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Let $X, Y \sim iid U(0,1)$ and $c_1, c_2 \in \mathbb{R}$. In the linear combination $Z = c_1X+c_2Y$, we know that the probability density function of $Z$ depends on the relationships of $c_1$ and $c_2$.

For example, when $c_1 > c_2 > 0$, then the pdf of $Z$ is given by

$$ f(z) = \mathbf{1}_{0 \leq z \leq c_2} \dfrac{z}{c_1c_2} +\mathbf{1}_{c_2\leq z \leq c_1} \dfrac{c_2}{c_1c_2} + \mathbf{1}_{c_1 \leq z \leq c_1+c_2} \dfrac{c_1+c_2-z}{c_1c_2}.$$

The range of the support in this case is: $[0, c_1+c_2]$.

If $c_1 >0$ but $c_2<0$, then the support would be $[-c_2,c_1]$.

By symmetry, if $c_1 < c_2 < 0$ then the support would be $[-(c_1 + c_2),0]$.

I'm not really concerned about the pdf, just the support. Lastly, what will happen to the support of more complicated transformations like: support for pdf of $\sqrt{Z^2 -1}$ and $Z + \sqrt{Z^2 - 1}$? Is there a quick way to find out?

I am also interested in the supports so that the last two transformations are real or complex.

Your insights are appreciated.

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  • $\begingroup$ I am not sure I buy this whole $c_{1} > c_{2} > 0$ thing, since $X$ and $Y$ are both $\sim\text{i.i.d.}\mathcal{U}(0,1)$, this makes the ordering of $c_{1}$ and $c_{2}$ arbitrary, right? So instead of $c_{1} > c_{2} > 0$, don't you mean $c_{1} > 0$ and $c_{2} > 0$? Same for the $<0$ case. $\endgroup$
    – Alexis
    Commented Apr 24, 2015 at 5:16
  • $\begingroup$ Yes, ordering wouldn't matter. $\endgroup$
    – cgo
    Commented Apr 24, 2015 at 7:08
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    $\begingroup$ Also "If $c_1 >0$ but $c_2<0$, then the support would be $[-c_2,c_1]$." Should be "If $c_1 >0$ but $c_2<0$, then the support would be $[\mathbf{c_2},c_1]$," since $c_{2} \times -1 > 0$ in that case, but the support should span 0. Same thing on the following line: the support should just be "$[(c_{1}+c_{2}),0]$," with no negative sign. $\endgroup$
    – Alexis
    Commented Apr 24, 2015 at 8:07
  • $\begingroup$ In the last part, are you asking about cases where $T(Z)$ (e.g. $T(Z)=\sqrt{Z^2-1}\,$) is monotonic, or just defined everywhere in the domain of $Z$, or are you asking about cases where the transformation may not even be defined? (Also, please address Alexis' last comment; your question still has errors.) $\endgroup$
    – Glen_b
    Commented Apr 29, 2015 at 1:05

1 Answer 1

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$\DeclareMathOperator{\support}{support}$The general question here is a very hard problem, for the following reason.

Let $f(x_1, \dots, x_n)$ be any function, and let $X_1, \dots, X_n$ be Gaussian random variables. Then the r.v. $f(X_1, \dots, X_n)$ has support at 0 if and only if the equation $f(x_1, \dots, x_n) = 0$ has real-valued solutions. So determining the support of a function of random variables is at least as hard as finding the zeros of the relevant functions. But for even fairly simple classes of functional expressions, finding their zeros is undecidable (there is no algorithm!) For instance, by Richardson's Theorem, even if $f$ is restricted to be a function of one argument using the constants $\pi$ and $\ln 2$, addition, multiplication, and the functions $\sin, \exp$ and absolute value--the question of whether $f(X)$ has support at 0 is in general undecidable.

You can make some progress on specific sub-cases. For instance, if $X$ has support on $[a, b]$ and $f$ is continuous and monotone increasing in that range, then $f(X)$ has support on $[f(a_X), f(b)]$.

Similarly, if the random variables $X$ and $Y$ have joint support on the entire rectangle $[a_X, b_X] \times [a_Y, b_Y]$, and $f$ is a continuous function of two arguments, monotone increasing in both on that rectangle, then $f(X, Y)$ has support on $[f(a_X, a_Y), f(b_X, b_Y)]$ (and there are similar versions if $f$ is monotone decreasing in one or both arguments instead).

Finally, to tackle non-monotone functions of random variables, you can chop up their domain into rectangular chunks where they are monotone, and then calculate the support of the function conditional on the arguments lying within that chunk, and then take the union of those supports at the end: for instance, $\support(X^2) = \support(X^2 \mid X \ge 0) \cup \support(X^2 \mid X \le 0)$. But the process of conditioning in this way can get very thorny if there are lots of sub-expressions and dependence between the arguments.

Clever application of the above three principles will probably get you fairly far, but there's no general answer for arbitrary functions.

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  • $\begingroup$ Could you provide the definition you are using of "support"? You seem to equate it to the range of the random variable (as a measurable function $X:\Omega\to\mathbb{R}$), whereas the appropriate definition of support in this context involves the probability distribution and is not necessarily the same thing. $\endgroup$
    – whuber ♦
    Commented Apr 24, 2015 at 20:21
  • $\begingroup$ @whuber: no, sorry, I'm just using $\min$ in a perhaps non-standard way (meaning "the minimum value taken on by $A$ with probability > 0" rather than the minimum of the pdf or something). Does that clarify things? Your comment is vague enough that I'm having a hard time figuring out whether I'm missing something or just didn't explain myself well. $\endgroup$
    – Ben Kuhn
    Commented Apr 24, 2015 at 23:03
  • $\begingroup$ (It's true that I omitted some corner cases here, like variables with gaps in the middle of their support or what happens when your variables have support on all of $\mathbb{R}$, but I don't think that complicates things terribly.) $\endgroup$
    – Ben Kuhn
    Commented Apr 24, 2015 at 23:04
  • $\begingroup$ The discussion of Diophantine equations seems not to pay any attention to the underlying probability distribution--that's confusing. Furthermore, your algorithm is not generally correct. For instance, suppose the supports of $A$ and $B$ are $[-101,-100]\cup[100,101]$. Then the support of $A+B$ is quite a bit smaller than the value $[-202,202]$ given by your algorithm! $\endgroup$
    – whuber ♦
    Commented Apr 24, 2015 at 23:09
  • $\begingroup$ @whuber: (1) it does mention the probability distribution--the underlying probability distribution of the $X_i$ needs to have support along the entire real line. Other than that it doesn't matter. I can flesh out the argument if you're still confused. (2) As I mentioned in my comment right above yours, I omitted the case of variables with non-convex support. I'll edit in the extra condition this entails (continuous functions and support equal to an interval). $\endgroup$
    – Ben Kuhn
    Commented Apr 24, 2015 at 23:33

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