# "Bridging" uniform and "mass" distributions

Foreword. The original formulation of this problem was inaccurate; chamomille and Didier Piau came up with a simple example which would not solve the problem in its accurate formulation. Sorry for my inaccuracy. Below is an edited version.

My goal is to find a family X(a, b) of random variables (continuously) depending on two non-negative parameters a and b . The family should have the following properties:

(1) X(a, b) take values in the unit interval [0, 1] for all a, b;

(2) For dependent random variables Y(a, b) defined as 1/(a+b*X(a, b)) the expected values E[Y(a, b)] exist;

(3) When b/a is close to 0, the distribution of X(a, b) is close to uniform on [0, 1];

(4) When a/b is close 0, the distribution of X(a, b) is close to “mass“ distribution (that is, X(a, b) equals 1 with probability 1).

So my goal is to find a family of random variables parameterized by a and b to “bridge” the uniform and “mass” distributions.

I tried different parameterizations but was not able to find a parameterization satisfying all conditions (1)-(4).

• I do not understand why you are not happy with the choice $X(a,b)=1$ for all $a,b>0$. Apr 26 '11 at 16:39
• Thank you, camomille. I agree. In my original post I missed an additional condition (required for the purpose of my investigation) that would not permit this obvious choice. My fault. Sorry and thank you again.
– Max1
Apr 26 '11 at 18:14

These conditions are met if $X(a,b)\in[0,1]$ almost surely for every $a$ and $b$ and if $X(a,b)\to1$ in probability when $a\to0$. Hence $X(a,b)=1$ almost surely solves the problem. If one wants to avoid Dirac masses, $X(a,b)$ uniform on $[1/(1+a),1]$ (or on any interval $[c(a),1]$ with $c(a)$ in $[0,1)$ and $c(a)\to1$ when $a\to0$) will do.
Edit This is to answer the edited version of the question. A solution is to consider $X(a,b)$ uniform on $[b/(a+b),1]$ or on any interval $[k(a/b),1]$ where the function $k(\ )$ is such that $k(r)$ in $[0,1]$ for every $r\ge0$, $k(r)\to1$ when $r\to0$ and $k(r)\to0$ when $r\to+\infty$, for example $k(r)=1/(1+r)$ or $k(r)=\mathrm{e}^{-r}$.
Second edit Still another solution, such that the support of $X(a,b)$ is the full interval $(0,1)$ for every $(a,b)$, is to assume that $X(a,b)$ has density $c(b/a)x^{c(b/a)-1}$ for $x$ in $(0,1)$, where the function $c(\ )$ is such that $c(r)\to1$ when $r\to0$ and $c(r)\to0$ when $r\to+\infty$. A realization is $X(a,b)=U^{1/c(b/a)}$ with $U$ uniform on $(0,1)$, for example $$X(a,b)=U^{a/(b+a)}.$$