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If random variables $x$ and $y$ have probability distributions $f(x)$ and $f(y)$ that each sum to 1, indexed by $i$,

\begin{array} {|r|r|}\hline i & f(x) & f(y) & f(x,y) \\ \hline 1 & 0.1 & 0.1 & \\ \hline 2 & 0.2 & 0.4 & \\ \hline 3 & 0.4 & 0.3 & \\ \hline 4 & 0.2 & 0.15 & \\ \hline 5 & 0.1 & 0.05 & \\ \hline \end{array}

How do I compute corresponding probability samples for their joint distribution $f(x,y)$ for the final column in the table?

Dave
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develarist
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    $f(y)$ sums to 2.8, not 1. What are $i$? What is a *probability sample*? Without the copula, there is no way to obtain the joint distribution from the marginals. – Richard Hardy Nov 01 '20 at 16:09
  • @RichardHardy I took the observations to be paired. – Dave Nov 01 '20 at 16:23
  • @RichardHardy thanks i've corrected the error in decimal places. $i$ are just the index positions so that observations across $x$ and $y$ can be paired based on index position – develarist Nov 01 '20 at 16:43
  • @Dave can the joint probability samples be calculated with the data given? – develarist Nov 01 '20 at 16:44
  • no, the joint there is trivariate for two marginals, and there are only 2 outcomes (binary) which is unlikely to happen with (bell-shaped) histograms – develarist Nov 01 '20 at 18:03

1 Answers1

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Now that I understand your chart better, no, I do not believe you have enough information to write the joint density. For example, if $X$ takes a value of $1$, the joint density could say that $Y$ is assured of taking a value of $1$...or it could say that $Y$ cannot take a value of $1$.

EDIT

If you are willing to assume independence, there is an answer. Remember the definition of independence.

$$P(X= x, Y= y) = P(X=x)P(Y= y)$$

You are assuming independence for all $x,y\in\{1,2,3,4,5\}$

Multiply out the $5\times5$ grid to get your 25 probability values. While I suspect you get what I mean, I will give a few examples.

$$P(X= 1, Y= 1) = P(X=1)P(Y= 1) = (0.1)(0.1) = 0.01$$

$$P(X= 1, Y= 2) = P(X=1)P(Y= 2) = (0.1)(0.4) = 0.04$$

$$P(X= 2, Y= 1) = P(X=2)P(Y= 1) = (0.2)(0.1) = 0.02$$

$$P(X= 4, Y= 5) = P(X=4)P(Y= 5) = (0.2)(0.05) = 0.01$$

Do this for the remaining $21$ pairs.

Dave
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  • even if $f(x)$ and $f(y)$ were probability values in their histograms? – develarist Nov 01 '20 at 17:44
  • @develarist, what else are they supposed to be? – Richard Hardy Nov 01 '20 at 17:51
  • @develarist You’d still have no insight into the dependence structure, and what I wrote applies. If you want to assume independence, I can give you the joint distribution. – Dave Nov 01 '20 at 17:53
  • @Dave sure, i forgot there could be independence – develarist Nov 01 '20 at 17:56
  • Thanks for the edit. so it is always the case that if $x$ is an $m$-long vector of probabilities and $y$ is an $n$-long vector, then $f(x,y)$ will be an $m\times n$ matrix that cannot be represented as a vector whatsoever like how I set it up to be in the question? – develarist Nov 01 '20 at 18:07
  • To me, a matrix means some kind of linear transformation is going on, so I don't really want to call it a matrix. You're on the right track, though, and there are 3d plotting tools in software that expect this kind of $m\times n$ array. (One of my big plotting tools requires it, though I can't remember if it's base R's `persp` or 3d plotting in Python/matplotlib. It could be both.) – Dave Nov 01 '20 at 18:10
  • @Dave the last line $P(X= 4, Y= 5) = P(X=1)P(Y= 1) = (0.1)(0.05) = 0.005$ has minor typo(s) but not sure how to fix it – develarist Nov 05 '20 at 10:37