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Let $X_1,...,X_n$ be i.i.d random variables. $X_i$ ~ $Unif(-1,1)$

Let $Y_1,...,Y_n$ be i.i.d random variables. $~~$$Y_i$ ~ $Unif(-1,1)$

Let us define a new random variable $Z_i$. $Z_i=1$ if $(X_i,Y_i)$ lies within the unit-disk. And $Z_i=0$ if $(X_i,Y_i)$ lies outside the unit disk.

. What is the mean and variance of $\bar{Z_n}$? What about $4\bar{Z_n}$?

. Is $4\bar{Z_n}$ close to $\pi$? Find $P(|4\bar{Z_9}-\pi|>.01)$. Use Chebyshev's theorem to find bound on this probability. What about $P(|4\bar{Z}_{100}-\pi|>.01)$?

. Instead of finding a bound as in the last example, find the $P(|4\bar{Z_9}-\pi|>.01)$ approximately using Central Limit Theorem.

I am having trouble using Chebyshev's theorem. I think my $Var(4\bar{Z}_{9})$ might be wrong but unsure why.

$$Z_i \sim Bern(\frac{\pi}{4})$$

$$E(\bar{Z_n})=\dfrac{\pi}{4} ,~ Var(\bar{Z_n})=(\dfrac{\pi}{4n}-\dfrac{\pi^2}{16n})$$

$$E(4\bar{Z_n})=\pi ,~ Var(4\bar{Z_n})=\dfrac{\pi}{n}(4-\pi)$$

Since $P(|X-E(x)|>k)\leq\dfrac{Var(X)}{k^2}$

$$P(|4\bar{Z_9}-\pi|>.01)\leq \dfrac{(4\pi - \pi^2)}{9(.01^2)}$$

But this number is clearly incorrect. Where did I go wrong?

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  • $\begingroup$ Why is it clearly incorrect? It looks correct. $\endgroup$ – E-A Sep 3 '18 at 23:13
  • $\begingroup$ the bound evaluates to almost 3000 @E-A. Does this matter? $\endgroup$ – 1337 Sep 3 '18 at 23:16
  • $\begingroup$ No; it would be an issue if it were, say, negative. This is after all a "bound"; it does not need to be a good one. This bound will get better as n goes to infinity. (i.e. when you put n instead of 9) $\endgroup$ – E-A Sep 3 '18 at 23:19
  • $\begingroup$ Am I correct in saying $Z_i \sim Bern(\dfrac{\pi}{4})$ @E-A? $\endgroup$ – 1337 Sep 3 '18 at 23:22
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    $\begingroup$ As always I would like to point out that independence assumptions are important in probability. This question has no answer unless $X_i$'s are independent of $Y_i$'s. $\endgroup$ – Kavi Rama Murthy Sep 3 '18 at 23:34

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