Can Gaussian mixture models with different amount of populations have the same pdf Concretely, denote $f_{\mu, \sigma}$ for the pdf of a normally distributed random variable with mean $\mu$ and standard deviation $\sigma$. Is it possible that
$$\displaystyle \sum_{i = 0}^n p_i f_{\mu_{1i}, \sigma_{1i}}(x) = \displaystyle \sum_{j = 0}^k q_j f_{\mu_{2j}, \sigma_{2j}}(x) \ \ \ \ \ \sum p_i = \sum q_i = 1; p_i, q_i > 0$$
for $n \neq k$? Intuitively, I'd say yes because if two normals have the same mean, the weighted sum looks like a bell curve with the same mean and different standard deviation, but I have not been able to proof it.
 A: The answer is no: the mixture determines its components uniquely.
There's an elementary way to show it.  The idea is that each component of the mixture determines how the density behaves at extreme values and we can use this asymptotic behavior to identify (and remove) the components one by one.
To carry out this program, consider a Normal mixture
$$f(x) = \sum_{i=1}^n \frac{p_i}{\sigma_i} \phi\left(\frac{x-\mu_i}{\sigma_i}\right)$$
where
$$\phi(z) = \frac{1}{\sqrt{2\pi}} e^{-z^2/2}$$
is the standard Normal density, the $p_i\ne 0$ and $\mu_i$ are any real numbers, the $\sigma_i$ are positive real numbers, and the ordered pairs $(\mu_i,\sigma_i)$ are all distinct.  Re-order the indexes $i$ if necessary so that
$$\sigma_1 \le \sigma_2 \le \ldots \le \sigma_n$$
and, whenever $\sigma_i = \sigma_j,$ then $\mu_i \le \mu_j.$ This ordering (the lexicographic ordering) is unique.  We're going to examine the last component first (to reduce the length of the sum) and to that end this ordering has us first visit the most spread-out components and, within them, those that are positioned at higher values.
To study what happens to $f$ as $x$ grows large, consider 
$$\eqalign{&\frac{\sigma_n}{\phi\left(\frac{x-\mu_n}{\sigma_n}\right)} f(x) \\&=
p_n + \sum_{i=1}^{n-1} p_i \frac{\sigma_n}{\sigma_i} \exp\left(\frac{(\sigma_i^2-\sigma_n^2)x^2 - 2x\left(\sigma_n^2\mu_i - \sigma_i^2\mu_n)\right) - \sigma_n^2\mu_i^2 + \sigma_i^2\mu_n^2}{2\sigma_i^2\sigma_n^2}\right).
}$$
Because every $\sigma_i^2-\sigma_n^2 \le 0,$ the limiting values of the exponentials in the sum are all $0$ unless the coefficient of the $x^2$ term is zero, which is only the case when $\sigma_i = \sigma_n.$  But in those cases the limiting values are still zero unless the coefficient of $x$ is zero, which (since now $\sigma_i^2=\sigma_n^2$) occurs only when $\mu_i=\mu_n.$  But we have arranged at the outset that this never happens: there is no $i\ne n$ for which $(\mu_i,\sigma_i)=(\mu_n,\sigma_n).$  Thus,
$$\lim_{x\to\infty}  \frac{\sigma_n}{\phi\left(\frac{x-\mu_n}{\sigma_n}\right)} f(x)  = p_n.$$
Had we used any value other than $\sigma_n$ in this analysis, the limit would have been either $0$ or diverging to $\pm \infty;$ and, having used $\sigma_n,$ using any other value in place of $\mu_n$ also would have produced a limit of $0$ or $\pm\infty.$  In other words, $(\mu_n,\sigma_n)$ is the only parameter for which we can achieve a finite nonzero limit and that limit determines $p_n.$
This shows that any mixture of $n$ distinct Normal densities determines its last component (in the lexicographic ordering of components).  Subtracting off this component yields a Normal mixture with one less component, instantly giving an inductive proof of the result:

Let $p_i\ne 0,q_j\ne 0,\sigma_i\gt 0,\tau_j\gt 0,\mu_i,$ and $\nu_j$ be any real numbers for $1\le i \le n$ and $1\le j \le k.$ If every number $x$ in a set with no upper bound satisfies $$ \sum_{i=1}^n \frac{p_i}{\sigma_i} \phi\left(\frac{x-\mu_i}{\sigma_i}\right) = \sum_{i=1}^k \frac{q_i}{\tau_i} \phi\left(\frac{x-\nu_i}{\tau_i}\right)$$ and each sum involves distinct Normal components and is ordered lexicographically, then $n=k$ and for each $1\le i\le n,$ $p_i=q_i,$ $\mu_i=\nu_i,$ and $\sigma_i=\tau_i.$

(I phrased this in a way that indicates how this analysis can be generalized to some other families of distributions, including discrete distributions.)
Notice we never had to assume the $p_i$ were positive or that they had to sum to unity. In fact, we didn't even have to assume $n$ and $k$ are finite!  In the countably infinite case, all that's needed to carry the demonstration through is that the sets of $\sigma_i$ and $\tau_j$ are bounded above, have only one accumulation point each, and that each set of $\mu_i$ and $\nu_j$ has at most one accumulation point in the extended reals.
