0
$\begingroup$

[I am posting this question here and not in Mathematics Stack Exchange because I will be using conventions as they are usually used in statistical physics (especially the notation is more involved in pure statistics).]

Given a multivariable characteristic function $G(k_1,k_2)$, what is the marginal characteristic function $G(k_1)$?

First some definitions. For continuous stochastic variables $x_1$ and $x_2$, the characteristic function is the Fourier transform of the probability distribution (I drop any pre-factors for simplicity).

$G(k) = \int_{-\infty}^{\infty} dx\ e^{ikx}\ p(x)$

$G(k_1,k_2) = \int_{-\infty}^{\infty} dx_1\ \int_{-\infty}^{\infty} dx_2\ e^{ik_1x_1}\ e^{ik_2x_2}\ p(x_1,x_2)$

The marginal probability for the variable $x_1$ is then $p(x_1) = \int_{-\infty}^{\infty} dx_2\ p(x_1,x_2)$

Using the above three definitions, I want to show that (if true)
$$G(k_1) = \int_{-\infty}^{\infty} dk_2\ G(k_1,k_2)$$


My attempt at a proof:

I start by writing $G(k_1)$ as the inverse Fourier transform: $$ G(k_1) = \int_{-\infty}^{\infty} dx_1\ e^{ik_1x_1}\ p(x_1) $$ Now I write $p(x_1)$ as the marginal probability of the multivariate probability distribuion $p(x_1,x_2)$: $$ G(k_1) = \int_{-\infty}^{\infty} dx_1\ e^{ik_1x_1}\ \int_{-\infty}^{\infty} dx_2\ p(x_1,x_2) $$ Now I need to get a $G(k_1,k_2)$ in there somehow, so I multiply by $1=e^{-ik_2x_2}e^{ik_2x_2}$: $$ \begin{align} G(k_1) &= \int_{-\infty}^{\infty} dx_1\ e^{ik_1x_1}\ \int_{-\infty}^{\infty} dx_2\ e^{-ik_2x_2}e^{ik_2x_2}\ p(x_1,x_2)\\ &= \int_{-\infty}^{\infty} dx_2\ e^{-ik_2x_2}\ \int_{-\infty}^{\infty} dx_1\ e^{ik_1x_1}\ e^{ik_2x_2}\ p(x_1,x_2) \end{align} $$ which looks almost like what I need, except for the extra integrand $e^{-ik_2x_2}$, which makes it impossible to transform the integral into anything I know. Similar approaches (in which I substitute my known definitions largely end up with the same problem). How can I proceed? Which other approach can I use to show what I need?

$\endgroup$
1
  • $\begingroup$ The conjecture is wrong. Integrating over $k_2$ gives a Dirac delta.. $\endgroup$
    – lcv
    Nov 24, 2023 at 10:51

2 Answers 2

0
$\begingroup$

I think I figured it out, although I am not 100% on the validity of the last step:


Begin with the definition of the marginal probability: $$ p(x_1) = \int_{-\infty}^{\infty} dx_2\ p(x_1,x_2) $$ which I now multiply by $e^{ik_1x_1}$ and integrate over $x_1$: $$ \int_{-\infty}^{\infty} dx_1\ p(x_1)\ e^{ik_1x_1} = \int_{-\infty}^{\infty} dx_1\ \int_{-\infty}^{\infty} dx_2\ p(x_1,x_2)\ e^{ik_1x_1} $$ The LHS is now the defintion of the characteristic function of $p(x_1)$, $G(k_1)$. The RHS is almost the definition of the characteristic function of $p(x_1,x_2)$, but it's missing a factor of $e^{ik_2x_2}$. This factor is identically $1$ if $k_2=0$. So I insert it under the condition that $k_2=0$ (Is this a valid step?): $$ G(k_1) = \int_{-\infty}^{\infty} dx_1\ \int_{-\infty}^{\infty} dx_2\ p(x_1,x_2)\ e^{ik_1x_1}\ e^{ik_2x_2} $$ Now the RHS is the definition of the characteristic function of $p(x_1,x_2)$ evaluated at $k_2=0$, $\left.G(k_1,k_2)\right|_{k_2=0}$. So finally: $$ G(k_1) = \left.G(k_1,k_2)\right|_{k_2=0} $$

$\endgroup$
0
$\begingroup$

I think it's simpler: given the characteristic function

$$G(k_1, k_2) = \int dx_1 dx_2 e^{ik_1 x_1} e^{ik_2 x_2} p(x_1, x_2) \, ,$$

setting $k_2=0$ gives

$$G(k_1, 0) = \int dx_1 e^{ik_1 x_1} \int dx_2 p(x_1,x_2) = \int dx_1 e^{ik_1 x_1} p(x_1) \, ,$$

where $p(x_1)$ is the marginal distribution for $x_1$. Hence $G(k_1, 0)$ is the marginal characteristic function for $x_1$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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