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variance |
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| Jan23-07, 11:37 AM | #1 |
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variance
How do I approach finding variance of sample mean of Poisson distribution?
thanks. |
| Jan23-07, 02:35 PM | #2 |
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Recognitions:
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| Jan23-07, 03:12 PM | #3 |
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I did: Var(X) = E(X2) - (E(X))2
the problem is that this is given for rv X, not X' or X-bar, that's where i get lost. I know that last term is lambda2. Actually what I need to find for this problem is E(X'2), i.e. E(X'2) = Var(X') + lambda2 |
| Jan23-07, 09:17 PM | #4 |
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variance
ok, last question, pleeeease some one look
i can't find much on this anywhere and the book does not say much... i don't have intuition for these things....can I say that sample variance is sum(lambda)/n? I found something that said sample variance is Sum(Var[X]) of whatever it is the RV divided by n.... |
| Jan27-07, 08:51 AM | #5 |
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According to my old lecture notes, the sample variance for (X1, ..., Xn) is defined with [tex]\frac{1}{n} \sum_{i=1}^n(X_{i}-\overline{X})^2[/tex], where [tex]\overline{X} = \frac{1}{n}\sum_{i=1}^n X_{i}[/tex] , and, in your case X~P(lambda).
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| Jan27-07, 03:27 PM | #6 |
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Yes, the Poisson distribution depends on a single parameter and has the property that both mean and standard distribution are equal to that parameter.
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