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/']
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The probability density for the step $n$ is given by
$\ds{{\rm p}\pars{\vec{r}\_{n}} \equiv {\delta\pars{r\_{n} - 1} \over 2\pi}}$. The probability density $\pp\_{N}\pars{\vec{r}}$ of arriving at $\vec{r}$ after $N$ steps is given by:

> 
> \begin{align}
> \pp\_{N}\pars{\vec{r}}
> &\equiv
> \int\dd^{2}\vec{r}\_{1}\,{\rm p}\pars{\vec{r}\_{1}}\ldots
> \int\dd^{2}\vec{r}\_{N}{\rm p}\pars{\vec{r}\_{N}}
> \delta\pars{\vec{r} - \sum\_{\ell = 1}^{N}\vec{r}\_{\ell}}
> \\[3mm]&=
> \int\dd^{2}\vec{r}\_{1}\,{\rm p}\pars{\vec{r}\_{1}}\ldots
> \int\dd^{2}\vec{r}\_{N}{\rm p}\pars{\vec{r}\_{N}}
> \int\exp\pars{\ic\vec{k}\cdot\bracks{\vec{r} - \sum\_{\ell = 1}^{N}\vec{r}\_{\ell}}}
> \,{\dd^{2}\vec{k} \over \pars{2\pi}^{2}}
> \\[3mm]&=
> \int\expo{\ic\vec{k}\cdot\vec{r}}
> \bracks{\int\dd^{2}\vec{R}\,{\rm p}\pars{\vec{R}}\expo{-\ic\vec{k}\cdot\vec{R}}}^{N}
> \,{\dd^{2}\vec{k} \over \pars{2\pi}^{2}}
> =
> \int\expo{\ic\vec{k}\cdot\vec{r}}
> \bracks{\int\_{0}^{2\pi}\expo{-\ic k\cos\pars{\theta}}\,{\dd\theta \over 2\pi}}^{N}
> \,{\dd^{2}\vec{k} \over \pars{2\pi}^{2}}
> \end{align}
> 

However
$$
\int\_{0}^{2\pi}\expo{-\ic k\cos\pars{\theta}}\,{\dd\theta \over 2\pi}
=
\int\_{-\pi}^{\pi}\expo{\ic k\cos\pars{\theta}}\,{\dd\theta \over 2\pi}
=
{1 \over \pi}\int\_{0}^{\pi}\expo{\ic k\cos\pars{\theta}}\,\dd\theta
=
{\rm J}\_{0}\pars{k}
$$
where ${\rm J}\_{\nu}\pars{k}$ is the
$\nu$-$\it\mbox{order Bessel Function of the First Kind}$.

> 
> \begin{align}
> \pp\_{N}\pars{\vec{r}}
> &=
> \int\expo{\ic\vec{k}\cdot\vec{r}}{\rm J}\_{0}^{N}\pars{k}
> \,{\dd^{2}\vec{k} \over \pars{2\pi}^{2}}
> =
> {1 \over 2\pi}\int\_{0}^{\infty}\dd k\,k\,{\rm J}\_{0}^{N}\pars{k}\int\_{0}^{2\pi}
> \expo{\ic kr\cos\pars{\theta}}\,{\dd\theta \over 2\pi}
> \end{align}
> $$
> \color{#ff0000}{\pp\_{N}\pars{\vec{r}}
> =
> {1 \over 2\pi}\int\_{0}^{\infty}{\rm J}\_{0}\pars{kr}{\rm J}\_{0}^{N}\pars{k}k
> \,\dd k}
> $$
> 

The probability ${\rm P}\_{N{\Huge\circ}}$ that it returns to the unit circle after $N$ steps is given by:
\begin{align}
\color{#0000ff}{\large{\rm P}\_{N{\Huge\circ}}}
&=
\int\_{r\ <\ 1}\pp\_{N}\pars{\vec{r}}\,\dd^{2}\vec{r}
=
\int\_{0}^{\infty}\overbrace{\bracks{\int\_{0}^{1}{\rm J}\_{0}\pars{kr}r\,\dd r}}
^{{\rm J}\_{1}\pars{k}/k}
{\rm J}\_{0}^{N}\pars{k}k\,\dd k
\\[3mm]&=
\color{#0000ff}{\large\int\_{0}^{\infty}{\rm J}\_{1}\pars{k}
{\rm J}\_{0}^{N}\pars{k}\,\dd k}
\end{align}
We compute a few values with Wolfram Alpha:
$$
\begin{array}{rclrcl}
{\rm P}\_{0{\Huge\circ}} =\int\_{0}^{\infty}{\rm J}\_{1}\pars{k}
\,\dd k & = & 1\,,\quad
{\rm P}\_{1{\Huge\circ}} =\int\_{0}^{\infty}{\rm J}\_{1}\pars{k}
{\rm J}\_{0}\pars{k}\,\dd k & = & \half
\\
{\rm P}\_{2{\Huge\circ}} =\int\_{0}^{\infty}{\rm J}\_{1}\pars{k}{\rm J}\_{0}^{2}\pars{k}
\,\dd k & = & {1 \over 3}\,,\quad
{\rm P}\_{3{\Huge\circ}} =\int\_{0}^{\infty}{\rm J}\_{1}\pars{k}
{\rm J}\_{0}^{3}\pars{k}\,\dd k & = & {1 \over 4}
\\
{\rm P}\_{4{\Huge\circ}} =\int\_{0}^{\infty}{\rm J}\_{1}\pars{k}{\rm J}\_{0}^{4}\pars{k}
\,\dd k & = & {1 \over 5}\,,\quad
{\rm P}\_{5{\Huge\circ}} =\int\_{0}^{\infty}{\rm J}\_{1}\pars{k}
{\rm J}\_{0}^{5}\pars{k}\,\dd k & = & {1 \over 6}
\end{array}
$$

It $\tt\large seems$ the exact result is
$\ds{%
{\rm P}\_{N{\Huge\circ}}
=
\int\_{0}^{\infty}{\rm J}\_{1}\pars{k}{\rm J}\_{0}^{N}\pars{k} \,\dd k
=
{1 \over N + 1}}$

Mathematica can solve this integral and it yields the $\ul{\mbox{exact result}}$ $\color{#0000ff}{\Large{1 \over N + 1}}$ when $\color{#ff0000}{\large\Re N > -1}$.
This is not a complete answer, but is kind of long for a comment. 

For $t=2$, draw a picture. It's pretty easy to see the answer is $1/3$.

For the $t=3$ case, let the angles be $\theta\_1$, $\theta\_2$, and $\theta\_3$. By symmetry, you can assume without loss of generality that $\theta\_1 = 0$. So you need to find the probability that
$\|\langle 1, 0 \rangle + \langle \cos\theta\_2, \sin\theta\_2 \rangle +\langle \cos\theta\_3,\sin\theta\_3\rangle \|<1$. A little computation shows that this is equals 
$P(1+\cos\theta\_2+\cos\theta\_3+\cos(\theta\_2-\theta\_3)<0)$, where $\theta\_2$ and $\theta\_3$ are chosen independently from the uniform distribution on $[0,2\pi]$. That's as far as I got.

This is the probability that the particle is in the unit disc *at time* $t=3$ (not $t$ equals 2 or 3).