# How can I show that the maximum is 0?

I am trying to solve this exercise (I don't know if it's from a book, feel welcome to credit it if you've been it before). It says that X,Y are two independent, equal random variables. If their (common) PDF ($$p(x)=p_X(x)=p_Y(y)$$) is even, show that $$p_{X+Y}$$ is also even and then that its maximum is taken at 0.

So far I've tried to show what it asks, I took the tranform $$x=v$$ and $$y=u-v$$ and that way I got that: $$p_U(u) = \sum_{v\in\mathcal{R}_V} p(v)\cdot p(u-v)$$ which is also even as a sum of products of even functions.

But now, I am asked to show that it is also max at $$0$$. I have this: $$p_U(u) = \sum_{v\in\mathcal{R}_V} p(v)\cdot p(u-v) = \sum_{v\in\mathcal{R}_V} p(v)\cdot p(v-u) \leq \sum_{v\in\mathcal{R}_V} p(v)\cdot p(v) = \sum_{v\in\mathcal{R}_V} p^2(v)$$

But that's all I can do. I'm totally confused about this exercise. Does anyone have an idea on how I can work on this further or start again?

You are almost there, you just need to notice that the right-hand side is $$p_U(0)$$ :)
$$\begin{split} p_U(u) &= \sum_{v\in\mathcal{R}_V} p(v)\cdot p(u-v)\\ &= \sum_{v\in\mathcal{R}_V} p(v)\cdot p(v-u) &\,\,&\text{(as p is even)}\\ &\leq \sqrt{\sum_{v\in\mathcal{R}_V} p(v)^2}\sqrt{\sum_{v\in\mathcal{R}_V} p(v-u)^2}& \,&\text{(by Cauchy-Schwarz)}\\ &\leq \sqrt{\sum_{v\in\mathcal{R}_V} p(v)^2}\sqrt{\sum_{v\in\mathcal{R}_V} p(v)^2}& \,&\text{(change of index v\rightarrow v+u)}\\ &\leq \sum_{v\in\mathcal{R}_V} p(v)^2& \,&\text{(rearranging)}\\ &\leq\sum_{v\in\mathcal{R}_V} p(v)p(-v)& \,&\text{(as p is even)}\\ &\leq p_U(0) \end{split}$$
Note: Some people get confused that I keep using $$\leq$$ between two lines that are equal (like the last one and the one before). The reason I do so is because $$\leq$$ is relative to the left-hand side, ie. $$p_U(u)$$.