I am working on a problem and am a bit confused:
The problem:
Find the joint CDF of the function: $$ \begin{split} p(x,y) &= {1\over8} (x+y)\\ 0 \le X &\le 2\\ 0 \le Y &\le 2\\ \end{split} $$ I am not exactly sure how we calculate this CDF, do we need to set up an integral?
I am new to this so if anything is unclear please let me know.
A CDF tells you the probability that $X < x$ and $Y < y$. So for any $a$ such that $0 \leq a \leq 2$ and for any $b$ such that $0 \leq b \leq 2$, the cumulative distribution function can be taken by the integral
$$P(X < a, Y < b) = F(a,b) = \int_{0}^{a}\int_{0}^{b}\frac{1}{8}(x+y) \ dy \ dx.$$
If $a > 2$ and $b \in [0,2]$, then you just have to take the integral with respect to $y$ so that
$$P(X < a, Y < b) = F(a,b) = \int_{0}^{b}\frac{1}{8}(x+y) \ dy.$$
Similarly, if $b > 2$ and $a \in [0,2]$, then you just have to take the integral with respect to $x$ so that
$$P(X < a, Y < b) = F(a,b) = \int_{0}^{a}\frac{1}{8}(x+y) \ dx.$$
Finally, if both $a$ and $b$ are greater than 2, then you know the probability that your $X$ and $Y$ are below $a$ and $b$ respectively is 1.
If either of your $a$ or $b$ is less than 0, then there is a 0% chance that you will pick an $X$ and $Y$ that satisfies that. You can then use this information to make a piecewise function to complete your CDF.