# Finding maximum likelihood function for 2 normally distributed samples

1. Apr 2, 2012

### Gullik

1. The problem statement, all variables and given/known data
The question is about how to combine to different samples done with 2 different methods of the same phenomena.

Method 1 gives normally distributed variables $X_1,X_2,...X_{n_1}$, with $\mu$ and $\sigma^2_1$

Method 1 gives normally distributed variables $Y_1,Y_2,...,Y_{n_2}$, with $\mu$ and $\sigma^2_2$

All measurements can be considered independent.

Define $\overline{X}=1/n_1*\Sigma^{n_1}_{i=1}X_i$ and $\overline{Y}={1/n_2}*\Sigma^{n_2}_{j=1}Y_j$

Both the samples are taken from the same population so they have the same $\mu$

The question is set up a likelihood function for the $n_1 + n_2$ measurements, and show that the maximum likelihood estimator is given by

$\mu_{mle}={(\sigma^2_2n_1\overline{X}+\sigma_1^2n_2\overline{Y})/(\sigma_2^2n_1+\sigma^2_1n_1)}$

2. Relevant equations

The likelihood function for 1 sample of a normally distributed function.

$L(\mu)=L(X_1,X_2,...X_n;\mu,\sigma^2)=1/((2\pi\sigma^2)^{n/2})*e^{-1/2*\Sigma^{n}_{i=1}(X_i-\mu)/\sigma}$

3. The attempt at a solution

The main problem is that I don't know how I should handle the 2 samples instead of 1, I'm not sure how those two should combine.

Edit; I think I came up with the solution while I was writing up this which took quite some time since I don't know latex. I'll see when I try it out.

Since the measurement is presumed independent does the likelihood function become something like?

$L(\mu)=L(X_1,X_2,...X_{n_1};\mu,\sigma^2_1)*L(Y_1,Y_2,...,Y_{n_2};\mu,\sigma_2^2)$

Last edited: Apr 2, 2012