# Marginal characteristic function from a multivariate charateristic function

[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?

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

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: $$G(k_1) = \left.G(k_1,k_2)\right|_{k_2=0}$$

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 $$x_1$$.