# Looking for a certain kind of a distribution

Is there any probability distribution supported on a compact or a half-open interval (of $\mathbb{R}$) such that if a vector $\vec{x} \in \mathbb{R}^n$ is sampled by sampling its coordinates like that then there is a closed form expression for the distribution of $\langle \vec{a} , \vec{x}\rangle$ (as a function of $\vec{a}$)?

The closest example I know of is the Gaussian distribution which is supported on the whole of $\mathbb{R}$.

(1) supported on half planes of $\mathbb{R}^n$, you may want to look at folded Gaussian distributions.

(2) supported on a compact surface like $\mathbb{S}^n$, you may want to look at projected Gaussian distributions. If $\vec{a}$ is fixed you will get a degenerated version of projected Gaussian.

• Thanks! Could you share your intuition about why you think the PDF of $\langle \vec{a}, \vec{x} \rangle$ has a closed form expression with coordinates of $\vec{x}$ coming from say the folded normal? Aug 1 '17 at 22:51
• @gradstudent You are basically projecting the folded normal $vec{x}$ onto the direction defined by $\vec{a}$ Aug 1 '17 at 22:55

If each coordinate is distributed exponentially with parameter $\lambda$, then there are explicit expressions for the distribution of the inner product. E.g.:

In two dimensions where $\vec{a}=(a,b)$: $$P[\langle \vec{a},\vec{x} \rangle < d] = \frac{f(a)-f(b)}{a-b}\ \text{ with }\ f(u) = \max(u,0)-|u|e^{\min(0,-\lambda d/u)}$$

In three dimensions where $\vec{a}=(a,b,c)$, with $a,b,c,d>0$:

$$P[\langle \vec{a},\vec{x} \rangle < d] = \frac{g(a)}{(c-a)(a-b)}+\frac{g(b)}{(a-b)(b-c)}+\frac{g(c)}{(b-c)(c-a)}$$ $$\text{ with } g(u)=u^2 (e^{-\lambda d/u}-1)$$

• Thanks! Are these obtained by some brute force computation like just doing iterated convolutions or are there tricks to quickly generalize this to $n-$dimensions? Aug 2 '17 at 15:58
• Brute force will get you formulas, but not nice formulas. It took substantial tinkering with Mathematica results to get something readable. Maybe tinkering with the n=4 case will make a generalization obvious. Try Probability[ a x + b y < d, {x [Distributed] ExponentialDistribution[lambda], y [Distributed] ExponentialDistribution[lambda]}] and the like, but beware that this command for two dimensions gives a result with 13 cases before tinkering! Aug 2 '17 at 16:10

In general, you are just asking about a weighted sum of i.i.d. variables from distribution $D$, with weights $\alpha_1,\dots,\alpha_n$. The Gaussian distribution is the only one that is rotationally invariant when coordinates are sampled i.i.d., so I'd expect it to be the only one that depends only on $\|\alpha\|$ rather than on all of the $\{\alpha_i\}$.

So one direction to look is for distributions where weighted sums belong to some other known distribution, stable distributions being a special case. (I guess the Levy is supported only on half the line, so it should fit what you're looking for...but it may not be a very "nice" distribution.)

Another approach is via the moment generating function or characteristic function. Suppose $S = \sum_{i=1}^n X_i$, where $X_i = \alpha_iY_i$ and all $Y_i$ are i.i.d. Then

\begin{align*} \mathbb{E} e^{tS} &= \prod_{i=1}^n \mathbb{E} e^{tX_i} \\ &= \prod_{i=1}^n \mathbb{E} e^{t \alpha_i Y_i} \\ &= \prod_{i=1}^n f(t \alpha_i) \end{align*} where $f$ is the MGF of the distribution. For example, with exponential$(\lambda)$ variables, the MGF is $\frac{\lambda}{\lambda-t}$, and if $Y_i$ is exponential(1) then $\alpha_i Y_i$ is exponential$(1/\alpha_i)$, so its MGF is $\frac{1}{1-\alpha_i t}$. So for exponentials, the MGF of $\langle \vec{\alpha}, \vec{X} \rangle$ is $$\prod_{i=1}^n \frac{1}{1-\alpha_i t} .$$ Now whether you can recover a distribution from the MGF is another question, I don't have expertise in this but can point to https://www.quora.com/Given-an-MGF-of-a-random-variable-how-does-one-derive-the-PDF-or-PMF and https://math.stackexchange.com/questions/655302/gamma-distribution-out-of-sum-of-exponential-random-variables