Functional Equation Logic Dispute

Question

I have a dispute about functional equation logic with my professor

We have a function $\alpha (t) \, t \geq0$ that represents a probability density, for which we know: $$P(t_1 \leq t \leq t_2)= \int_{t_1}^{t_2} \alpha (t) dt$$ $$ P(0 \leq t \leq t_1)=P(t_0 \leq t \leq t_0 + t_1 |t \geq t_0) $$

After doing some manipulating we can arrive at.
$$\alpha '(t_0+t_1)= - \alpha(t_0) \alpha(t_1) \, \forall t_0,t1 \in \mathbb{R^+_0}$$

My argument is that if there exists a function that fulfills this property, so for which this statement is true for all $t_0$, then the function must certainly fulfill that statement for $t_0=0$

$$\alpha '(t_1)= -\alpha(0) \alpha(t_1)$$

This is the functional equation for the negative exponential!

My argument is that this condition is so strong, that it forces $\alpha(t)$ in general to be an expontential function, despite having only looked at one case. This single case is so strong, that it already forces the shape of the function.

He says that this only proves it for the case $t_0=0$. I am genuinely curious as to who is right, it seems like a pretty important logical point.

Answer 1

Being given functional/differential equation

$$f'(x+y)= - f(x)f(y)\, \ \ \forall x,y \in \mathbb{R^+_0}\tag{1}$$

Assuming $f$ is twice differentiable :

$$f''(x+y)=\begin{cases}-f'(x)f(y) \ \ (\text{diff. wrt } x)\\-f(x)f'(y) \ \ (\text{diff. wrt } y)\end{cases}$$

As a consequence :

$$\frac{f'(x)}{f(x)}=\frac{f'(y)}{f(y)}$$

$x,y$ being independent variables, by a classical argument, the previous fraction must be a constant $k$.

Therefore, $f$ must indeed verify functional equation :

$$f'(x)=k f(x)$$

As a consequence:

$$f(x)=Ae^{kx}$$

with $k<0$ due to the presence of minus sign in (1).

Answer 2

You have already noted that $\alpha(0)>0$.

On the other hand, it follows from $\alpha '(t)= -\alpha(0) \alpha(t)$ that $$ \alpha(t)=\alpha(0)e^{-\alpha(0)t} $$ for all $t\ge0$. Substituting in $\alpha '(t_0+t_1)= - \alpha(t_0) \alpha(t_1)$ gives $$ -\alpha(0)^2e^{-\alpha(0)(t_0+t_1)}=-\alpha(0)e^{-\alpha(0)t_0}\alpha(0)e^{-\alpha(0)t_1}, $$ which means that your function $\alpha(t)$ also satisfies the first identity for $t_0\ne0$.

Summing up, you do have an exponentially decaying function $\alpha(t)=\alpha(0)e^{-\alpha(0)t}$.

Answer 3

I guess the problem is behind interpreting of what we know about $\alpha(t)$.

In particular, great point by lulu about second condition.

First "condition": $$P(t_1 \le t \le t_2)=\int_{t_1}^{t_2}\alpha(t)dt \quad \forall t,t_1,t_2\in\Re_0^+\tag{1}\label{1}$$ Since in this context it's just another way to say that $\alpha(t)$ represents a probability density for probability distribution $P$ you shouldn't be able to get something specific about $\alpha(t)$ from the first "condition".

