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Thread: comparing two Poissonians

  1. #16
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    If you do not want to approximate asymptotic LRT, you may via the following
    LRT test:

    Let X_1 \sim Poisson (\theta_1), X_2 \sim Poisson (\theta_2)

    H_0 : \theta_1 = \theta_2 vs H_1 : \theta_1 \neq \theta_2

    \sup_{\theta \in \Theta_0} L(x_1, x_2) 
= e^{-\frac{x_1+x_2} {2}} \frac {(\frac {x_1 + x_2} {2})^{x_1}} {x_1!}
e^{-\frac{x_1+x_2} {2}} \frac {(\frac {x_1 + x_2} {2})^{x_2}} {x_2!}

    \sup_{\theta \in \Theta} L(x_1, x_2) 
= e^{-x_1} \frac {x_1^{x_1}} {x_1!} e^{-x_2} \frac {x_2^{x_2}} {x_2!}

    \Lambda(x_1, x_2) = \frac {\sup_{\theta \in \Theta_0} L(x_1, x_2)}
{\sup_{\theta \in \Theta} L(x_1, x_2) } 
= \frac {(\frac {x_1 + x_2} {2})^{x_1 + x_2}} {x_1^{x_1}x_2^{x_2}}

    Note \Lambda(x_1, x_2) \in [0, 1]
    and we reject H_0 when \Lambda(x_1, x_2) < c
    \Lambda(0, 10) = \frac {1} {2^{10}} = \frac {1} {1024}

    You can choose the corresponding critical value by setting
    \Pr\left\{\Lambda(X_1, X_2) \leq c \bigg|H_0\right\} = \alpha
    where the significance level \alpha is prespecified

    or you evaluate the P-value
    = \Pr\left\{\Lambda(X_1, X_2) \leq \frac {1} {1024}\bigg|H_0\right\}

    But that is hard to evaluate so taking the approximation seems much easier

  2. #17
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    Getting back to the business about asymptotic distribution of the LRT, I was thinking that the convergence ought to depend on how large lambda is, and not the number of sampling volumes.

    Doesnt poisson become more "normalish" as lambda gets bigger, or am I mistaken in that regard.

    I think increasing lamda and increasing sample size are related cuz

    X1 + X2 ~ Poisson(2*lambda) is like doubling lambda.

    It suggests that convergence of the lrt depends on both number of samples and the lambda?

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    It does depend on lambda. Because of the stationarity of the Poisson process, samples for a year could be considered 12 samples, one from each month. In that case, it seems like it would depend more on the observation count (which of course depends on lambda).

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    Quote Originally Posted by squareandrare View Post
    It does depend on lambda. Because of the stationarity of the Poisson process, samples for a year could be considered 12 samples, one from each month. In that case, it seems like it would depend more on the observation count (which of course depends on lambda).
    My understanding is that once the unit of time is set (ie months or years), lamba is then independent of the number of observations, correct? Though I can see how the convergence of the lrt depends on both number of samples and the lambda.

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    Lambda is proportional to the time scale. Lambda will be 12x larger using years as opposed to months.

    I guess the best way to say it is that the convergence of the LRT depends on the total number of counts observed, whether it be a large count in a single time period or small counts over multiple time periods.
    Last edited by squareandrare; 06-23-2010 at 12:16 PM.

  6. #21
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    well put

    It reminds me of chi square distribution. If you add a bunch up it becomes more normalish, but this is equivalent to sampling once from a distribution with n degrees of freedom.

    With poisson you could draw one one-year sample to get X ~ poisson(12*lambda) or 12 one months and add em up.

    I consider this dead horse to be beaten.

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