# Question about characteristic function with independence assumption

Let $X$ be a random vector taking values in $\mathbb R^2$ with probability density $p(x) = p_1(x_1)p_2(x_2)$, i.e. the components of $X$ are independent.

Let $V$ be an open set in $\mathbb S^1$, the $1$-sphere. Now, given Radon projections of $X$ along $v \in V$, i.e. the densities of the random variables $\langle v,X \rangle$, can we reconstruct $p_1$ and $p_2$?

We have for the characteristic function \begin{align*} \varphi_{\langle v,X \rangle}(t) = \varphi_{X_1}(s v_1)\varphi_{X_2}(s v_2). \end{align*} The left-hand sideis given for specific $v \in V$ and is the product of the characteristic functions of the marginal densities. The question is, if $v \in V$ and $V$ open in $\mathbb S^1$ suffices to determine the factors (i.e. the marginal densities).

To determine the distributions it is sufficient to have two vectors in $S^1$, with the first coordinate not having the same absolute value, and the assumptions of the Hamburger moment problem.

Claim 1: Suppose $(v_1,v_2), (w_1,w_2) \in S^1$ with $v_1^2 \neq w_1^2$. Then $v_1^k w_2^k - v_2^k w_1^k \neq 0$ for any $k \in \mathbb{N}$.

Proof: Assume $v_1^k w_2^k = v_2^k w_1^k$ for some $k \in \mathbb{N}$ and lead this to a contradiction. If $k$ is even, we can take the $k/2$-th root to obtain $v_1^2 w_2^2 = v_2^2 w_1^2$. Now using that we are on the sphere, we get $v_1^2(1-w_1^2) = (1-v_1^2)w_1^2$ and this is equivalent to $v_1^2 = w_1^2$ wich is a contradiction to the assumption. If $k$ is odd, then the assumption holds only if $v_1 w_2 = v_2 w_1$ which is a special case of the even case. q.e.d.

Claim 2: Suppose $(v_1,v_2), (w_1,w_2) \in S^1$ with $v_1^2 \neq w_1^2$. Let $X_1, X_2 \in \mathbb{R}$, independent, satisfy the assumptions of the Hamburger moment problem. Then the distributions of $v_1 X_1 + v_2 X_2$ and $w_1 X_1 + w_2 X_2$ determine the distributions of $X_1$ and $X_2$.

Proof: We want to deduce the distributions of the moments of $X_1$ and $X_2$ iteratively. This suffices since both distributions satify the assumptions of the Hamburger moment problem. Suppose we know the moments from order $1$ to order $k-1$ for some $k \in \mathbb{N}$. By assumption we know the distribution of $$Y:= v_1 X_1 + v_2 X_2 \, , \quad Z:= w_1 X_1 + w_2 X_2 \, .$$ Consider $\mathbb{E}[Y^k]$ and $\mathbb{E}[Z^k]$: \begin{align} m_{Y,k} &= \mathbf{E}(Y^k) = v_1^k \mathbf{E}(X_1^k) + \sum_{l=1}^{k-1} {k\choose l} v_1^l v_2^{k-l} \mathbf{E}(X_1^{l}) \mathbf{E}(X_2^{k-l}) + v_2^k \mathbf{E}(X_2^k) \, ,\\ m_{Z,k} &= \mathbf{E}(Z^k) = w_1^k \mathbf{E}(X_1^k) + \sum_{l=1}^{k-1} {k\choose l} w_1^l w_2^{k-l} \mathbf{E}(X_1^{l}) \mathbf{E}(X_2^{k-l}) + w_2^k \mathbf{E}(X_2^k) \, , \end{align} Using Claim 1 we can eliminate the $\mathbf{E}(X_1^k)$ term in a linear combination of the two previous lines and deduce an expression for $\mathbf{E}(X_2^k)$ only depending on $m_{Y,k}, m_{Z,k}$ and moments of $X_1$ and $X_2$ of exponents less than $k$. Thereafter we can calculate $\mathbf{E}(X_1^k)$. This provides the moments of order $k$ and we can continue iteratively. q.e.d.

For necessity, two points with $v_1^2 = w_1^2$ is not enough: Thanks to Zeno44 for the good counterexample he provided; it extends to other cases with $v_1^2 = w_1^2$.

It is not clear to me what is the intuitive reason for the condition $v_1^2 \neq w_1^2$. The above "necessity" part is not the whole truth, since there is still the moment problem assumption. For this I can only provide the following comment.

Regarding the moment problem: Example 6.4 in an article by Svante Janson, http://arxiv.org/abs/math/0605642, tells you that it might be difficult to have the result holding without moment assumptions. However, he has no independence assumptions, so maybe one cannot easily find a counterexample in your case.

• But your proof of the second claim has a mistake. You say at the end that you can use Claim 1. But what if $\frac{v_2}{v_1}$ is negative? Claim 1 only considers positive $a,b >0$. Here is an example: Consider $X_1 \sim N(0,\sigma_1^2)$ and $X_2 \sim N(0,\sigma_2^2)$ as well as $v = (1,1)$ and $w=(1,-1)$. Then $\langle v,X_1 \rangle = X_1 + X_2 \sim N(0,\sigma_1^2 + \sigma_2^2)$ and $\langle w,X_1 \rangle = X_1 + X_2 \sim N(0,\sigma_1^2 + \sigma_2^2)$. You cannot reconstruct $\sigma_1$ and $\sigma_2$ uniquely here. – user45183 Apr 17 '14 at 13:07
• @Zeno44 you are right. I corrected Claim 1 so that it can be used. a,b>0 was not necessary, only that $a$ and $b$ are not the same. $a^n-b^n$ equals zero (for real $a$ and $b$) only when $a$ and $b$ are equal. – Thomas Rippl Apr 19 '14 at 14:28
• but "$a^n-b^n$ equals zero (for real $a$ and $b$) only when $a$ and $b$ are equal" clearly is not true. Consider $a=1, b=1$ and $n=2$. This is exactly my counter example in my comment above :) – user45183 Apr 20 '14 at 19:10
• @Zeno44: again you are right. Edited the answer accordingly. – Thomas Rippl Apr 22 '14 at 13:04