Having a random variable $X$ with a given density function $f_{X} : [a .. b] \rightarrow\mathcal{R}_{+}$ and a point c, $c \in [a .. b]$ I am curious if the following problem has a solution:

Find two independent random variables $X_1$ and $X_2$ such that $X = X_1 + X_2$ with the additional property that the density of $X_1$ is defined over the interval $[a .. c]$ and the density of $X_2$ is defined over the interval $[c .. b]$.

Could you also please point me out to relevant literature to read about this topic. Thanks, Bogdan.

Based on your observations the solution I have in mind is the following:

  1. Write $X = X_1 + X_2 - c$, where $X_1$ and $X_2$ are defined over the interval $[a ..c]$ and $[c ..b]$ respectively.
  2. Assume $X_2$ to be normal with mean $\mu = \frac{b + c}{2}$ and $\sigma = 1$.
  3. Compute the density of $X_1$ by solving the deconvolution problem knowing the density of $X$ and the density of $X_2$.

The question I have in mind now is that for what kind of densities of X one can find the density of X1 assuming normal density for X2.

Thanks, Bogdan.

  • 4
    $\begingroup$ A "trivial" observation: If $X_1$ takes values in $[a,c]$ and $X_2$ takes values in $[c,b]$, then whether they are independent or not, $X_1 + X_2$ takes values in $[a+c,b+c] \neq [a,b]$. Have I misunderstood your question? If not, and there is a way to clarify your question, e.g., by shifting the range of $X_2$ appropriately, I'd be interested in seeing it. Cheers. $\endgroup$
    – cardinal
    Commented Apr 3, 2013 at 20:45
  • $\begingroup$ If the first is on $[a,c]$ you need that second one to be defined on $[0,b-c]$ so that the convolution is on $[a,b]$. If you know either $X_1$ or $X_2$, or sufficient characteristics of one of them, you have a problem in deconvolution. In specific situations problems like yours may be over- or under-determined. It may be useful to work with characteristic functions (c.f. Fourier transforms), which reduce your deconvolution problem to one more akin to factorization - writing a function as a product of functions with some particular characteristics $\endgroup$
    – Glen_b
    Commented Apr 4, 2013 at 2:02
  • $\begingroup$ $X_1$ and $X_2$ are fixed to the intervals $[a ..c]$ and $[c ..b]$. Probably the independence is not needed. I had the impression will make the problem simpler (not impossible). Do you think the problem can have a solution if the variables are dependent ? $\endgroup$
    – Bogdan
    Commented Apr 4, 2013 at 7:43
  • 1
    $\begingroup$ The current version of the question in nonsensical. The OP wants $X_2$ to have density with support $[c,b]$ and also $X_2$ to be a normal random variable. These two requirements cannot be satisfied simultaneously. For the original version of the question, as @Aksakal's answer points out, if $X_1 \in [a,c]$ and $X_2 \in [c,b]$, then $X_1+X_2 \in [a+c,b+c]$ (independence is not needed to say this) and $[a+c,b+c] \neq [a,b]$ as the OP wants unless $c=0$. If $c=0$, then, $X_1,X_2$ can have any joint density (independence not needed) and $X_1+X_2$ will have density with support $[a,b]$. $\endgroup$ Commented Dec 22, 2014 at 23:58

2 Answers 2


Since $f_{X_1 + X_2} = f_{X_1} * f_{X_2}$ (easily shown using MGFs and independence) you basically have to verify $$\int^{b}_{a} f_{X}(x) dx = \int^b_a \int^{c}_{a} f_{X_1}(\xi)f_{X_2}(x-\xi) d\xi dx = 1$$ With your assumptions $$f_{X_1}(x_1) = \chi_{[a, c]}(x_1) f_{X_1}(x_1)$$ $$f_{X_2}(x_2) = \chi_{[c, b]}(x_2) f_{X_2}(x_2)$$ The convolution integral which defines $f_X(x)$ is $$f_X(x)=\int^{+\infty}_{-\infty} f_{X_1}(\xi)f_{X_2}(x-\xi) d\xi$$ $$=\int^{+\infty}_{-\infty} \chi_{[a, c]}(\xi) f_{X_1}(\xi) \chi_{[c, b]}(x-\xi)f_{X_2}(x-\xi) d\xi$$ $$=\int^{c}_{a} f_{X_1}(\xi) \chi_{[c, b]}(x-\xi) f_{X_2}(x-\xi) d\xi$$ Now, $\chi_{[c,b]}(x-\xi) = \chi_{[x-b, x-c]}(\xi)$ because the indicator function is nonzero for $c \leq x-\xi \leq b$ that is $x-b < \xi < x-c$ so the integral is nonzero when $[x-b, x-c] \cap [a, c] \neq \emptyset$ but since $x \in [a, b]$ by your hypotheses, then you should check the intersection of the two supports.

