1

Is there any way to prove that $N$ Gaussian distribution functions are linearly independent if and only if the means are different. For example, if f1(x)=\frac{1}{\sqrt{2\pi\sigma^2}}\exp{\frac{-(x-\mu1)^2}{2\sigma^2} \vdots fN(x)=\frac{1}{\sqrt{2\pi\sigma^2}}\exp{\frac{-(x-\muN)^2}{2\sigma^2} $f_1,...,f_N$ they are all linearly independent iff the means $(\mu_1, ... ,\mu_N)$ are not equal.

user204238
  • 11
  • What do you mean by linear independence of Gaussian distribution functions? That the densities, or cdf, are linearly independent? That the rv are linearly independent? – André Nicolas Jan 01 '15 at 15:10
  • I mean the densities f1...fN when you look at them as a function of a given RV lets say x. Are f1(x)... fN(x) LI iff the means are different – user204238 Jan 01 '15 at 15:20
  • Linearly independent over what? The densities of the standard normal and the normal with mean $0$ and variance $4$ are linearly independent over the reals. – André Nicolas Jan 01 '15 at 15:26
  • They are not linear independent functions because you can find c for which f1+cf2=0 – user204238 Jan 01 '15 at 15:32
  • You cannot. The second has density of the shape $ke^{-x^2/8}$. – André Nicolas Jan 01 '15 at 15:39
  • Right I should have said same variance with different means. f1 ... fn they all have the same variance – user204238 Jan 01 '15 at 15:55

1 Answers1

1

Assume without loss of generality that $(\mu_n)$ is decreasing. The simplest approach could be to divide any null linear combination $\sum\limits_na_nf_n=0$ by the gaussian density with parameters $(0,\sigma^2)$, yielding $$\sum_na_n\exp(x\mu_n-\mu_n^2/2)=0.$$ When $x\to\infty$, every exponential term except the first one is negligible with respect to the first one since $\exp(x\mu_1)\gg\exp(x\mu_n)$ for every $n\ne1$, hence $a_1=0$. Lather, rinse, repeat.

Did
  • 279,727