# Confusion in deriving the distribution of sum of two uniform variates

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

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.

• 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? Feb 12 at 6:18
• Please add self-study as a tag. Feb 12 at 7:20
• For an account of several ways to deal with this issue, see stats.stackexchange.com/a/43075/919.
– whuber
Feb 12 at 15:16

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$$\mathbb I_{(0,1)}(u-v)\mathbb I_{(0,1)}(v)=\mathbb I_{(\max(0,u-1),\min(u,1))}(v)$$
• What does the symbol $\longmapsto$ mean here? Feb 12 at 8:06
• This is a traditional mathematical notation for functions and reads "maps $v$ to $\mathbb I$..." Feb 12 at 8:45