The following question is crossposted from MathStackExchange upon recommendation from the MSE community and a lack of responses on my post over there.
Consider the following problem from a course on statistical inference:
If we generate a sample $x_i$ for $i \in$ {$1 ... n$} from $ p(x_i) = \sum_{k=1}^2 w_kp(x_i| \mu _k, \sigma^2_k)$ where $p(x_i| \mu _k, \sigma ^2_k)$ are Gaussian densities and $w_2 = 1 - w_1$ (where we assume that all the parameters are unknown)
Find the log-likelihood function for the sample $x_{1:n}$.
The log-likelihood is defined as:
$$ l( \theta | x_{1:n}) = \log(p(x_{1:n})) = \log( \Pi _{i=1}^n p(x_i)) = \sum _{i=1}^n \log(p(x_i))$$
Now substituting in for $p(x_i)$ we get:
$$ l( \theta | x_{1:n}) = \sum_{i=1}^n \log \Big{(} \sum_{k=1}^2 w_k p(x_i | \mu _k , \sigma^2_k) \Big{)} $$
We can now substitute in the Gaussian densities which gives us:
$$ l( \theta | x_{1:n}) = \sum_{i=1}^n \log \Big{[} w_1(2 \pi \sigma ^2 _1)^{-1/2} \exp \Big{(}\frac{(x - \mu _1)^2}{2 \sigma ^2 _1} \Big{)} + w_2 (2 \pi \sigma ^2 _2)^{-1/2} \exp \Big{(}\frac{(x - \mu _2)^2}{2 \sigma ^2 _2} \Big{)} \Big{]} $$
The following solution aligns with my working, however, there is a proportionality relation on the last line that is unclear to me.
Why does the final line hold? I can't see a clear and obvious way to get from the point at which I have substituted the Gaussian densities into the sum to it being proportional to the given double summation.
I understand that we can exclude any multiplicative constants, however, it seems as though the summation has been taken out of the logarithm at some stage in order to derive a proportionality relation. Although I'm not sure if this holds and if it does, why does it hold?
Edit
It seems as though this is a mistake as Whuber presents a counterexample in the comments. Here is the second part of the question:
Presumably, this is also incorrect for similar reasons to the above. If this does end up being incorrect is there any way to salvage this?
