Characteristic function problem Suppose $X_1$ and $X_2$ are independent random variables and suppose also that $X_1$      and $X_1-X_2$ are independent. Show that
$$\mathbb{P}_{X_1}[X_1=c]=1$$
for some constant $c$.
What I get so far is that $\phi(t)\phi(-t)=1$. Then I have problems transforming $\phi(-t)$ to the form of $\phi(t)$. From there, we can get $f(x)$ by integrating $\phi(t)$, but I am not sure I was thinking correctly. 
 A: Hint: Unless you are specifically asked to give a
proof via characteristic functions, consider 
instead that if $X_1$ and $X_1-X_2$ are independent, then
they are also uncorrelated (that is, their covariance is $0$). Thus, from
the bilinearity of the covariance function and the knowledge that 
$\operatorname{cov}(X,X) = \operatorname{var}(X)$, we have that
$$\operatorname{cov}(X_1, X_1-X_2)
= \operatorname{cov}(X_1,X_1) - \operatorname{cov}(X_1,X_2) =
\operatorname{var}(X_1) - 0 = \operatorname{var}(X_1) = 0.$$
So, what can we say about $X_1$ from this result?

Edit: added in response to @whuber's comment.
I don't have a complete answer when $X$ and $Y$ do not
have a variance and so cannot be said to have a covariance,
but the following is an attempt in that direction
Suppose that $X$ and $Y$ are random variables and
$a$ is a real number such that either $P\{X = a\} > 0$
or $X$ admits of a density at $a$, and this density is continuous
and positive at $X = a$.  Then, consider the following.


*

*The conditional distribution of $X-Y$ given that 
$X=a$ is the same as the conditional distribution of $a-Y$ given 
that $X = a$. 

*In the special case that $X$ and $Y$ are independent (and thus 
so are $a-Y$ and $X$ independent
random variables) the conditional distribution of
$a-Y$ given $X = a$ is the same as the unconditional
distribution of $a-Y$. 

*From 1. and 2. we have that for independent random
variables $X$ and $Y$, the conditional distribution of
$X-Y$ given that $X=a$ is the same as the unconditional
distribution of $a-Y$.  This unconditional distribution depends on
the value of $a$

*On the other hand, if $X$ and $Z$ are independent random
variables, then the conditional distribution of $Z$ given
that $X=a$ is the same as the unconditional distribution
of $Z$, and this unconditional distribution in no way depends
of what $a$ is.

*In the problem at hand, $X=X_1$ and $Y = X_2$ are independent
random variables and thus from 3. we have
that the conditional distribution of
$X_1-X_2$ given that $X_1 = a$ is the same as the unconditional
distribution of $a-X_2$. This distribution depends on the value of $a$;
different choices of $a$ result in different distributions.

*On the other hand, with $X=X_1$ and $Z = X_1-X_2$ being given
to be independent random variables, 4. tells us that the 
conditional distribution of $Z = X_1-X_2$ given $X_1=a$ does not
depend on the value of $a$ at all!
$X_1$ cannot admit of a continuous density since then we would have
more than one $a$ and thus (as per 5.), different conditional distributions 
for $X_1-X_2$ in violation of 6. For the same reason, $X_1$ cannot 
take on two or more values with positive probability. In short,
the only way that 5. and 6. can peacefully co-exist is when
$X_1$ takes on only one value $a$ with probability $1$.
A: The question is on the right track.  Since such a problem evidently is intended to help one understand and work with characteristic functions, I will include many of the details that might otherwise be omitted from the solution.

