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This is my first question here, so please be patient!

For chance, I met these statistical functions...

s(t) = 0.8^t   {t>0}
s(t) = 0.9^t   {t>0}

They go from 1 (at t=0) towards 0 (for x going to +inf)

Calculating the integral from 0 to k and from k to +inf and finding the point k0 where the two areas are balanced (equal) we get:

Solve[Integrate[(.9^(x)),{x,0,k}]==Integrate[(.9^(x)),{x,k,Infinity}],{k}]

{{k->6.57881}}

What surprised me is that at x=2.40942enter image description here

0.9^x is exactly 0.5

Solve[.9^(x)==.5,{x}]
{{k->6.57881}}

So, dividing the domain [0,+inf) in two such that the left area and the right area below the curve are the same, is equivalent to dividing the codomain (0,1] in two at half

Same applies for 0.8, for 0.75, for 0.5 and, I'd guess, any other base from 0 to 1.

My questions are:

  • why is this? What is the mathematical reason?
  • does it have a phisical world or statistical/logical meaning?
  • you can probably adjust many functions to satisfy some similar property, but are there other functions that naturally satisfy this property, is it a known/studied group of functions?

Thanks.

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    $\begingroup$ What's the meanig of "Same applies for 0.8, for 0.75, for 0.5 "? $\endgroup$ – popi Jun 5 '19 at 11:17
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    $\begingroup$ So you're studying functions of the form $f(x) = a^x = \exp{bx}$ for some $b<0$. It's easy to integrate, the integral is $F(x) = \frac{1}{b}\exp{bx}$. The definite integral from 0 to infinity is then $F(\infty) - F(0) = 0 - \frac{1}{b} = -\frac{1}{b}$... And so on. You could just analyse this algebraically. $\endgroup$ – Matti P. Jun 5 '19 at 11:21
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    $\begingroup$ This is because of the scalability or the self-similarity of the exponential function. $\endgroup$ – tomi Jun 5 '19 at 11:21
  • $\begingroup$ @popi I meant to say that the same property that is true for 0.9^x is true for 0.8^x and 0.75^x and as proved below for any b^x with b<1 $\endgroup$ – FrancescoMM Jun 5 '19 at 16:52
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Great question!

Your functions all have the property that they are of the form $$ f(x) = a b^x $$ for some constants $a$ and $b$ (with $b < 1$). It's tempting to think that something special is going on with this class of "exponential" functions.

Alas, that's not true. Consider a function like $$ h(x) = \begin{cases} c & 0 \le x < 1 \\c/2 & x = 1\\ c/3 & 1 < x < 4 \\ 0 & \text{otherwise} \end{cases} $$ for any $c > 0$.

It, too, has your property: the maximum value is $c$, and if we split at height $c/2$, then half the area is to the left and half the area is to the right of the split-point. We can adjust the maximum value by changing $c$, and can move the split-point by considering $k(x) = h(x-a)$, which has its "split" at $x = a+1$. So there are a lot of these functions! (And it's pretty clear how to construct others like them)

You might object that my function is not continuous, but it shouldn't be too hard to add some sloping bits to connect the straight parts and still maintain the property you describe.


As for "why do your functions have this property?", let's look at $f(x) = a b^x = a \exp(x \ln b)$. Its maximum (for $b$ positive) is at $x = 0$, and is $a$; the half-max is at the point where $a b^x = a/2$, so $b^x = 0.5$, so $x = \log_b \frac12$.

The area to the left of this split point is \begin{align} A_1 &= \int_0^{\log_b \frac12} a \exp(x \ln b)dx \\ &= a \int_0^{\log_b \frac12} \exp(x \ln b)dx \\ &= a \left(\left. \frac{\exp(x \ln b)}{ \ln b} \right|_0^{\log_b \frac12}\right) \\ &= a \left(\left. \frac{b^x}{ \ln b} \right|_0^{\log_b \frac12}\right) \\ &= a \left(\frac{b^{{\log_b \frac12}}}{ \ln b} - \frac{b^0}{ \ln b} \right) \\ &= a \left(\frac{\frac12}{ \ln b} - \frac{1}{ \ln b} \right) \\ &= a \left(\frac{-\frac12}{ \ln b} \right) \end{align} which makes sense because for $b < 1$, we have $\ln b < 0$, so this number $A_1$ is positive.

The other area is \begin{align} A_2 &= \int_{\log_b \frac12}^\infty a \exp(x \ln b)dx \\ &= a \left(\left. \frac{\exp(x \ln b)}{ \ln b} \right|_{\log_b \frac12}^\infty\right) \\ &= a \left(\left. \frac{b^x}{ \ln b} \right|_{\log_b \frac12}\right)^\infty \\ &= a \left( \lim_{c \to \infty} \frac{b^c}{ \ln b} - \frac{b^{{\log_b \frac12}}}{ \ln b} \right) \\ &= a \left( 0 - \frac{b^{{\log_b \frac12}}}{ \ln b} \right) \\ &= a \left(\frac{-\frac12}{ \ln b} \right) \end{align} which is the same as $A_1$. So your exponential functions do indeed have this unusual property.

It's pretty cool that you happened to notice this -- that's how discoveries get made. It's too bad that it's not as special as it might have first appeared.

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  • $\begingroup$ Thanks, that is exactly what I was expecting, great and very clear explanation! Considering these functions represent the probability of a user subscribed to a service where 20% (0.8) or 10% (0.9) leave the service every year do you think this very nice property can represent any real world (statistical or logical) property? $\endgroup$ – FrancescoMM Jun 5 '19 at 16:27
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Not an answer to the asked question, but a generalization.

As the answer by @JohnHughes pointed out, there are many more functions $f()$ where the area is divided into halves at exact the point the height is divided into halves. For simplicity, lets assume $f(x) = 0$ when $x \le 0$ and $f$ is decreasing. Then your requirement is simply:

$${f(x) \over f(0)} = {\int_x^\infty f(y) dy \over \int_0^\infty f(y) dy} \,\,\,\,\, \text{ at some point $x$ where both ratios $= \frac12$}$$

Since you specified only what must happen at one point, this leaves a lot of leeway for what can happen at other points.

However, your original functions are actually exponentials, and for them, the property holds much more generally, indeed for any value of the ratio, i.e.

$${f(x) \over f(0)} = {\int_x^\infty f(y) dy \over \int_0^\infty f(y) dy} \,\,\,\,\, \text{ at every point $x > 0$}$$

You can easily verify that the above holds for any $f(x) = A e^{-bx}$ by integrating. You can also easily try it out by modifying your code snippets to try other ratios besides $\frac12$.

If I'm not mistaken, the latter requirement (that the ratios be equal for all $x>0$) is only satisfied by the exponential function. And I'm guessing this is the "self-similarity" that the comment by @tomi refers to.

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  • $\begingroup$ Thanks, great addition, thia is even broader! $\endgroup$ – FrancescoMM Jun 5 '19 at 18:45
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    $\begingroup$ This is great! Nice generalization of the constraint to get a unique solution. $\endgroup$ – John Hughes Jun 6 '19 at 0:34

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