I am trying to find the J.D.F. of L and M, where $L = min(X,Y)$ and $M = max(X,Y)$, and $X,Y$ are i.i.d. Geom($p$) R.V. Here's my attempt and question.

We can use LTOP and compute the J.D.F, $P(L=l, M=m)$. Let $A_{1}, A_{2}, A_{3}$ be the events that $X>Y, X<Y, X=Y$ such that $A_{1} \cup A_{2} \cup A_{3} = S$, where $S$ is the sample space. Now, using the LTOP we have,

$P(L=l,M=m) = P(L=l,M=m|A_{1})*P(A_{1})+P(L=l,M=m|A_{2})*P(A_{2})+P(L=l,M=m|A_{3})*P(A_{3})$.

Now, let's compute all the terms. The $1^{st}$ and $2^{nd}$ terms are equal by symmetry.

Based on the above,

$P(L=l,M=m) = \begin{cases} 2P(L=l,M=m|X<Y)*P(X<Y), \ \ \ l \neq m\\ P(L=l,M=m|X=Y)*P(X=Y) \ \ \ \ l = m \end{cases}$

$P(L=l,M=m) = \begin{cases} 2P(X=l)P(Y=m)*P(X<Y), \ \ \ l \neq m\\ P(X=l)P(Y=l)*P(X=Y) \ \ \ \ l = m \end{cases}$

Now, the terms are fairly trivial to compute using the Geometric distribution. I came across multiple solutions to a similar question (not the same which is why I am asking this question) and my answer matches except the extra terms for $P(X<Y)$ and $P(X=Y)$, which result from an application of LTOP. So, what's wrong with my approach? Am I thinking about LTOP in the wrong way? Please help. Thanks!

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    $\begingroup$ Use the joint probability rather than the products. Re: $\mathsf P(E\mid F)\mathsf P(F)=\mathsf P(E,F)$ $$\begin{align}\mathsf P(L{=}\ell,M{=}m)&=\mathsf P(L{=}\ell,M{=}m,X<Y)+\mathsf P(L{=}\ell,M{=}m,X{=}Y)+\mathsf P(L{=}\ell,M{=}m,X>Y)\\&=\mathsf P(X{=}\ell,Y{=}m,\ell<m)+\mathsf P(X{=}\ell,Y{=}\ell,\ell=m)+\mathsf P(X{=}m,Y{=}\ell,\ell<m)\end{align}$$ $\endgroup$ Commented Apr 23, 2020 at 14:46
  • $\begingroup$ @GrahamKemp This did it. Thanks! I guess the mistake in my approach arises from the fact that once you break down the terms, P(X<Y) should be equal to 1 if X=l and Y=m (by definition). But, what confuses me is that I am conditioning on the information that P(X<Y) rather than the other way around. So, technically I am not even sure if P(X<Y) = 1. All this sounds circular and confusing to me. Anyway, your argument definitely makes the solution much more intuitive to me. I guess it's never a good idea to read too much into the symbols. I am definitely doing something wrong here, not sure what. $\endgroup$ Commented Apr 23, 2020 at 22:15
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    $\begingroup$ $X<Y$ is the event that $X=L, Y=M, L<M$. $$\{L=\ell, M=m, X<Y\}=\{X=\ell,Y=m, \ell<m\}$$ $\endgroup$ Commented Apr 23, 2020 at 23:02
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    $\begingroup$ Now $\mathsf P(X<Y)\neq 1$ but also $\mathsf P(L=\ell, M=m\mid X<Y)\neq\mathsf P(X=\ell,Y=m)$. $\endgroup$ Commented Apr 23, 2020 at 23:14
  • $\begingroup$ @GrahamKemp Thanks again! It makes much more sense now. I think I was looking at the term P(X<Y) totally out of context, i.e. we already know that L < M. So, like you said X < Y is the event X=L, Y=M, L < M. $\endgroup$ Commented Apr 24, 2020 at 12:03

1 Answer 1


Consider, since $L=\min\{X,Y\}$ and $M=\max\{X,Y\}$, then you have:

$$\begin{align}\mathsf P(L{=}\ell,M{=}m)&=\begin{cases}\mathsf P(L{=}\ell,M{=}m,X{<}Y)+\mathsf P(L{=}\ell,M{=}m,X{>}Y)&:& \ell<m\\\mathsf P(L{=}\ell,M{=}\ell,X{=}Y)&:& \ell=m\\0&:&\textsf{otherwise}\end{cases}\\[2ex]&=\begin{cases}\mathsf P(X{=}\ell,Y{=}m)+\mathsf P(Y{=}\ell,X{=}m)&:& \ell<m\\\mathsf P(X{=}\ell,Y{=}\ell)&:& \ell=m\\0&:&\textsf{otherwise}\end{cases}\\&~~\vdots\end{align}$$


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