For any random variable $Z$, then, and a real number $t$, let us write
$$\phi_Z(t) = \mathbb{E}(e^{i t Z})$$
for the characteristic function (cf) of $Z$.  For future reference, note that
$$|\phi_Z(t)| \le \mathbb{E}(|e^{i t Z}|) = \mathbb{E}(1) = 1.$$
It is well known--and easily shown by expanding the exponential into its Taylor series at $0$--that when $\phi_Z$ is differentiable at the origin, its derivatives (divided by suitable powers of $i$) are the moments of $Z$.  Regardless, $\phi_Z$ will be continuous at $0$ (essentially because its behavior there reflects the tails of the distribution of $Z$, which necessarily approach asymptotic values and therefore don't change $\phi_Z(t)$ much for very small $|t|$).
When two variables $(Z_1,Z_2)$ are independent, the characteristic function of a linear combination $\alpha_1 Z_1 + \alpha_2 Z_2$ is neatly related to the cfs of the variables themselves via
$$\eqalign{\phi_{\alpha_1Z_1+\alpha_2Z_2}(t)&=\mathbb{E}\left(e^{i t (\alpha_1Z_1+\alpha_2Z_2)}\right)=\mathbb{E}\left(e^{i t \alpha_1Z_1}e^{i t \alpha_2Z_2}\right)=\mathbb{E}\left(e^{i t \alpha_1Z_1}\right)\mathbb{E}\left(e^{i t \alpha_2Z_2}\right) \\
&=\phi_{Z_1}(\alpha_1 t)\phi_{Z_2}(\alpha_2 t).
}$$
The penultimate equality required independence of the random variables $e^{it\alpha_iZ_i}$, but that is assured by the independence of the $Z_i$.
Assuming $X$ and $Y$ are independent and that $X$ and $X-Y$ are independent, apply this result to a linear combination of $X$ and $X-Y$, say $\alpha X + \beta (X-Y)$, computing it in two ways, using both independence assumptions:
$$\eqalign{\phi_X((\alpha+\beta)t)\phi_Y(-\beta t) &= \phi_{(\alpha+\beta)X-\beta Y}(t)=\phi_{\alpha X+\beta(X-Y)}(t)=\phi_X(\alpha t)\phi_{X-Y}(\beta t) \\
&=\phi_X(\alpha t)\phi_X(\beta t)\phi_Y(-\beta t).
}$$
This was the key move.  The desired conclusion follows directly, but some care is needed in the analysis.
The tricky part is that we cannot just divide both sides by $\phi_Y(-\beta t)$, because for many values of $t$ it could be zero.  However, since $\phi_Y(0)=\mathbb{E}(1)=1$ and $\phi_Y$ is continuous at $0$, then for any real $\beta$, $\phi_Y(-\beta t)$ must be nonzero in a neighborhood of zero.  Within this neighborhood (which expands as  $|\beta|$ decreases) we may cancel the common $\phi_Y(-\beta t)$ terms appearing at both ends of the preceding series of equations, yielding
$$\phi_X((\alpha+\beta)t) = \phi_X(\alpha t)\phi_X(\beta t).$$
(Setting $\alpha=-\beta=1$ gives $1=\phi_X(0)=\phi_X(t)\phi_X(-t)$, which is the partial result reported in the question.)
Another way to express this is to take logarithms.  For any real number $\gamma$ define the cumulant generating function (cgf) as
$$\psi(\gamma) = \log(\phi_X(\gamma)).$$
Since $\phi_X$ is continuous and positive near $0$, $\psi$ is continuous near $0$.  Consider the special case $t=1$ and assume both $|\alpha|$ and $|\beta|$ are small enough that they and $|\alpha+\beta|$ lie in the neighborhood of values $t$ where $\phi_Y(-\beta t)$ is nonzero.  It is immediate that
$$\psi(\alpha+\beta) = \psi(\alpha) + \psi(\beta);\ \psi(0) = 0.$$
It is easy to show that the only such functions are the linear ones.  Thus, there must exist a number $\gamma$ for which
$$\psi(t) = \gamma t$$
for all sufficiently small $t$.  Equivalently, in this neighborhood
$$\phi_X(t) = e^{\gamma t}.$$
Since $|\phi_X(t)| \le 1$ for (sufficiently small) positive and negative values of $t$, $\gamma$ must be a purely imaginary number.  Writing it as $\gamma = i c$ for a real number $c$ yields
$$\phi_X(t) = e^{i c t}.$$
Since this is differentiable at the origin, we can now obtain all the moments of $X$.  In particular,
$$\mu_1=\mathbb{E}(X) = \frac{1}{i}\frac{d}{d t} e^{i c t} \vert_{t=0} = c$$
and
$$\mu_2=\mathbb{E}(X^2) = \frac{1}{i^2}\frac{d^2}{d t^2} e^{i c t} \vert_{t=0} = c^2.$$
Therefore the variance of $X$ is
$$\text{Var}(X) = \mu_2 - \mu_1^2 = 0$$
and the desired result follows immediately: $X$ must be almost everywhere a constant (equal to the value of $c$ that emerged in the analysis).
