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Asked to compute estimator for the following function,

$$\theta = \int_0^\infty \mathrm{e}^{-x^2}~\mathrm{d}x$$

which can be solved by transforming the limits to 0 to 1 and solving the following expectation using Monte Carlo,

$$ E\left[\frac{1}{U^2}~\mathrm{e}^{-(-1+1/U)^2}\right]$$

where $U$ is uniformly distributed

The estimator is, $$\hat\theta = \frac{1}{N} \sum_0^n \frac{1}{U} \space \mathrm{e}^{-(-1+1/U)^2} $$

How do I go about computing the sample variance and then the error? How can I compute the variance without knowing the population mean?

Are there any formulas to choose the best $N$?

takeshi kovacs
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2 Answers2

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You probably mean to use the identity $$\theta= E\left[U^{-2}\mathrm{e}^{-(-1+1/U)^2}\right],$$ where $U$ is uniformly distributed, and the sequence of estimators $(\hat\Theta_n)_{n\geqslant1}$ defined as $$\hat\Theta_n = \frac{1}{n} \sum_{k=1}^nV_k,\qquad V_k= U_k^{-2}\mathrm{e}^{-(-1+1/U_k)^2}. $$ Then, $$ E(\hat\Theta_n)=E(V_1)=\int_0^1u^{-2}\mathrm e^{-(1-u)^2/u^2}\mathrm du=\int_0^\infty\mathrm e^{-x^2}\mathrm dx=\theta, $$ by the change of variable $x=(1-u)/u$, $u=1/(x+1)$, and $$ \mathrm{var}(\hat\Theta_n)=\frac{\sigma^2}n,\qquad\sigma^2=\mathrm{var}(V_1). $$ Note that $$ E(V_1^2)=\int_0^1u^{-4}\mathrm e^{-2(1-u)^2/u^2}\mathrm du=\int_0^\infty(x+1)^2\mathrm e^{-2x^2}\mathrm dx, $$ by the same change of variable. Thus, $$ \theta=\frac12\sqrt\pi,\quad E(V_1^2)=\frac12+\frac5{16}\sqrt{2\pi},\quad\sigma^2=\frac12+\frac5{16}\sqrt{2\pi}-\frac14\pi. $$

Are there any formulas to choose the best $N$?

Sorry? The larger $n$ is, the more accurate the estimate $\hat\Theta_n$ is.

Did
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You can also calculate $(\int_0^\infty e^{-x^2}dx)^2=(\int_0^\infty e^{-x^2}dx)(\int_0^\infty e^{-x^2}dx)=(\int_0^\infty e^{-x^2}dx)(\int_0^\infty e^{-y^2}dy)=(\int_0^\infty\int_0^\infty e^{-x^2}e^{-y^2}dxdy)=\int_0^\infty\int_0^\infty e^{-(x^2+y^2)}dxdy$.

Now we use polar coordinates, and let $x=r\cos(\theta),y=r\sin(\theta)$, $r^2=x^2+y^2$ ($0\leq r\leq\infty$), $\,$ $(0\le\theta \leq2\pi)$. now we have:

$\int_0^\infty\int_0^\infty e^{-(x^2+y^2)}dxdy=\int_0^{2\pi}\int_0^\infty re^{-r^2}drd\theta=(\int_0^{2\pi}1d\theta)(-\frac{1}{2}e^{-r^2}|^\infty_0)=\pi$.

Taking the square root we get $\int_0^\infty e^{-x^2}dx)^2=\sqrt{\pi}$

I hope that may help.

Ellya
  • 11,783