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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.

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    $\begingroup$ 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
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    $\begingroup$ 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

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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.

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  • $\begingroup$ 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
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    $\begingroup$ 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
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    $\begingroup$ 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
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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 $[a\dots c],[c\dots b]$.

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  • $\begingroup$ 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

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