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Suppose we want to estimate $\pi$. We adopt the following strategy. We generate $n$ iid observations $Z_1,Z_2,...,Z_n$ from the unit disk $D=\{(x,y):x^2+y^2<1\}$ and let $R=\{(x,y):-\dfrac{1}{\sqrt{2}}\leq x,y\leq \dfrac{1}{\sqrt{2}}\}$ be a square inside $D$. Let $T$ be the number of times a generated observation falls inside $R$. Let $I=\max\{1\leq i\leq n:Z_i\in R\}$ and define $I=0$ if no observation falls inside $R$.

I am supposed to find an unbiased estimator of $\pi$ using $T$ and $I$. I have made certain observations. It is clear that $T\sim Bin(n,\dfrac{2}{\pi})$. I have also found that $P(I=k)=P(Z_k\in R,Z_j\notin R,j>k)=\dfrac{2}{\pi}(1-\dfrac{2}{\pi})^{n-k}$ if $1\leq k\leq n$ and $P(I=0)=(1-\dfrac{2}{\pi})^n$.

$E(I)$ is turning out to be a pretty nasty thing to compute. How can we use $T$ and $I$ to compute a function $f(T,I)$ with $E(f(T,I))=\pi$?

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    $\begingroup$ Define $f(T,I) = \pi$ for all $T, I$. $\endgroup$ – Michael Sep 2 '18 at 5:31
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    $\begingroup$ A natural idea is to consider $\hat\pi=2n/T$... except that $P(T=0)\ne0$. Any idea to adapt this? $\endgroup$ – Did Sep 2 '18 at 5:41
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    $\begingroup$ You need to define more clearly what we're allowed to use. I take it it's implied that we're not allowed to use $\pi$; otherwise @Michael's answer is clearly optimal. Are we allowed to use $n$? If so, you can drop $I$, since $T$ is a sufficient statistic. $\endgroup$ – joriki Sep 2 '18 at 6:16
  • $\begingroup$ You can get properly sized parentheses, braces etc. that adjust to their content by preceding them with \left and \right. $\endgroup$ – joriki Sep 2 '18 at 6:17
  • $\begingroup$ Of course we cannot use $\pi$, we are estimating it! We are allowed to use $n$. But I don't see why $T$ is sufficient. $\endgroup$ – Landon Carter Sep 2 '18 at 6:46
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I assume $\{Z_i\}_{i=1}^{n}$ are i.i.d. vectors uniformly distributed over the unit disc in $\mathbb{R}^2$.

If you are seeking something more than my $f(T,I)=\pi$ answer, then I think you are on a wild goose chase. That is because you suggest an answer can be given for any positive integer $n$, which would include $n=1$.

In the case $n=1$ we have $T=I\sim \mbox{Bernoulli}$ with $P[T=1]=p=2/\pi$. If you want a function $f(T)$ such that $E[f(T)]=\pi$, you must find $f(0)$ and $f(1)$, such that $$ (1-2/\pi)f(0) + (2/\pi)f(1) = \pi$$ Thus $$ (\pi-2)f(0) + 2f(1)=\pi^2 $$ Now if both $f(0)$ and $f(1)$ are rational numbers, it would contradict the fact that $\pi$ is transcendental!

So we are forced to make irrational choices for $f(0)$ and/or $f(1)$, in which case why not just choose $f(0)=f(1)=\pi$?


More generally, for any positive integer $n$ we seek a rational-valued function $f(t,i)$ such that $$ \sum_{t=0}^n \sum_{i=0}^n f(t,i)(2/\pi)^t(1-2/\pi)^{n-t} c(t,i)=\pi$$ where $c(t,i)$ is the (integer) number of ways to have $t$ successes with the last success at index $i$. Multiplying this equation by $\pi^n$ gives $$ \sum_{t=0}^n \sum_{i=0}^n f(t,i) 2^t(\pi-2)^{n-t}c(t,i) = \pi^{n+1}$$ Hence the equation $$ \sum_{t=0}^n \sum_{i=0}^n f(t,i)2^t(x-2)^{n-t}c(t,i)=x^{n+1}$$ is a polynomial equation with rational coefficients for which $\pi$ is a root, again contradicting the fact that $\pi$ is transcendental!


On the other hand I observe:

i) With an infinite number of observations $\{Z_i\}_{i=1}^{\infty}$, the expected number of observations until the first one lands in the square is $1/p = \pi/2$.

ii) If we define a larger square $\{(x,y):-1\leq x,y\leq 1\}$ that contains the unit disc, and generate a random vector that is uniformly distributed over that square, then the probability this vector lies in the unit disc is $\pi/4$.

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  • $\begingroup$ Thank you for the answer. I am aware of the last two alternatives you mentioned. This is actually an exam question. $\endgroup$ – Landon Carter Sep 2 '18 at 15:56

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