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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 $\lim_{N\rightarrow\infty} \hat{c_k} = c_k$?

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You don't want the $2 \pi$ in the denominator of your formula for $\hat{c_k}$. – Robert Israel Sep 27 '12 at 20:03
up vote 1 down vote accepted

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 $c_k$.

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The Strong Law of Large Numbers guarantees that $\bar{x_N}\rightarrow E\left[x\right]$, where $E\left[x\right]$ is the expected value of $f\left(x\right)$, but does it hold when there is a transformation such as that proposed above? I.e. $c_k=g\left(f\left(x\right)\right)$ and $\hat{c_k}=h\left(X\right)$. My notation is a bit sloppy, but you get what i mean? If by the SLLN, $\bar{X_N}\rightarrow E\left[X_N\right]$ does it necessarily hold that $g\left(\bar{X_N}\right)\rightarrow g\left(E\left[X_N\right]\right)$? – okj Sep 27 '12 at 20:54
Apply the SLLN directly to $\exp(ikx_n)$. It says $\dfrac{\exp(ikx_1) + \ldots + \exp(ikx_N)}{N} \to E[\exp(ikx)]$. – Robert Israel Sep 27 '12 at 22:46

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