Let $E:\mathbb{R} \to \mathbb{R}$ be an infinitely continuously differentiable function and $E$ is not zero function 
such that $$E(u+v)=E(u)E(v).$$
Show that $E(x)=e^{ax}$ for some $a\in \mathbb{R}$.

My partial answer:

The function $x\mapsto e^{ax}$ satisfies the properties immediately.
For $y=0$, we have $E(x)=E(x)E(0)$. Thus, $E(0)=1$. Let $y\in \mathbb{R}$ is fixed, then 
$$E'(x+y)=E(y)E'(x)$$.

I don't know how to continue the answer. 
Could you give me some hint? 
Thanks.
['https://math.stackexchange.com/questions/612405', 'https://math.stackexchange.com', 'https://math.stackexchange.com/users/27594/']
This is a classic problem, first solved by the Dutch mathematician J.C. Kluyver in 1905. He derives the result given in the [answer](https://math.stackexchange.com/a/612395/87355) of Felix Marin, that the probability to return to the unit disc after $n$ steps is $1/(n+1)$:

![](https://ilorentz.org/beenakker/MO/Kluyver1.png)

![](https://ilorentz.org/beenakker/MO/Kluyver2.png)

For a purely geometric derivation, see [arXiv:1007.4870](https://arxiv.org/abs/1007.4870). (There the result is attributed to Rayleigh, but it appears Rayleigh only considered the large-$n$ limit.) See also this [MO posting.](https://mathoverflow.net/a/347336/11260)
You can create a probility dristribution for the location after one step (the unit circle with a certain constant prbability). After two steps, I think that should be doable too. You can also carculate for each point the probability of entering de unit disk in the next step. Then, by combining the probability distribution ($P\_1(x,y)$) and the probability of ending in the disk $P\_2(x,y)$, you get
$$
\int P\_1(x,y)P\_2(x,y) \,dx\,dy
$$