So let's consider second condition: $$P(0\le t \le t_1)=P(t_0+t_1\le t \le t_0|t\ge t_0) \quad \forall t,t_0,t_1\in\Re_0^+\tag{2}\label{2}$$ Let's rewrite to perform "more functional" approach: $$P(t \le t_1)=\frac{P(t_0\le t \le t_0+t_1)}{P(t\ge t_0)}$$ Notice that $P(t\ge t_0)=0$ is pointless, so let's multiply both parts by it: $$P(t \le t_1)P(t\ge t_0)=P(t_0\le t \le t_0+t_1)=\\P(t \le t_1)(1-P(t\le t_0))=P(t \le t_0+t_1)-P(t \le t_0)=\\P(t \le t_1)-P(t \le t_1)P(t\le t_0)=P(t \le t_0+t_1)-P(t \le t_0)$$ Denoting $f(z)=P(t \le z)$, $x=t_1$, $y=t_0$ and getting: $$f(x)-f(x)f(y)=f(y+x)-f(y)\\f(x+y)=f(x)+f(y)-f(x)f(y)\tag{3}\label{3}$$ Considering $0\le f(x)\le1$ and solving (in a similar way) we get $f(x)=1-e^{Cx},\,C<0$ and consequently $\alpha (t)=f'(t)=-Ce^{Cx}$ (we can get the same even with constraints $x>0,\,y>0$ in \eqref{3}).

However, if you refer to $t_0=0$ as to the case it seems like $t_0$ is meant to be some fixed value (and if so $t_0=0$ is pointless in \eqref{2}).

With the last interpretation $t_0=0$ is pointless (as it transforms \eqref{2} into identity) and similarly to the previous approach we can get: $$f(x+t_0)=f(x)+f(t_0)-f(x)f(t_0),\,f(t_0)>0\tag{4}\label{4}\\f(x)=1-g(x)\\1-g(x+t_0)=(1-g(x))+(1-g(t_0))-(1-g(x))(1-g(t_0))\\1-g(x+t_0)=1-g(x)+1-g(t_0)-1+g(x)+g(t_0)-g(x)g(t_0)\\g(x+t_0)=g(x)g(t_0)$$ If $g(t_0)=0$ then $f(t_0)=1-g(t_0)=1$ and \eqref{4} transforms into $f(x+t_0)=1\,\, \forall x\ge 0$ what is already satisfied inasmuch as $f$ stands for probability distribution and reaches value $1$ at $t=t_0$. So any probability distribution function $f$ that satisfies $f(t_0)=1$ and \eqref{1} will also satisfy both conditions (for instance $f(t)=\sin (\frac{\pi t}{2t_0})$ if $t\le t_0$ otherwise $f(t)=1$ and corresponding $\alpha (t)=f'(t)$).

If $g(t_0)>0$ we can use the following substitution: $$g(x)=h(x)g(t_0)^{x/t_0}\\h(x+t_0)g(t_0)^{(x+t_0)/t_0}=h(x)g(t_0)^{x/t_0}g(t_0)^{t_0/t_0}=h(x)g(t_0)^{(x+t_0)/t_0}\\h(x+t_0)=h(x)$$ Returning back to $f$ we get $f(x)=1-h(x)(1-f(t_0))^{x/t_0}$ where $h$ is periodic with period $t_0$ and such that $f$ satisfies \eqref{1} ($h(0)=1$ will also come in hand).

To provide another non-exponential example for this "broader" case let's take for convenience $t_0=1$, $f(t_0)=0.5$ and $h(x)=1+\frac{\sin^2 (\pi x)}{8}$, so $f(t)=1-(1+\frac{\sin^2 (\pi t)}{8})0.5^t$ and correspondent $\alpha (t)=f'(t)=\ln (2)(1+\frac{\sin^2 (\pi t)}{8})0.5^t-\pi \sin (2\pi t)0.5^{t+3}$ (you can check that $\alpha (t)\ge0$, $f(t)\rightarrow 1$ as $t\rightarrow \infty$, so that condition \eqref{1} is also satisfied).

In short, your claim is provable ($\alpha(t)$ is and only exponential) if we assume condition \eqref{2} $\forall t_0,t_1\in \Re$.

From the other hand, if $t_0$ is fixed value (and by the way the same concerning $t_1$ thanks to \eqref{3} being symmetrical) non-exponential examples of $\alpha(t)$ can be provided as well.