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?


1 Answer 1


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. $p_U(u)$.

  • 1
    $\begingroup$ OH!!! I see! Thank you! $\endgroup$
    – Tita
    Jan 6, 2022 at 20:48
  • $\begingroup$ You're welcome! $\endgroup$ Jan 6, 2022 at 21:20

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