Estimating p with Bernoulli Sample

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The discussion focuses on estimating the parameter p in a Bernoulli random sample, where p is constrained between 0 and 0.5. The initial method of moments estimator, p = (sum Xi)/n, risks exceeding 0.5 if all observations are 1. A proposed adjustment suggests using p = min((sum Xi)/n, 0.5) to ensure the estimate remains within the valid range. While this adjustment may artificially inflate the frequency of estimating p as 0.5, it enhances the accuracy of the estimation given the known constraints. Ultimately, the consensus is that the adjusted estimator is more reliable despite concerns about its artificiality.
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A Bernoulli random sample :

X=1 , Pr(X=1)=p;
X=0 , Pr(X=0)=1-p;

taken X1, ..., Xn

and if it is known that 0=< p =<0.5

find a method of moments estimator of p

If I just take estimator of P =(sum Xi )/n than the estimate p may be bigger than 0.5 as in extreme case all Xi =1 .
What should I do , take estimator of p = Max (0.5*n , Sum Xi) /n ?
 
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chicory said:
A Bernoulli random sample :

X=1 , Pr(X=1)=p;
X=0 , Pr(X=0)=1-p;

taken X1, ..., Xn

and if it is known that 0=< p =<0.5

find a method of moments estimator of p

If I just take estimator of P =(sum Xi )/n than the estimate p may be bigger than 0.5 as in extreme case all Xi =1 .
What should I do , take estimator of p = Max (0.5*n , Sum Xi) /n ?

Just simply state that fact, that is

P= sum Xi /n if sum Xi /n < 0.5
P= 0.5 otherwise.
 
But... is it too artificial?
Pr( estimator (p) = 0.5 ) is slightly bigger ...?
 
Artificial? You like better min( sum Xi/n, 0.5) ?

No, p will not be generally bigger and you will rarely get estimations of 0.7 or 0.8, remember that they already assure you that the real p is no higher than 0.5

Adjusting the estimation makes the value p=0.5 more common but that is irrelevant and only makes the estimation more accurate.
 
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Thank you so much !
 
You're welcome :smile:
 
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