Finding Variance of Sample Mean of Poisson Distribution

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To find the variance of the sample mean of a Poisson distribution, start with the definition of variance, which is Var(X) = E(X²) - (E(X))². For a Poisson random variable X with parameter lambda, the mean and variance are both equal to lambda. The variance of the sample mean X' can be expressed as Var(X') = Var(X)/n, where n is the sample size. Additionally, E(X'²) can be calculated using the relationship E(X'²) = Var(X') + lambda². Understanding these relationships clarifies how to approach variance calculations for the sample mean in a Poisson context.
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How do I approach finding variance of sample mean of Poisson distribution?
thanks.
 
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EvLer said:
How do I approach finding variance of sample mean of Poisson distribution?
thanks.

You can certainly start with looking at the definition of variance.
 
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
 
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ok, last question, pleeeease some one look :redface: 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...
 
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According to my old lecture notes, the sample variance for (X1, ..., Xn) is defined with \frac{1}{n} \sum_{i=1}^n(X_{i}-\overline{X})^2, where \overline{X} = \frac{1}{n}\sum_{i=1}^n X_{i} , and, in your case X~P(lambda).
 
Yes, the Poisson distribution depends on a single parameter and has the property that both mean and standard distribution are equal to that parameter.
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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