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I am interested in deriving an expression for the probability of a value $X$ being larger than a value $Y$. More specifically: I want to calculate an expression for $P(X>Y|I)$ and I know the probability densities $P(X|I)$ and $P(Y|I)$. The $I$ is some background knowledge I have concerning both variables.

Let us assume that $X$, $Y$ are real-valued and absolutely continuous values in $(x_{min},x_{max})$ and $(y_{min},y_{max})$ respectively. The intervals might be infinite,but do not have to be. Here is what I arrived at (steps shown at the end of the question):

$$P(X>Y|I) = \int_{x_{min}}^{x_{max}} dX \, P(X|I) \int_{y_{min}}^{\min(X,y_{max})} dY \, P(Y|X,I)$$

where the comma (,) notation means a logical and.

My questions

  • Is this expression correct?
  • What other assumptions about my $X$,$Y$ have I made that I am unaware of?

Calculation To arrive at the expression I wrote

\begin{eqnarray}P(X>Y|I) &=& \int_{x_{min}}^{x_{max}} dX \int_{y_{min}}^{y_{max}} dY \, P(X>Y,X,Y|I) \\ &=& \int_{x_{min}}^{x_{max}} dX \, \int_{y_{min}}^{y_{max}} dY P(X>Y|X,Y,I)\cdot P(Y|X,I)\cdot P(X|I) \end{eqnarray} and now I can write $P(X>Y|X,Y,I)=\theta(X-Y)$ where $\theta$ is the step function. From this follows:

$$P(X>Y|I) = \int_{x_{min}}^{x_{max}} dX \, P(X|I) \int_{y_{min}}^{y_{max}} dY \, P(X|I) \cdot \theta(X-Y)\cdot P(Y|X,I) $$

Using the $\theta$ function to constrain the upper integration border, the expression above should follow... I hope.

Notes: Related Question on math/SE

I am aware of this related question on math/SE but I am specifically interested in an expression where $X$,$Y$ need not be independent.

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  • $\begingroup$ When it comes to your first question ("Is this expression correct?"), I would say yes. $\endgroup$
    – inmybrain
    Commented Feb 13, 2020 at 7:51

1 Answer 1

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If the random variables (conditional) are not necessarily independent, but you know their marginals (conditional on I), you can try to model them using a copula. This has been studied in a Bayesian context in:

Bayesian Inference for $P(X<Y)$ Using Asymmetric Dependent Distributions

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