# Joint Probability Distribution Function.

I have just started learning Probability Theory. I came across this result that : Suppose X and Y are two non-independent random variables defined on same probability space and their individual probability density functions are given, i.e $f_X(x)$ and $f_Y(y)$, then they are not sufficient to completely determine joint probability distribution function, i.e $f_{X,Y}(x,y)$. Can you give an example to illustrate the above fact?? Also what extra information we may want to calculate joint PDF??

• We need to know if $X$ and $Y$ are independent or not. See How to find Joint PDF given PDF of Two Continuous Random Variables – WaveX Feb 24 '18 at 16:50
• @WaveX If they are independent, then joint pdf is just the product of individual Pdfs , so i don't need answer for that. Now i have edited the question. Thnx. – Jay Patel Feb 24 '18 at 16:58
• A good example for a discrete case can be found here. The main problem with only having marginal pdf's is that there isn't a unique joint pdf for a set of marginals. – WaveX Feb 24 '18 at 17:07
• Possible duplicate of Independence and Joint PDF – Arnaud Mortier Mar 3 '18 at 21:30

Define $Q(x,y) \subseteq \mathbb{R}^2$ as a square (including interior points) in $\mathbb{R}^2$ with side length $1$, sides parallel to the $x$- and $y$-axes and center at $(x,y)$.

Case 1: Let $X,Y: \Omega \rightarrow \mathbb{R}$ be real-valued continuous random variables with joint distribution function $f_{X,Y}(x,y) = 1/2$ for $(x,y) \in Q(1/2,1/2) \ \cup Q(3/2,3/2)$ and $f_{X,Y}(x,y) = 0$ otherwise. Then the individual densitiy of $X$ is $f_{X} = 1/2$ for $x \in [0,2]$ and $f_{X} = 0$ otherwise. The individual densitiy of $Y$ is $f_{Y} = 1/2$ for $y \in [0,2]$ and $f_{Y} = 0$ otherwise. $X$ and $Y$ are not independent.

Case 2: Let $X,Y: \Omega \rightarrow \mathbb{R}$ be real-valued continuous random variables with joint distribution function $f_{X,Y}(x,y) = 1/2$ for $(x,y) \in Q(1/2,3/2) \ \cup Q(3/2,1/2)$ and $f_{X,Y}(x,y) = 0$ otherwise. Then the individual densitiy of $X$ is $f_{X} = 1/2$ for $x \in [0,2]$ and $f_{X} = 0$ otherwise. The individual densitiy of $Y$ is $f_{Y} = 1/2$ for $y \in [0,2]$ and $f_{Y} = 0$ otherwise. $X$ and $Y$ are not independent.

The above two cases show, that the individual densities of two random variables are not enough to determine the joint densitiy, since in both cases you have the same individual densities but different joint density.

Suppose $X$ and $Y$ each take values 0 or 1.

Model 1: Points $(0,0)$ and $(1,1)$ each have probability $1/2.$

Model 2: Points $(0,0), (0,1), (1,0),$ and $(1,1)$ each have probability $1/4.$

Both models have $P(X=0) = P(X=1) = 1/2$ and $P(Y=0) = P(Y=1) = 1/2.$ So the two models have the same marginal distributions. However, in Model 2 $X$ and $Y$ are independent, and in Model 1 they are not.

• Suppose we know that to which real number does my random variable $X$ and $Y$ maps elementary outcomes of my sample space, then along with the information of marginal pdfs i can construct joint pdf?? – Jay Patel Feb 25 '18 at 10:38
• If you mean what I think you mean, then Yes. But in practical applications the joint distribution is usually specified by independence or by an appropriate sequence of conditional distributions. – BruceET Feb 25 '18 at 17:03

For any continuous RVs $X$ and $Y$ with marginal CDFs $F_X(x)$ and $F_Y(y),$ it's always a possibility that we have $$Y =F_{Y}^{-1} (F_X(X)).$$ A good exercise is to check that this is consistent, i.e. if $X$ has CDF $F_X$ then $Y$ as defined above has CDF $F_Y.$ And it's pretty clear that $X$ and $Y$ are not independent (except perhaps in some corner cases).