# Characteristic function of the sum of random variables

1. The problem statement, all variables and given/known data I am trying to understand the very last equality for (let me replace the tilda with a hat) $$\widehat{P_X(K)}=\widehat{P(k_1=k_2=\cdots=k_N=k)} \tag 1$$

1. Relevant equations I also thought that the following imaginary exponential delta identity may be useful, due to the equality of the $k_i$, but see comments below:

$$\int dk \, \exp(ikx) = \delta(x=0)$$

1. The attempt at a solution

So it sees to me the goal is something like expressing $\widehat{P_{X}(K)}$ in terms of $P(\widehat{k}_i)$ ?

So these are given by $\widehat{P(k_1)\cdots P(k_n)}= \int d^N \vec{x} \, p(\vec{x}) \exp\left( -i \sum_j x_j k_j\right)$

I thought I'd first try to look at the simplified case of independent random variables to understand $(1)$ but still can't seem to get it.

So in this case $p(\vec{x}) = \prod_i p(x_i)$

And then we have

(If I am correct in that the notation is that $\prod_i dx_i = d^N (\vec{x})$) $$\widehat{P(k_1)\cdots P(k_n)}= \int \prod_i dx_i \, p(x_i) \exp\left( -i \sum_j x_j k_j\right)$$ and then you can seperate the integrals and so we have:

$$\widehat{P(k_1)\cdots P(k_n)}= \widehat{P_{x_1}(k_1)} \cdots \widehat{P_{x_n}(k_n)} \tag 2$$

Now if I consider the independent case in $\widehat{P_{X}(K)}$ I have:

\begin{align} \widehat{P_{X}(K)} & = \int \prod_i dx_i \, p(x_i) \exp\left( -i k \sum_j x_j \right) \\[10pt] & = \int dx_1 \, p(x_1) e^{-ix_1 k} \int dx_2 \, p(x_2) e^{-ix_2 k} \cdots \int dx_N \, p(x_N) e^{-ikx_N} = \widehat{P_{x_1}(k)} \cdots \widehat{P_{x_n}(k)} \end{align}

So if I compare this to $(2),$ and can reason( I'm not sure you can) that it does matter whether you have $k_i$ or $k$, this is just the label of the fourier transform, but look at the lower notation that gives the distribution, that $\widehat{P_{x_1}(k)}= \widehat{P_{x_1}(k_1)}$ and then I have $k_1=k$ and can do the same for each $k_i$ etc.

Without independence instead i have:

$$\int d^N \vec{x} \, p(\vec{x}) \exp\left( -i k \sum_j x_j\right)$$

and I can't see how you can make any conclusions without knowing what $p(\vec{x})$ is?

Many thanks

• Please see my edits for proper MathJax usage. In particular, the whole point of the $\exp$ notation is so that instead of writing things like $e^{\sum_{i=1}^n \left( \frac{x_i-\mu} \sigma \right)^2 },$ one can write $\displaystyle \exp \left( \sum_{i=1}^n \left( \frac{ x_i-\mu}\sigma \right)^2 \right),$ with no superscript. So writing $\exp^{\sum_{i=1}^n \left( \frac{x_i-\mu} \sigma \right)^2 }$ defeats the purpose. $\qquad$ – Michael Hardy Jun 16 '18 at 23:52
• sorry has your reply since been deleted, it isn't appearing on my computer atm? – yourlazyphysicist Jun 22 '18 at 14:59