
#1
Mar712, 11:32 AM

P: 297

1. The problem statement, all variables and given/known data
Assume [itex] z_1, ..., z_m[/itex] are iid,[itex] z_i = μ+\epsilon_i [/itex] [itex] \epsilon_i] [/itex]is N(0,σ^2) Show that f(z; μ) = g([itex] \bar{z}[/itex]; μ)h(z) where h(·) is a function not depending on μ. 2. Relevant equations 3. The attempt at a solution Now z is normal distributed with mean my and variance sigma^2 [itex] f(z,\mu) = \frac{1}{\sigma^2 \sqrt{2 \pi}} e^{\frac{(z\mu)^2}{2 \sigma^2}} [/itex] f(z; μ) = [itex]\prod_{i=1}^m \frac{1}{\sigma^2 \sqrt{2 \pi}} e^{\frac{(z_i\mu)^2}{2 \sigma^2}} [/itex] [itex] f(\bf{z},\mu) = (\frac{1}{\sigma^2 \sqrt{2 \pi}})^m \prod_{i=1}^m e^{\frac{(z_i\mu)^2}{2 \sigma^2}} [/itex] but how do I go from here to f(z; μ) = g([itex] \bar{z}[/itex]; μ)h(z) And am I in the right track? 



#2
Mar712, 12:53 PM

HW Helper
Thanks
P: 4,677

RGV 



#3
Mar712, 04:14 PM

P: 297

Thanks
I'm not sure what the notation mean, but I assumed g was a function of the sample mean and mu. But if it is the pdf of the sample mean. The pdf of [itex]\bar{z} [/itex] is the pdf to the normal distribution with mean [itex]\mu] [/itex] and variance [itex] \sigma^2/m [/itex] So [itex] g(\bar{z}) = \frac{1}{(\sigma^2/m )\sqrt{2 \pi}} e^{\frac{(\bar{z}\mu)^2}{2 (\sigma^2/m)}} [/itex] But I'm not sure where to go from here. 



#4
Mar712, 05:04 PM

HW Helper
Thanks
P: 4,677

Statistics, calculate the distribution[tex] g = \frac{1}{\sqrt{2\pi}\sigma/\sqrt{n}}\exp{\left[\left(\frac{z_1+z_2+\cdots+z_n}{n}\mu \right)^2/(2 \sigma^2 /n)\right]}. [/tex] You are supposed to show that f(z,μ)/g does not have μ in it. PS: your expressions for the normal distributions are a bit wrong: you should have[itex] \frac{1}{\sigma \sqrt{2 \pi}}, \text{ not } \frac{1}{\sigma^2 \sqrt{2 \pi}}[/itex] in front. RGV 



#5
Mar812, 01:44 PM

P: 297

Thanks for all the help.
To those who want to see the rest of the calculation it is in the attachment. 


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