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Given X and Y as two independent random variables following $U(0,1)$ and We are required to obtain the distribution of $Z = X + Y$. The answer is given as follows:

$$f(z) = \begin{cases} z & \text{for $0 < z < 1$} \\ 2-z & \text{for $1 \le z < 2$} \\ 0 & \text{otherwise.} \end{cases}$$

I generally get confused in these types of problems when there is a split in the support of the distribution of interest. Here, I know one thing for sure that $Z$ will have values ranging from 0 to 2. I am not able to understand how do we think through the splitting of these z values into 0 to 1 and then 1 to 2.

Let me show you how would I have approached this problem.

Let $$U = X + Y \text{ and } V = Y,$$

then I can write $$X = U - V \text{ and } Y = V.$$

I am trying to find the joint distribution of U and V by jacobian transformation. The value of jacobian will be 1.

$|J| = 1$

Now, we can write:

$f(u,v) = g(x,y)|J| = g(x,y) = g(u-v,v) = 1,\text{ for } 0<u-v<1, 0<v<1$

Now, to get the marginal of $U$, I would need to integrate this joint with respect to $v$. We can see that v takes value from 0 to 1 but at the same time $v>u-1$. So, we can integrate the above joint density with respect to $v$ from $u-1$ to 1 and we get $f(u) = 2-u$.

Now, I have a question here that where exactly in the above steps, I would have thought of splitting that support of z and why? Any help would be appreciated.

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  • $\begingroup$ I was thinking about this. One thing that I came up with is: since $0<u-v<1$, it implies that $v<u<1+v$. Also, $v$ itself is between 0 and 1. I can write $0<v<u<1+v<2$. Hence, from this new modified limits, I can see two different ranges of values that v can take. One is $0<v<u$ and another is $u-1<v<1$. Is this reasoning correct for this problem? $\endgroup$
    – userNoOne
    Commented Feb 12, 2021 at 6:18
  • $\begingroup$ Please add self-study as a tag. $\endgroup$
    – Xi'an
    Commented Feb 12, 2021 at 7:20
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    $\begingroup$ For an account of several ways to deal with this issue, see stats.stackexchange.com/a/43075/919. $\endgroup$
    – whuber
    Commented Feb 12, 2021 at 15:16

1 Answer 1

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The joint density is $$f(u,v)=\mathbb I_{(0,1)}(u-v)\mathbb I_{(0,1)}(v)$$ For a fixed value of $u$, the function $$v\longmapsto \mathbb I_{(0,1)}(u-v)\mathbb I_{(0,1)}(v)$$ can be rewritten $$v\longmapsto \mathbb I_{0\le u-v\le 1}\mathbb I_{(0,1)}(v) =\mathbb I_{u-1\le v\le u}\mathbb I_{(0,1)}(v)=\mathbb I_{(u-1,u)}(v)\mathbb I_{(0,1)}(v)$$ Meaning that, conditional on $U=u$, $V$ is between $0$ ans $1$ AND between $u-1$ and $u$. Thus$$ \mathbb I_{(0,1)}(u-v)\mathbb I_{(0,1)}(v)=\mathbb I_{(\max(0,u-1),\min(u,1))}(v)$$

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  • $\begingroup$ What does the symbol $\longmapsto$ mean here? $\endgroup$
    – userNoOne
    Commented Feb 12, 2021 at 8:06
  • $\begingroup$ This is a traditional mathematical notation for functions and reads "maps $v$ to $\mathbb I$..." $\endgroup$
    – Xi'an
    Commented Feb 12, 2021 at 8:45

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