I have some random variable, $x$, distributed according to a probability density function (pdf), $f\left(x\right)$. The Strong Law of Large Numbers (SLLN) implies that, for an expected value, given by:
$ E\left[x\right]=\int x f\left(x\right)\text{d}x $
if I take $N$ samples from $f\left(x\right)$, call them $x_j$ (where $j=1,2,\dots,N$), then the sample average will converge to the expected value as $N\rightarrow\infty$:
$\bar{x}_N = \frac{1}{N}\sum_{j=1}^N x_j \rightarrow E\left[x\right]\ \ \text{as}\ \ N\rightarrow\infty$
Now say that I have another random variable defined by a transformation of $x$:
$y=g\left(x\right)$
with corresponding pdf, $h\left(y\right)$. To use a specific example, let $y=e^{-ikx}$.
- Is $y$ a random variable?
- What is the relationship between the two pdfs, $f\left(x\right)$ and $h\left(y\right)$? E.g. is $h\left(y\right)=f\left(g\left(x\right)\right)$? It seems like no, but it seems like there ought to be a straightforward relationship.
- Is there any meaning to the following expression?:
$\int y f\left(x\right) \text{d}x=\int e^{-ikx} f\left(x\right) \text{d}x$
The reason for $\left(3\right)$ is that I want to know if the following is true:
$ \bar{y}_N=\frac{1}{N}\sum_{j=1}^N y_j=\frac{1}{N}\sum_{j=1}^N e^{-ikx_j}\rightarrow\int e^{-ikx} f\left(x\right) \text{d}x$
As $N\rightarrow\infty$. The answer given in this post seems to suggest that this last expression holds, but the SLLN implies $ \bar{y}_N=\frac{1}{N}\sum_{j=1}^N y_j\rightarrow\int e^{-ikx} h\left(y\right) \text{d}y$, and the two seem contradictory, your thoughts?