# How to calculate joint distribution probability samples from the probabilities of the two univariate distributions? [duplicate]

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?

• $f(y)$ 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 1 '20 at 16:09
• @RichardHardy I took the observations to be paired. – Dave Nov 1 '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 1 '20 at 16:43
• @Dave can the joint probability samples be calculated with the data given? – develarist Nov 1 '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 1 '20 at 18:03

## 1 Answer

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.

• even if $f(x)$ and $f(y)$ were probability values in their histograms? – develarist Nov 1 '20 at 17:44
• @develarist, what else are they supposed to be? – Richard Hardy Nov 1 '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 1 '20 at 17:53
• @Dave sure, i forgot there could be independence – develarist Nov 1 '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 1 '20 at 18:07