# finding probability of X and Y by given joint PMF

Let random variables X and Y have the joint PMF $\mathsf p_{X,Y} (x,y)$ given below.

$$\mathsf p_{X,Y} (x,y) = \begin{cases}0.05 & : x=1,2,3,4 \land y=1,2,3,4\\ 0 & :\text{ otherwise }\end{cases}$$

Let A denote the event that max$(X,Y) < 3$. Find the conditional PMF $\mathsf p_{X,Y|A} (x,y)$.

This is the question I asked. I have missed some lecture, so I have to learn it by myself, I need to learn this probability of $X$ and $Y$ ( $\mathsf p_{X,Y} (x,y)$ ) If you guys give me some good examples (near to life or can be imaginable examples) i would really appreciate.

• You need some restriction on $Y$, for example with it taking five possible values such as $0,1,2,3,4$ or $1,2,3,4,5$, so that the total probability is $1$, e.g. $0.05\times 4 \times 5$ – Henry Oct 26 '14 at 21:39
• I could make guesses about the values of $y$ for which the formula holds. However, these were undoubtedly specified in the problem, there were $5$ of them. Like Henry, I would guess it said $y=0$ to $4$ or $y=1$ to $5$. Some details of the answer will depend on the allowed values of $y$. – André Nicolas Oct 28 '14 at 17:13
• now i give restriction on $Y$ – Andy Oct 28 '14 at 17:16
• @Henry Can you explain now? – Andy Oct 29 '14 at 0:39
• @Andy, that is still not a valid pmf. The total probability does not equal 1. – Graham Kemp Oct 29 '14 at 0:52

You just need to apply the definition of conditional probability and the law of total probability \begin{align} \mathsf p_{X,Y\mid A}(x,y) & = \mathsf P(X=x\cap Y=y\cap \max(X,Y)<3) \\[1ex] & = \frac{\mathsf P(X=x\cap Y=y\cap \max(x,y)<3)}{\mathsf P(\max(X,Y)<3)} \\[2ex] & = \dfrac{ \mathsf p_{X,Y}(x,y)\,\operatorname{\bf 1}_{\max(x,y) < 3} }{ \mathop{\sum\sum}\limits_{k,h\mid\max(k,h) < 3}\,\mathsf p_{X,Y}(k,h) } \\[1ex] & = \begin{cases} \mathsf p_{X,Y}(x,y)\big/(\mathop{\sum\sum}\limits_{k,h\mid\max(k,h) < 3}\,\mathsf p_{X,Y}(k,h)) & :\max(x,y)<3, (x,y)\in\{...\} \\ 0 & : \text{elsewhere} \end{cases} \end{align}
Note, to continue you will need the correct support and probability mass for $\mathsf p_{X,Y}$, which isn't what you have given in the original post (as the total probability of that does not sum to 1).