Is there a probability density function satisfying the following conditions?

Question

I find myself in need of the solution of this problem in finding a probability density function. I had asked this question in Math Stack Exchange but I did not get an answer so I am posting it here.

Let $x$ and $y$ be positive reals satisfying all the following conditions:

1. $f(x,y)$ is strictly increasing for $x \ge 0$
2. $f(0,y) = 0$
3. $f(1,y) = 1$
4. $f(x,y) \to \infty$ as $x \to \infty$
5. For every $y > 0$, $$ \int_{0}^{\infty} e^{-y f(x,y)}dx = 1. $$

My questions are:

1. Is there a exist a function satisfying the all above conditions?

2. Are there infinitely many functions satisfying the all above conditions?

Preferably I would like to avoid artificially constructed function like piece-wise models. However this is not a strict constraint and if non-piecewise solutions are not available then piecewise solutions will do.

Condition 1-4 are easy to satisfy but condition 5 has made the solution elusive for me. However for my actual problem, I can do with a weaker form of condition 5 which is for all $0 < y < 5$, we must have $\int_{0}^{\infty} e^{-y f(x,y)}dx = 1$.

Answer 1

Your main problem is that you use an overly complicated notation $e^{-yf(x,y)}$ for a generic strictly positive probability density $p(x,y)$. Your conditions are just

1) $p$ is decreasing in $x$ 2) $p(0,y)=1$. 3) $p(1,y)=e^{-y}$

4) is redundant because any positive decreasing integrable function satisfies it automatically. Thus, all you need is a way to draw a decreasing probability density through the points $(0,1)$ and $(1,e^{-y})$ with $y>0$. Now the only thing that restricts you is your imagination.

However, this also makes me think that you screwed the problem statement up somewhere because it is now obvious that you can find a family like that so that every resulting distribution stays in an arbitrarily small neighborhood of the uniform distribution on $[0,1]$, which is sort of ridiculous (why would you need all this "family story" then?). So I'll stop here until you confirm that what you wrote is, indeed, what you meant.

Edit (a nonsensical example): Denote $a=10^{10^{10}}$. Put $$ p(x,y)=\begin{cases} 1-(1-e^{-y})x^{a-1}, &x\le 1 \\ e^{-y}\exp\left(-\frac{a}{e^y-1}(x-1)\right), &x\ge 1 \end{cases} $$ You will, probably, tell me that this has nothing to do with what you had in mind and I'll agree completely. However, all formal conditions are met. The thing I'm trying to squeeze from you is what exactly is wrong with using this monster in your problem.

Answer 2

I don't think this is a research question, but here you go. Treat the problem separately for each value of $y$. So we want a function $f(x)$ such that $f(0) = 0$, $f(1) = 1$, it increases to infinity as $x \to \infty$, and $\int_0^\infty e^{-yf(x)}\, dx = 1$.

The function $f(x) = x^a$ satisfies all the conditions but the last, for any positive value of $a$. If you take $a \to 0$ then the integral goes to $\int_0^\infty e^{-y}\, dx = \infty$. So it's possible to get a value that's too big. On the other hand, if we let $f(x)$ go from $1$ to $\infty$ arbitrarily fast on the interval $[1,\infty)$ then the integral over that region can be made as small as we want. Whereas the smallest the integral can be over the interval $[0,1]$ is with something like $f(x) = x^a$ for $a$ small, so the integral is approximately $\int_0^1 e^{-y}\, dx = e^{-y}$ on that region. This is less than 1 for any $y > 0$, so by making $f$ grow sufficiently fast for $x > 1$ you can also get the total integral to be less than 1. Since you can get a value that's too large and a value that's too small, somewhere in between you can get exactly the right value. It should be obvious that if there is one way to do it there will be infinitely many ways to do it. I don't think there's going to be any nice formula though.