Vtyjkyuk

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 $R=(x,y):-dfrac1sqrt2leq x,yleq dfrac1sqrt2$ be a square inside $D$. Let $T$ be the number of times a generated observation falls inside $R$. Let $I=max1leq ileq n:Z_iin R$ 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 $Tsim Bin(n,dfrac2pi)$. I have also found that $P(I=k)=P(Z_kin R,Z_jnotin R,j>k)=dfrac2pi(1-dfrac2pi)^n-k$ if $1leq kleq n$ and $P(I=0)=(1-dfrac2pi)^n$.

$E(I)$ 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 $E(f(T,I))=pi$?

I assume $Z_i_i=1^n$ are i.i.d. vectors uniformly distributed over the unit disc in $mathbbR^2$.

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 $T=Isim mboxBernoulli$ 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
$$ sum_t=0^n sum_i=0^n f(t,i) 2^t(pi-2)^n-tc(t,i) = pi^n+1$$
Hence the equation
$$ sum_t=0^n sum_i=0^n f(t,i)2^t(x-2)^n-tc(t,i)=x^n+1$$
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 $(x,y):-1leq x,yleq 1$ 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 $pi/4$.