An example solution would be, e.g. given $X_1 \sim U([-1, 0]), X_2 \sim U([0,1])$, $$f_X(x) = \begin{cases} 1-x & 0<x<1 \\ x+1 & -1<x<0 \\ 0 & \text{otherwise} \end{cases}$$

In case you are interested, try this Mathematica code out for the previous example and another case easily derived.

  • $\begingroup$ but this sound incorrect since if $f_{X_1}$ is the density of $X_1$ on $[a .. c]$ then it integrates to 1 and similar for $X_2$. This will give us 1 + 1 = 1. $\endgroup$
    – Bogdan
    Commented Apr 3, 2013 at 14:37
  • 1
    $\begingroup$ Can you please explain to me why these are the limits of the integration? $\endgroup$
    – Bogdan
    Commented Apr 3, 2013 at 14:47
  • $\begingroup$ Since $X = X_1 + X_2$ its PDF is the convolution of $f_{X_1}$ and $f_{X_2}$, in general $f_X(x) = \int^{+\infty}_{-\infty} f_{X_1}(\xi) f_{X_2} (x - \xi) d \xi$ provided that $X_1, X_2$ are independent. Since you have the additional information that $f_{X_1}$ is defined in $[a, c]$ then the integral in $d\xi$ is in $\xi \in [a, c]$ since that is the support of the PDF of $X_1$. Then you can integrate over the whole domain of $f_X(x)$ which is $[a, b]$, and since $f_X (x)$ is a PDF the integral on its support must be 1. $\endgroup$
    – V.C.
    Commented Apr 3, 2013 at 14:53
  • $\begingroup$ But we also have $f_{X_2}$ in the formula which is defined only on $[c .. b]$ and we have only one $d$ξ and two integrals ... isn't this a contradiction? If we have a double integration we should have two dξ $\endgroup$
    – Bogdan
    Commented Apr 3, 2013 at 14:56
  • 1
    $\begingroup$ With the edit your calculations look more rigorous, but what are they accomplishing? The question is not to verify that $X_1$ and $X_2$ have the desired properties, but given $X$ (and possibly $c$, too) to find such $X_1$ and $X_2$. $\endgroup$
    – whuber
    Commented Apr 3, 2013 at 17:28

Would you agree that this is always true?

$a+c\le (X_1+X_2) \le c+b$

If you do, as I hope, then you'd agree that unless $f_{X}(x\in [a .. c]) =f_{X}(x\in [c .. b])=0$ it is impossible to come up with $X=X_1+X_2$.

So, my answer is that in a general case you can't decompose $X$ into independent $X_1+X_2$ with domains $[a\dots c],[c\dots b]$.

  • $\begingroup$ What does your notation "$f_{X} : [a .. c] $" mean? Efforts to interpret it in the way the O.P. used it lead only to mathematical nonsense. $\endgroup$
    – whuber
    Commented Dec 22, 2014 at 21:54
  • $\begingroup$ @whuber, I clarified the notation. $\endgroup$
    – Aksakal
    Commented Dec 22, 2014 at 21:58
  • $\begingroup$ Right--so how are you using it? At any rate (reading the long-standing post by V.C.), suppose $X_1$ has a uniform distribution on $[a,c]=[-1,0]$ and $X_2$ independently has a uniform distribution on $[c,b]=[0,1]$. Then $X=X_1+X_2$ has a triangular distribution on $[a,b]=[-1,1]$. So if we begin with this $X$, by construction the problem has a solution. That would seem to contradict your claim. $\endgroup$
    – whuber
    Commented Dec 22, 2014 at 21:59

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.