0
$\begingroup$

Assume that I have some probability density function, $f\left(x\right)$. If I want to approximate it using a Fourier series I can use the identity:

$c_{k}=\frac{1}{2\pi}\int_{-\pi}^{\pi} f\left(x\right)e^{-ikx}\text{d}x$

If I don't know what my true pdf is, but I have $N$ samples, $\left\{x_1,x_2,\dots,x_N\right\}$ from my distribution, I can get an approximation for my Fourier coefficients by the following:

$c_k\approx\hat{c_k}=\frac{1}{2\pi N} \sum_{n=1}^{N} e^{-ikx_n}$

My question is, how can I prove whether or not $\lim_{N\rightarrow\infty} \hat{c_k} = c_k$?

  • 0
    You don't want the $2 \pi$ in the denominator of your formula for $\hat{c_k}$.2012-09-27

1 Answers 1

1

Assuming the $x_i$ are independent, the Strong Law of Large Numbers says that for any $f$ that is integrable on $[-\pi, \pi]$, with probability $1$ the limit as $N \to \infty$ is $c_k$.

  • 0
    Apply the SLLN directly to $\exp(ikx_n)$. It says $\dfrac{\exp(ikx_1) + \ldots + \exp(ikx_N)}{N} \to E[\exp(ikx)]$.2012-09-27