It says in my random processes book that if a random process $X_t$ with stationary independent increments has value $0$ at the start ($X_0 = 0$) then it is completely determined by it's first order distribution.
And that $E(X_t) = at$ where $a = E(X_1)$. From this it makes sense to me that
$$ X_t = tX_1 $$
But I don't know how to make it a formal proof, I tried for natural numbers,
$$ X_t = (X_t - X_{t/2} ) + (X_{t/2} - X_0) = 2X_{t/2} $$
And similarly for arbitrary $n\in \mathbb{N}$,
$$ X_t = nX_{t/n} $$
I don't know how to extend this to real numbers, i tried like this, if $n \to \infty$, then $nt$ "converges" (I know this cannot work, this is where I am stuck) to a natural number and so
$$ X_t = ntX_{1/n} $$
Then because $nX_{1/n} = X_1$, it follows that $X_t = tX_1$
How would one really prove this, i.e. extend the natural numbers argument to real numbers?
Note that the equality
$$X_t = 2 X_{t/2}$$
does not hold. It follows from stationarity of the increments that
$$X_{t}-X_{t/2} \stackrel{d}{\sim} X_{t/2},$$
i.e. they are equal in distribution and not almost surely. (Recall that $X \stackrel{d}{\sim}Y$ does not imply $X+Y = 2X$.) This means that we can use the equation
$$X_t = \underbrace{(X_t-X_{t/2})}_{\sim X_{t/2}} + (X_{t/2}-X_0) \tag{1}$$
whenever we want to draw conclusions about the distribution of $X_t$. E.g. $(1)$ implies
$$\mathbb{E}X_t = \mathbb{E}(X_t-X_{t/2}) + \mathbb{E}(X_{t/2}) = 2 \mathbb{E}(X_{t/2}).$$
One of the most basic examples for processes with stationary and independent increments is the Poisson process. Just draw a picture how the typical sample path of a Poisson process looks like - then you will see that $X_t = t X_1$ does not hold.