Moment generating function of the sum of uncorrelated random variables

Question

Suppose that $\theta_1,\theta_2,\theta_3 \sim\mathcal{U}_{(0,2\pi)}$ are i.i.d. random variables uniformly distributed on $(0,2\pi)$. We define the random variables:

$c_{12} = \cos(\theta_1-\theta_2)$ and $c_{13} = \cos(\theta_1-\theta_3)$.

We note that $c_{12},c_{13}$ are uncorrelated but not indipendent. What we can say about the moment generating function of $c_{12}+c_{13}$? In particular is it true that, in this case:

$$ M_{c_{12}+c_{13}}(t) = \mathbb{E}(e^{t(c_{12}+c_{13})}) = \mathbb{E}(e^{tc_{12}})\mathbb{E}(e^{tc_{13}}) = M_{c_{12}}(t)M_{c_{13}}(t) \mbox{ ?} $$

Motivation

By using the following code in R (which simply uses Monte Carlo simulation), I see that the covariance between $e^{c_{12}}$ and $e^{c_{13}}$ is approximately zero, as the covariance between $c_{12}$ and $c_{13}$.

it = 10000
theta1 = runif(it,0,2*pi)
theta2 = runif(it,0,2*pi)
theta3 = runif(it,0,2*pi)
cos12 = cos(theta1-theta2)
cos13 = cos(theta1-theta3)
cos23 = cos(theta2-theta3)
exp12 = exp(cos12)
exp13 = exp(cos13)
exp23 = exp(cos23)
cov(cos12,cos13)
cov(exp12,exp13)

Answer 1

I do not have a proof, but a computation using Mathematica shows that the MGF of $c_{12}$ is $$M_{c_{12}}(t) = \mathcal I_0(t),$$ where $\mathcal I_n(z)$ is the modified Bessel function of the first kind; and $$M_{c_{12} + c_{13}}(t) = \mathcal I_0^2(t).$$ So this implies that the relationship $$M_{c_{12} + c_{13}}(t) = M_{c_{12}}(t) M_{c_{13}}(t),$$ since $c_{12}$ and $c_{13}$ are exchangeable and have the same MGF. Note that because of periodicity, $$\begin{align} M_{c_{12}}(t) &= \frac{1}{4\pi^2} \int_{\theta_2 = 0}^{2\pi} \int_{\theta_1 = 0}^{2\pi} e^{t \cos(\theta_1 - \theta_2)} \, d\theta_1 \, d\theta_2 \\ &= \frac{1}{4\pi^2} \int_{\theta_2 = 0}^{2\pi} \int_{\psi = -\theta_2}^{2\pi-\theta_2} e^{t \cos(\psi+\theta_2-\theta_2)} \, d\psi \, d\theta_2 \\ &= \frac{1}{\pi} \int_{\psi = 0}^{\pi} e^{t \cos \psi} \, d\psi, \end{align}$$ and from Equation (5), this is $I_0(t)$.

I suspect a substantially similar approach applies for the calculation of the MGF of the sum, but I have not attempted it at this time.