# Finding the distribution of iid variables X, Y given distribution of X-Y

Say I know the distribution of $X-Y$, but I do not know the distributino of $X$ (or $Y$), but I know that they are statistically independent, and I know they have the same distribution. Is the problem of finding the distribution well-defined, as in will there only be one solution?

An example could be the Logit model, which can be derived from a random utility framework with extreme value type 1 shocks. Given the RUM setup, and given the final logistic distribution on $X-Y$, is finding the distribution on the shocks in the RUM well-defined?

My intuition tells me that is it, as I assume that $X$ and $Y$ are iid. Of course if the distributions could differ, the situation is different.

What I eventually want to get at is: are there numerical strategies for finding the original distributions given the distribution on the differences?

Edit Forgot something: assume mean zero in the "original" distributions

• Do you know about characteristic functions? – Dilip Sarwate Apr 29 '16 at 13:53
• Here is a related question, with some answers: mathoverflow.net/questions/157573/… – kjetil b halvorsen Apr 29 '16 at 14:27
• – kjetil b halvorsen Apr 29 '16 at 17:42
• Concerning uniqueness of solutions, here's a little thought experiment. Let $X_1,X_2,Y_1,Y_2$ be iid and suppose $X=X_1^2+X_2^2$ and $Y=Y_1^2+Y_2^2$. Another way to obtain $X-Y$ is to subtract $U=-X_2^2+Y_2^2$ from $V=X_1^2-Y_1^2$. The variables $U$ and $V$ differ from $X$ and $Y$ in that they are symmetric about $0$ whereas $X$ and $Y$ are non-negative. – whuber May 1 '16 at 18:25

We know two independent random variables $X,Y$ has a common distribution, and we know both have zero expectation (or some other convenient centering). We don't know what that common distribution is, but we do know the distribution of $D=X-Y$. Is it possible to determine the distribution function of $X$ and $Y$?

Let us assume that $X,Y,D$ has continuous distributions, with cumulative distribution function (cdf) and density $F,f$ (for $X,Y$) and $G,g$ (for $D$). First, note that $D$ always have a symmetric distribution: $$\newcommand\myeq{\mathrel{\stackrel{{\mbox{ D}}}{=}}} X-Y \myeq Y-X \myeq -(X-Y)$$ since $X\myeq Y$, where the symbol $\myeq$ means "have the same distribution". So for the problem to be well posed, it is necessary that we postulate a symmetric distribution for the difference $D$.

Now $$G(u) =P(X-Y \le u) = \int_{-\infty}^\infty P(X \le u+Y \mid Y=y) f(y)\; dy \\ = \int F(u+y) f(y) \, dy$$ and by differentiation wrt $u$ under the integral sign we get $$g(u) = \int f(u+y) f(y) \; dy$$ which is an integral equation. But, for a probability problem it is easier to formulate this in terms of the moment generating function (mgf) of the difference, assuming that it exists. If it do not exist, we can work in like manner using the characteristic function. So assume the difference have mgf $M(t)$ and $X,Y$ have common mgf $G(t)$. Since $D$ has a symmetric distribution, we have $$\DeclareMathOperator{\E}{\mathbb{E}} M(t) = \E e^{tD} = \E e^{-tD} = M(-t)$$ so that necessarily $M(t)=M(-t)$. We also find $$M(T)= \E e^{t(X-Y)} = \E e^{tX} \E e^{-tY} = G(t) G(-t)$$ giving the equation $M(t) = G(t) G(-t)$ which we can try to solve for $G(t)$. But this is not really enough, since nothing guarantees that a function $G(t)$ we find that way is a mgf for a probability distribution!

Let us look at some examples. If $D$ has a centered normal distribution with variance 2, its mgf is $M(t)= e^{t^2} = \exp\left( \frac12 t^2 + \frac12 t^2\right)$ so we find the solution $G(t) = e^{\frac12 t^2}$ which indeed is the mgf of the standard normal distribution.

If $D$ has a centered triangular distribution with density function $$f(x) =\begin{cases} 1-|x|,& |x| \le 1 \\ 0, & |x| > 1 \end{cases}$$ then its mgf is $M(t) = \frac{e^t-2+e^{-t}}{t^2}$ which can be factored as $M(t) = \left(\frac{e^{t/2} - e^{-t/2}}{t}\right)^2$ giving the factor $G(t) = \frac{e^{t/2} - e^{-t/2}}{t}$ which is indeed the mgf of the uniform distribution on the interval $(-1/2, 1/2)$.

Now let $D$ have the symmetric Laplace distribution with mgf (see wikipedia) $M(t)=\frac1{1-t^2}= \frac1{1-t} \cdot \frac1{1+t}=G(t)G(-t)$ with $G(t)=\frac1{1-t}$ which is indeed the mgf of an exponential distribution. In the first two examples, our solution was symmetric, but in this third example the solution is an asymmetric distribution. Is there in this case also a symmetric solution? we could try the alternative solution $$G(t) =\sqrt{\frac1{1-t^2}}$$ But, I do not know if this is a valid mgf of some probability distribution. If it is, we would have shown that this problem do not necessarily have a unique solution.

The OP did not tell us in which form he has information on the distribution of $D$. If he has an iid sample from $D$, maybe he could calculate an empirical estimate of the moment generating function, estimating $G(t)$ by taking the square root of the estimate of $M(t)$, and try to approxiamtely invert that by using a saddle point approximation, see How does saddlepoint approximation work?

If he wants a better answer we need to know in which form is his information.

• Was unable to log in to my account, but had the same question (about x=d -x and y=d -y). Makes sense so far! – pkofod Apr 30 '16 at 20:02

As a counter-example:

If $X \sim N(\mu, \sigma^2)$ and $Y \sim N(\mu, \sigma^2)$ are iid, then $Z = X - Y$ is $N(0, 2 \sigma^2)$.

Conversely, if we observe that $Z \sim N(0, 2 \sigma^2)$, then there are an infinite number of Normal parents $X \sim N(\mu, \sigma^2)$ that satisfy same ... i.e. while we can fix $\sigma$, there an infinite number of $\mu$ solutions.

• Thank you for the answer. Does it help if I'm willing to assume that the mean of the distribution is 0? I had actually forgotten that assumption in the question. – pkofod Apr 29 '16 at 11:48
• Well, yes of course that helps ... – kjetil b halvorsen Dec 2 '19 at 13:07