stationary independent increments: f(t+s) = f(t) + f(s), therefore f(t)=ct

Question

I don't get it... Why is the only solution to:

$$f(t+s) = f(t) + f(s)$$

equal to:

$$f(t) = ct$$

???

I have no idea how they came to this conclusion...

Answer 1

$X(t)$ has stationary increments if $X(t)-X(s)$ has same distribution as $X(t-s)$ for $t-s>0$.

$X(t)$ has independent increments if for $t_n>\ldots>t_0$ if the increments $X(t_n)-X(t_{n-1}),\ldots,X(t_1)-X(t_0)$ have the same distribution.

Stationary independent increments are used in equation (5.123).

Let us suppose the function $f$ is continuous (not enough details given for that, but let us suppose this). A function that fulfills $f(ax+by)=af(x)+bf(x)$ is said to be linear which probably is the part you missed.

In probabillity $f$ is deterministic but in statistics it is random when taking finite number of samples. Myself, I used to mess up possibly with statistics. In probability we get the same mean (the theoretical mean we get from intergration) everytime but in statistics we get always a random mean.

PS: Have a look at the difference between weak and strong stationarity.

Answer 2

As you mentioned, The part about the solution of the functional equation

$\bullet~$ Lemma: The only non-constant continuous function $f$ $\in$ $\mathbb{R}$ satisfying functional equation $$ f(s + t) = f(s) + f(t) \quad \text{for any } s , t \in \mathbb{R} $$ is the linear function $f(x) = c x~$ for any $x$ $\in$ $\mathbb{R}$ and some $c$ $\in$ $\mathbb{R}$.

$\bullet~$ Proof: Plugin $s = t = 0$, we have $$ f(0) = f(0) + f(0) \implies f(0) = 0 $$ Then plugin $s = t = x$, we have that $$ f(2x) = 2f(x) \quad \text{for any } x \in \mathbb{R} $$ Again, plugin $s = -t $, we have that $$ f(0) = f(t) + f(- t) \implies f(-t) = -f(t) \quad \text{for any } t \in \mathbb{R} $$

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$\bullet~$ Claim: For any $n \in \mathbb{Q},$ $f(n) = c n$.

$\bullet~$ Proof: Then we have a sub-claim

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$\bullet \bullet~$ Sub-Claim: The function $f$ satisfies $$ f(kt) = kf(t) \quad \text{for any } t \in \mathbb{R} ~\text{ and any } k \in \mathbb{N} $$

$\bullet \bullet~$ Proof: The proof is straightforward by induction.

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And we have that $f$ is an odd function. Then for any $n \in \mathbb{N},$ we have that $$ f(-nt) = n f(-t) = -n f(t) \quad \text{for any } t \in \mathbb{R} $$ Then we have that $f(k) = ck ~$ for any $k \in \mathbb{Z}$

The part of the proof of $\mathbb{Q},$ Is easy.

Hence we have that $f(t) = ct ~\text{ for any } t \in \mathbb{Q}$

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$\bullet~$ Claim: The solution function $f(t) = ct~$ holds for any $t \not\in \mathbb{Q}$.

$\bullet~$ Proof: We can construct a sequence of rationals convergent to an irrational point.

So let's consider $\{x_{n} \}_{n = 1}^{\infty} \in \mathbb{Q}$ such that $~x_{n} \to x~$ and $~x \not \in \mathbb{Q}.~$

Then we have that \begin{align*} &~f(x_{n}) = c x_{n} \quad\\ \implies & \lim_{n \to \infty} f(x_{n}) = \lim_{n \to \infty} cx_{n}\\ \implies & f\Big(\lim_{n \to \infty} x_{n}\Big) = c \lim_{n \to \infty} x_{n} \quad \text{as } f \text{ is continuous on } \mathbb{R}] \\ \implies & f(x) = cx \end{align*} As $x \not \in \mathbb{Q} $ was arbitrary, hence we can say that $f(x) = cx ~$ for any $x \not \in \mathbb{Q}.$

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Hence we have completed our proof of $f(x) = cx~$ for any $x \in \mathbb{R}$, when $f$ is continuous.