MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $\int_{0}^{\infty} e^{-y f(x,y)}dx = 1$.

share|cite|improve this question
Does condition 1 mean "for each $y$, the function $f(x,y)$ is strictly increasing as a function of $x$"? Otherwise it conflicts with conditions 2 and 3. – Nik Weaver Aug 14 '13 at 15:25
@NikWeaver: Yes that is correct – Nilotpal Sinha Aug 14 '13 at 18:21

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.

share|cite|improve this answer
That's a slicker way to look at it. I guess it's worth mentioning that if $p$ is decreasing from $1$ at $(0,y)$ to $e^{-y}$ at $(1,y)$, then for small $y$ you've almost used up the entire unit of mass already, and have to decrease fast for $x > 1$. – Nik Weaver Aug 14 '13 at 23:31
@Nik, I had the same formulation that you have given and I can draw several distributions through $(0,1)$ and $(1,e^{-y})$. However satisfying condition 5 is the problem. For example $p(x,y) = e^{-yx^a}$ satisfies the first four conditions. Now if conditions 5 is also going to be true the we find that $y$ must be exactly equal to $\Gamma(1+1/a)^a$. But the minimum value of $\Gamma(1+1/a)^a$ is $e^{-\gamma}$. So this solutions is good for $y > e^{-\gamma}$ but it does not hold for $0 < y < e^{-\gamma}$. Likewise I have not been able to find a function $p(x,y)$ which satisfies 5 for all $y > 0$. – Nilotpal Sinha Aug 15 '13 at 13:05
Looks like you still don't understand what I'm talking about: $p(x,y)=1$ for $x<1$ and $0$ for $x>1$ satisfies the "limit" requirements and I can create a family of continuous curves satisfying the conditions you posed formally and giving the distributions indistinguishable from this one for any practical purpose, which means that the problem statement is either flawed or incomplete because such solutions would obliterate any meaning of the $y$-dependence. What exactly are you after? – fedja Aug 15 '13 at 14:36
@Fedja: No I still don't understand. Can you exemplify with one working example of a solution that works for all $y > 0$. – Nilotpal Sinha Aug 15 '13 at 15:00
@Nilotpal Sure. I posted one. :) – fedja Aug 15 '13 at 15:40

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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.