Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $y=e^{-ikx}$.

  1. Is $y$ a random variable?
  2. 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.
  3. 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?

share|improve this question

1 Answer 1

up vote 2 down vote accepted

I always find it irritating when people use the same letter to refer both to the random variable and to the variable being integrated in expressions like your $\int xf(x)\,dx$ above. I prefer in most circumstances to use a capital $X$ to refer to the random variable and lower-case $x$ to refer to the bound variable in the integral. That way when one refers to $f_X(x)$, one knows what is meant, and an expression like $f_X(3)$ means the value of the density function at $3$.

That is the notation I will follow below.

If $X$ is a random variable then so is $g(X)$. One need not find $f_{g(X)}(x)$ in order to find $\mathbb{E}(g(X))$, since, according to the law of the unconscious statistician, one has $$ \mathbb{E}(g(X)) = \int_{-\infty}^\infty g(x) f_X(x)\,dx. $$ If $\mathbb{E}\left(|g(X)|^2\right)<\infty$, then the strong law of large numbers is applicable to the probability distribution of $g(X)$, so the sample average of an i.i.d. sample from that distribution will converge almost surely to $\mathbb{E}(g(X))$. (I use an absolute value since I don't want to require $g(X)$ to be real, so the square would not otherwise necessarily be positive.)

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.