A Crash Course in Statistics

The exercises sofar have shown, that it is not easy to distinguish one voice from another using just a sample of one of each of the voices. To resolve these problems we need to look at larger samples an averages. To start use the two voice samples y1 and y2 which you already have. After computing their Fourier transforms z1 and z2 and truncating these transforms to the same length compute the average

z=(z1+z2)/2;

Now compare z1, z2 and z3 (the Fourier Transform of your team partners voice) directly to the average as follows:

pdj=abs(z-zj).^2;
nrmj=sqrt(sum(pdj))

where j=1,2,3. Answer the following questions:

1. Which of the three values for nrmj is the smallest, which one is the largest.
2. Compute

   p=abs(z).^2;
   nrm=sqrt(sum(p))
   

   and compare the values of nrmj with nrm.
3. Would you expect the result of the previous exercises to be better or worse if the number of recordings of your own voice is increased? How many times should your own voice be sampled?

In statistics, we usually use the variance or the standard deviation of a sample. If the sample consists of $N$ single real valued measurements $x_1,x_2,\dots,x_N$ with a mean $\overline{x}$ then the standard deviation $S$ is computed by

$$S=\sqrt{\frac1{N-1}\sum_{k=1}^N (x_k-\overline{x})^2}.$$

Unfortunately, our measurements are neither real valued, nor single numbers, but rather complex valued vectors. This gets us to the following problems:

1. If the sample consists of $n$ single complex measurements $z_1,z_2,\dots,z_N$ with mean $\overline{z}$, how should the formula for $S$ be changed?
2. If the sample consists of $N$ complex vectors ${\bf z}_1,{\bf z}_2,\dots, {\bf z}_N$ with mean ${\bf\overline{z}}$, how should the standard deviation be computed?

In order to do statistical testing we need to have an idea, of how your voice samples are distributed. Unfortunately, we do not know that. However, we always have Chebyshev's Inequality. This says that at least 3/4 of all measurements from the same population fall within 2 standard deviations of the mean. Assuming that $\bf\overline{Z}$ represents your true ``average voice'' and that $S$ is the correct standard deviation of your voice, this says that

$$P(\Vert {\bf Z}-{\bf\overline{Z}}\Vert <2S)>\frac34.$$

for a sample voice ${\bf Z}$ spoken by you. So if a sample voice ${\bf Z}$ lies outside this ball of radius $2S$, the probability that it is your own is less that one quarter, and the probability that it is some one else's is greater than three quarters.