Hey flashylightsmeow.
Do you have a specific distribution in mind or is this from a general distribution?
Hi there, I'm just going through the statement: "The population mean is equal to the expected value of the mean of sample distribution of means taken from a population"
So I've been going through the following proof:
Let X1, X2, X3, ..., Xn be a simple random sample from a population with mean μ.
E(Xbar)
= E(1/n ∑ Xi)
= 1/n * E(∑Xi)
expectation is a linear operator so we can take the sum out side of the argurement
= 1/n * ∑ E(Xi)
there are n terms in the sum and the E(Xi) is the same for all i
= 1/n * nE(Xi)
= E(Xi)
E(Xbar) = μ
I cannot for the life of me figure out how E(Xi) = μ
Would someone kindly shine some light on this!!!
Many many thanks!
Ummm. The topic is about how regardless of the distribution of a population, the sample distribution of the means xbar will follow a normal distribution for a large number of samples. The book I'm using goes on to say that the expected value of the sample distribution of all the means (of the samples taken from the population) is equal to the population mean mu. Which is all well and good, but I would like to see it proved because I find it difficult to accept these things without proof.
The CLT states the distribution of the sample mean has a certain normal distribution in the limit that the sample size gets large enough. How large depends on the nature of the distribution.
The best proof I saw was to use the Moment Generating function and show that it tends to the MGF of a Normal distribution. Most standard statistics books should have the proof.
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