 = 1$ with probability $1$. Now we can apply the multivariate CLT like @LandonCarter says.

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    $\begingroup$ How could there be any assumptions on the joint distribution of $(X_n,Y_n)$? Can't you calculate it? Every piece of information is given $\endgroup$ – mathworker21 Aug 5 '18 at 5:51
  • $\begingroup$ Are the $\{U_n\}$ independent of the $\{a_k\}$? $\endgroup$ – angryavian Aug 5 '18 at 5:53
  • $\begingroup$ Yes sorry I forgot to add that assumption. The question has been edited. $\endgroup$ – Flowsnake Aug 5 '18 at 5:55
  • $\begingroup$ @mathworker21 Sorry I just realized that now. One can easily calculate the covariance between the $U_ka_k$ and $(U_k^2-1/2)a_k$ to be $0$. Thank you. $\endgroup$ – Flowsnake Aug 5 '18 at 18:09
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By multivariate CLT, $$(X_n/\sqrt{n}, Y_n/\sqrt{n})\stackrel{d}{\to}N((0,0),diag(1/3,7/60))$$so in the limit $X_n/\sqrt{n}$ and $Y_n/\sqrt{n}$ become jointly independent Normals.

So $P[|X_n|>Y_n\sqrt{20/7}]=P[\dfrac{Y_n/\sqrt{n}}{|X_n|/\sqrt{n}}<\sqrt{7/20}]\to P[N_1/|N_2|<\sqrt{7/20}]$ as $n\to\infty$ where $N_1\sim N(0,7/60)$ and $N_2\sim N(0,1/3)$ independent. Note $N_1=\sqrt{7/60}Z_1$ and $N_2=\sqrt{1/3}Z_2$ where $Z_1,Z_2\sim N(0,1)$ iid.

So the limiting probability (after some simplification) is $P[Z_1/|Z_2|<1]$. Maybe you can simplify this using the fact that $Z_1/Z_2$ is $Cauchy (0,1)$.

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  • $\begingroup$ Ah, I was a bit confused as to why the limiting joint distribution was independent. Then I calculated that the covariance between $U_ka_k$ and $(U_k^2-1/2)a_k$ is $0$. Thank you. $\endgroup$ – Flowsnake Aug 5 '18 at 18:07
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    $\begingroup$ By the way note that you can indeed simplify this as follows. Observe $T=Z_1/|Z_2|$ is symmetrical about 0 so the final limiting probability is $P[T<1]=1/2+P[0<T<1]=0.5(1+P(|T|<1))$ and now use that T is standard Cauchy. $\endgroup$ – Landon Carter Aug 5 '18 at 18:27

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