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Given $ 0 < p < 1$, do there exist probability distributions for random variables $ X,Y $ such that all three of following are true:

$$ P(X < Y) = p $$ $$ pdf (X) = f(x, p) $$ $$ pdf (Y) = f(y, 1-p) $$

Of course the trouble is trying to keep $f$ same.

Or equivalently, find an $f$ that satisfies the following equation for all $ 0 < p < 1 $ $$ \int_{x=0}^{\infty} f(x,p) \int_{y=x}^{\infty} f(y, 1-p) dy~dx = p$$

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The question makes little sense. If $p \ne 1/2$ take any two discrete random variables with $P(X<Y)=p$ and have $f$ interpolate between their probability mass functions. For $p=1/2$ it's slightly more interesting since $X$ and $Y$ must have the same distribution, but there are easy examples with $3$ possible values. – Robert Israel Dec 18 at 19:40
X and Y are not same distributions. If p < 1/2, X gives lower values than Y precisely with probability p. Consider X, uniform [0, w1], and Y uniform [0, w2], w1 < w2. then probability that P(X<Y) = W1/ 2W2. If it came out to be W1/(W1+W2) i was all set. but it didn't. – atul Dec 18 at 20:49
(ignore p < 1/2 in the comment) – atul Dec 18 at 20:50
Construct an 'f', for which that integral is true for all values of 0<p<1, – atul Dec 18 at 20:52
If $p=1/2$, $f(x,p) = f(x,1-p)$ so they are the same distributions. I didn't say they were the same when $p \ne 1/2$. – Robert Israel Dec 18 at 23:18
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I'll assume $X$ and $Y$ are to be independent, and instead of "given $p$" you mean you want a parametric family of probability density functions (apparently on $(0,\infty)$?) for all $p \in (0,1)$. This is very easy. For each $p \in (0,1/2]$ you take an arbitrary pdf $f(\cdot, p)$, and then for $p \in (1/2, 1)$ you just need some pdf $f(\cdot, p)$ so that $\int_0^\infty f(x,p) \int_x^\infty f(y,1-p)\ dy \ dx = p$. Since that integral is near $1$ if $f(\cdot,p)$ is concentrated near $0$ and near $0$ if $f(\cdot, p)$ is concentrated on very large values, by the Intermediate Value Theorem any continuous one-parameter family that goes between these two extremes will have a member that makes the integral equal $p$.

For a specific easily computed example, note that if $X$ and $Y$ are independent exponential random variables with rates $r_1$ and $r_2$ respectively, $P(X < Y) = r_1/(r_1 + r_2)$. So we could take $f(\cdot,p)$ to be the exponential density with rate $r(p)=p$, i.e. $f(x,p) = p e^{-px}$ for $x > 0$, $0$ otherwise.

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Hmm, i simplified it for 2 variables. I don't want to really distinct families which work just because they are in different range [0,1/2], (1/2, 1] I probably should give the many variable form, p1, p2, .. pn p(Xi < Xj, for all i!=j ) = pi But it gets harder to write down relationships across them. Imagine p1, p2.. pn are weight of picking it to come as early as possible in the output. After doing first pick, normalize remaining and do the second pick etc. – atul Dec 19 at 9:36
It might help if you stated (clearly and precisely) the problem you are really interested in. – Robert Israel Dec 19 at 9:43
Given $W_1, W_2, ..., W_n$ weights. Let us say you pick them proportional to their weights. The probability of any permutation of $1,2, ..., n $ is determined by these weights. I want to generate $x_1, x_2, ..., x_n $ from a distribution purely defined by $W_i$ and sort them, and ordering of indices of $x_i$ should match that of $1,2, ..., n $ I initially thought $X_i = U(0,W_i)$ works, but it does not. For two variables it becomes p1/2p2 instead of p1/(p1+p2) that I wished for. – atul Dec 19 at 13:53

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