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Estimation of area using CLT

  1. Oct 11, 2012 #1
    Your help/input is much appreciated! Thanks in advance! :smile:

    Object A is in the unit square 0<x<1, 0<y<1. Consider a random point distributed uniformly over the square; let Y=1 if the point lies inside A and Y=0 otherwise. How could A be estimated from a sequence of n individual points unif. distr. on the square?

    I got this part just taking Y as ~Bernoulli(p=A). Let [itex]Y_{i} = f(X_{i})[/itex], then [itex]A≈\frac{1}{n}\sum{Y_{i}}[/itex]

    How could the CLT be used to gauge the probable size of the error of the estimate? Denoting the estimate by [itex]\hat{A}[/itex], if [itex]A=0.2[/itex], how large should n be so that [itex]P(|\hat{A} - A| < .01) ≈ .99[/itex]?

    I can't figure out this one. Am I supposed to take [itex]\hat{A}[/itex] as the random variable or [itex]|\hat{A}-A|[/itex] as the RV? What is the standard deviation of [itex]\hat{A}[/itex] vs. [itex]|\hat{A}-A|[/itex]? My attempt below is only for the first part of the question in the previous paragraph.

    Taking [itex]\hat{A}[/itex] as the RV, I started out like this...

    [itex]E[\dfrac{Y_{1}+...+Y_{n}}{n}]=\hat{A}[/itex]
    Then [itex]Var(\hat{A})=Var( E[\dfrac{Y_{1}+...+Y_{n}}{n}] )=\frac{1}{n^{2}}Var( nE[Y_{i}] ) = Var( E[Y_{i}] ) = Var(A)[/itex] (since Yi ~ Bernoulli(A) = A)

    To estimate [itex]A[/itex] based on [itex]\hat{A}[/itex], pick a small error and a high probability of getting that small error and you could use CLT as follows:

    [itex]P(|\hat{A} - A| < .01) = .99[/itex]
    [itex]P(-.01 < \hat{A} - A < .01) = .99[/itex]
    [itex]P(\dfrac{-.01}{σ_{\hat{A}}} < \dfrac{\hat{A} - A}{σ_{\hat{A}}} < \dfrac{.01}{σ_{\hat{A}}}) = .99[/itex]
    [itex]P(\dfrac{-.01}{\sqrt{A}} < Z < \dfrac{.01}{\sqrt{A}}) = .99[/itex]
    [itex]\Phi(\dfrac{.01}{\sqrt{A}}) - \Phi(\dfrac{-.01}{\sqrt{A}}) = .99[/itex]
    [itex]2\Phi(\dfrac{.01}{\sqrt{A}}) - 1) = .99[/itex]
    [itex]\dfrac{.01}{\sqrt{A}}= 2.58[/itex]
    [itex]A = .000015023?[/itex]

    I'm definitely sure this is wrong... how the heck did I come up with an actual number for A?
     
  2. jcsd
  3. Oct 12, 2012 #2

    haruspex

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    This is a bit hard to follow because you confuse A with A-hat.
    [itex]\hat{A} = \sum Y_i/n, A = E(\hat{A}[/itex]).
     
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