# How to find limits of integration on a convolution of CRVs

In finding the convolution of two independent and continuous random variables, I am struggling with limits of integration. I cannot seem to figure out over what intervals the probability density function $f_{Z}(a) = f_{X+Y}(a)$ breaks out to.

The most basic example is where $f_{X}(a)$ and $f_{Y}(a)$ are both uniform over $[0,1]$ and independent. The intervals for $f_{Z}(a)$ are $(0,1)$ and $(1,2)$. But why? There are also more complicated cases such as two exponential R.V.s (say with parameters $\lambda$ and $2\lambda$) or an exponential and a uniform, (say $\lambda$ and $[0, 1]$), etc.

For reference: $$f_{x+y}(a) = \int_{-\infty}^{\infty} f_X(a-y)f_Y(y)~dy$$

Once I can set it up, the integration is (usually) no trouble.

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The breakpoints for the cases you mention occur where $a-y$ becomes negative. One can keep track by by using indicator functions. But students, particularly those with a good background in 2 variable calculus, can usually push things through with a good diagram. – André Nicolas Mar 20 '13 at 22:17
I love the diagram method!!! I just don't know how to draw it in this case. – Katarzyna Mar 21 '13 at 0:11
Thanks, everyone! I solved it! I just wanted to say that so others didn't work hard to add thoughtful responses that are not needed. I will edit this answer with more details next week for the benefit of the future struggler... busy studying for now. – Katarzyna Mar 21 '13 at 4:43

One easy way to think about this is to look at $f_X$ and $f_Y$ and figure out where exactly these pdfs vanish, thus not contributing anything to the convolution.

Let's look at the uniform case. Since the uniform is defined on $[0,1]$, obviously anything outside of this won't make sense. So now we have $$f_{X+Y}(a)=\int_{-\infty}^\infty f_X(a-x)f_Y(x) \mathrm{d}x=\int_{0}^1 f_X(a-x)f_Y(x) \mathrm{d}x$$

Now, $f_Y$ is "safe", since it is defined on $[0,1]$. We need to ensure the same thing for $f_X$, that is, we want $0 \leq a-x \leq 1$. From this, one gets $x \leq a$ and $x \geq a-1$. This is what breaks the integral into two pieces. The first piece tells you that you should have $$f_{X+Y}(a)=\int_0^a \mathrm{d}x = a$$ as long as $x \leq a$, but $x \in [0,1]$ according to the convolution domain, implying $a \in [0,1]$ as well.

From the second inequality we get $$f_{X+Y} = \int_{a-1}^1 \mathrm{d}x = 2-a$$ as long as $x \geq a-1$. Since $x \in [0,1]$, then $a \in [1,2]$. This reconciles with the obvious observation that the sum of two r.v. in $[0,1]$ must live in $[0,2]$.

There's your triangular distribution: $a$, when $a \in [0,1]$ and $2-a$ when $a \in [1,2]$.

If you want to convolve, say, two exponentials, then your convolution domain reduces to $[0, \infty)$. Your inequality should say that $0 \leq a-x < \infty$. Can you take it from here?

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Not to diminish your efforts, but I still don't understand completely. So, there are really two intervals to figure out-- one is the limits of integration and the second is the domain. I assume one comes from the other, but I don't know how to relate them. Also... the infinity case is not clear at all. For a start, how do I split the inequality? Does dividing by -1 make it $-\infty$ – Katarzyna Mar 21 '13 at 0:05
Also, in the case where the two domains are not the same, such as with an exponential and a uniform... I don't understand which to apply and where. – Katarzyna Mar 21 '13 at 0:10