Using MGFs to deduce the PDF

**chella182** · December 9th 2009, 04:08 PM

Suppose that $Y_1,Y_2,...,Y_n$ are a random sample from a normal distribution with mean $\mu$ and variance $\sigma^2$ . Use MGFs to deduce the PDF of $\bar{Y}$ .

**mr fantastic** · December 9th 2009, 04:51 PM

Originally Posted by chella182

Suppose that $Y_1,Y_2,...,Y_n$ are a random sample from a normal distribution with mean $\mu$ and variance $\sigma^2$ . Use MGFs to deduce the PDF of $\bar{Y}$ .

Are $Y_1,Y_2,...,Y_n$ independent? Read this: Moment-generating function - Wikipedia, the free encyclopedia (Sum of independent random variables).

**chella182** · December 10th 2009, 06:02 AM

Yeah I know that, but it doesn't say in the question that they're independent

**Moo** · December 10th 2009, 10:56 AM

Originally Posted by Wikipedia

Random sampling can also refer to taking a number of independent observations from the same probability distribution, without involving any real population.

I think they introduced a sample this way in my stats course

**mr fantastic** · December 10th 2009, 01:41 PM

Originally Posted by chella182

Suppose that $Y_1,Y_2,...,Y_n$ are a random sample from a normal distribution with mean $\mu$ and variance $\sigma^2$ . Use MGFs to deduce the PDF of $\bar{Y}$ .

Asked here first: http://www.mathhelpforum.com/math-he...ution-mgf.html

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