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Bernard
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The question is well defined for estimating $\frac 1 \pi$, but not for estimating $\pi$. In the first case If $l>d$, minimizeyou need to evaluate $cos^{-1}$ which requires knowledge of $\pi$, otherwise the variance of the estimator of $\frac 1 \pi$decreases with$ \frac l d$, it will give youso you'd "practically" settle for $\frac{2l^*}{d\pi} = \frac 1 2$$l=d$.

The question is well defined for estimating $\frac 1 \pi$, but not for estimating $\pi$. In the first case, minimize the variance of the estimator of $\frac 1 \pi$, it will give you $\frac{2l^*}{d\pi} = \frac 1 2$.

The question is well defined for estimating $\frac 1 \pi$, but not for estimating $\pi$. If $l>d$, you need to evaluate $cos^{-1}$ which requires knowledge of $\pi$, otherwise the variance of the estimator decreases with$ \frac l d$, so you'd "practically" settle for $l=d$.

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Bernard
  • 427
  • 2
  • 9

The question is well defined for estimating $\frac 1 \pi$, but not for estimating $\pi$. In the first case, minimize the variance of the estimator of $\frac 1 \pi$, it will give you $\frac{2l^*}{d\pi} = \frac 1 2$.