Statistics, calculate the distribution

[MaxManus]

Homework Statement

Assume $z_1, ..., z_m$ are iid,$z_i = μ+\epsilon_i$

$\epsilon_i]$is N(0,σ^2)

Show that
f(z; μ) = g($\bar{z}$; μ)h(z)
where h(·) is a function not depending on μ.

Homework Equations

The Attempt at a Solution

Now z is normal distributed with mean my and variance sigma^2

$f(z,\mu) = \frac{1}{\sigma^2 \sqrt{2 \pi}} e^{-\frac{(z-\mu)^2}{2 \sigma^2}}$

f(z; μ) = $\prod_{i=1}^m \frac{1}{\sigma^2 \sqrt{2 \pi}} e^{-\frac{(z_i-\mu)^2}{2 \sigma^2}}$

$f(\bf{z},\mu) = (\frac{1}{\sigma^2 \sqrt{2 \pi}})^m \prod_{i=1}^m e^{-\frac{(z_i-\mu)^2}{2 \sigma^2}}$

but how do I go from here to
f(z; μ) = g($\bar{z}$; μ)h(z)

And am I in the right track?

 

[Ray Vickson]

Homework Statement

Assume $z_1, ..., z_m$ are iid,$z_i = μ+\epsilon_i$

$\epsilon_i]$is N(0,σ^2)

Show that
f(z; μ) = g($\bar{z}$; μ)h(z)
where h(·) is a function not depending on μ.

Homework Equations

The Attempt at a Solution

Now z is normal distributed with mean my and variance sigma^2

$f(z,\mu) = \frac{1}{\sigma^2 \sqrt{2 \pi}} e^{-\frac{(z-\mu)^2}{2 \sigma^2}}$

f(z; μ) = $\prod_{i=1}^m \frac{1}{\sigma^2 \sqrt{2 \pi}} e^{-\frac{(z_i-\mu)^2}{2 \sigma^2}}$




$f(\bf{z},\mu) (\frac{1}{\sigma^2 \sqrt{2 \pi}})^m \prod_{i=1}^m e^{-\frac{(z_i-\mu)^2}{2 \sigma^2}}$

but how do I go from here to
f(z; μ) = g($\bar{z}$; μ)h(z)

And am I in the right track?

I assume the notation means that $g(\bar{z})$ is the pdf of the sample mean. If so, you need to find the function g. What properties of the normal distribution are important for doing that? What is the importance of the fact that the $z_i$ are independent?

RGV

 

[MaxManus]

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 $\bar{z}$ is the pdf to the normal distribution with mean $\mu]$ and variance $\sigma^2/m$
So
$g(\bar{z}) = \frac{1}{(\sigma^2/m )\sqrt{2 \pi}} e^{-\frac{(\bar{z}-\mu)^2}{2 (\sigma^2/m)}}$

But I'm not sure where to go from here.

 

[Ray Vickson]

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 $\bar{z}$ is the pdf to the normal distribution with mean $\mu]$ and variance $\sigma^2/m$
So
$g(\bar{z}) = \frac{1}{(\sigma^2/m )\sqrt{2 \pi}} e^{-\frac{(\bar{z}-\mu)^2}{2 (\sigma^2/m)}}$

But I'm not sure where to go from here.

So,
$$g = \frac{1}{\sqrt{2\pi}\sigma/\sqrt{n}}\exp{\left[\left(\frac{z_1+z_2+\cdots+z_n}{n}-\mu \right)^2/(2 \sigma^2 /n)\right]}.$$

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$\frac{1}{\sigma \sqrt{2 \pi}}, \text{ not } \frac{1}{\sigma^2 \sqrt{2 \pi}}$ in front.

RGV

 

[MaxManus]

Thanks for all the help.

To those who want to see the rest of the calculation it is in the attachment.