Any help with this question?
Let A be the set in $R^2$ bounded between the lines $y = 0$, $x = y$ and $x = 2$.
Let $(X, Y)$ be jointly continuous random variables with pdf $f(x, y) = cxy^2$ for $(x, y)$ in $A$.
I want to calculate the CDF, I know that:
- If x is between 0-2 then y is between 0-x
- $F_{X, Y} (x, y)=\int_{-\infty}^{y}\int_{-\infty}^{x}f_{X, Y}(u, v) du dv$
Thus I get (partially correct):
$$F_{X, Y} (x, y)=\int_{-\infty}^{y}\int_{-\infty}^{x}cuv^2 du dv$$
Now, Regarding the limits:
Relevant values for $X$ are in the range $0-2$ AND smaller than $x$.
Relevant values for $Y$ are in the range $0-x$ AND smaller than $y$.
So my final step will be:
$$F_{X, Y} (x, y)=\int_{0}^{min(y, x)}\int_{0}^{min(2, x)}cuv^2 du dv$$
How can I calculate such integral? I've never dealt with a case where the limits contain $min$ function.
Side question, for which values of $X, Y$ this will be the correct CDF? I can't say only for values inside $A$ since for x = 100 or x = 50 and y > 2 this is also correct...
Ie, I'm looking for the full CDF in all ranges of values
EDIT: I made a huge mistake in the first solution. Here is the correct one.
As I said before, you don't need to put the minimum operator inside the integral. Instead, you can put the appropriate indicator functions (within the minimum operator, if you which) in the final solution.
STEP 1:
Let's find the value of constant $c$. Here, you should first integrate w.r.t. the variable $y$.
$$ \int_{x=0}^{x=2} \int_{y=0}^{y=x}cxy^2 dy dx = 1$$ $$ \int_{x=0}^{x=2}cx \cdot \frac{1}{3}x^3 dx = 1$$ $$ \frac{1}{15}c \cdot 2^5 = 1 \implies c = \frac{15}{32}$$
STEP 2:
To obtain the expression of $F_{X,Y}(x_0,y_0)$, we must consider some cases:
Case 1. $ y_0 < 0$
In this case, $F_{X,Y}(x_0,y_0) = 0$.
Case 2. $ x_0\le y_0$ and $ 0 \le x_0 \le 2$
In this case, the region of interest is the region delimited by the brown triangle on the picture above, which will be denoted by $T(x_0)$.
$$F_{X,Y}(x_0,y_0) = T(x_0) = \int_{x=0}^{x=x_0} \int_{y=0}^{y=x}\frac{15}{32}xy^2 dy dx = \int_{x=0}^{x=x_0}\frac{5}{32}x^4 dx = \frac{1}{32}{x_0}^5$$
Case 3. $ x_0 > y_0$ and $x_0 \le 2$, $y_0 \ge 0$
In this case, the region of interest is the region delimited by the brown triangle (denoted by $T(y_0)$) plus the green rectangle (which will be denoted by $R(x_0,y_0)$).
$$F_{X,Y}(x_0,y_0) = T(y_0) + R(x_0,y_0) = \frac{1}{32}{y_0}^5 + \int_{x=y_0}^{x=x_0} \int_{y=0}^{y=y_0}\frac{15}{32}xy^2 dy dx$$
$$F_{X,Y}(x_0,y_0) = \frac{1}{32}{y_0}^5 + \left(\frac{5}{64}{x_0}^2{y_0}^3 - \frac{5}{64}{y_0}^5 \right) = \frac{5}{64}{x_0}^2{y_0}^3 - \frac{3}{64}{y_0}^5$$
Case 4. $ x_0 > y_0$ and $x_0 > 2$, $y_0 \ge 0$
This case is similar to case 3, but with the rectangle $R(2,y_0)$,
$$F_{X,Y}(x_0,y_0) = T(y_0) + R(2,y_0) = \frac{5}{64}{2}^2{y_0}^3 - \frac{3}{64}{y_0}^5 = \frac{5}{16}{y_0}^3 - \frac{3}{64}{y_0}^5 $$
Case 5. $x_0 < y_0$ and $x_0 > 2$
In this case, we have the region of interest is equal to $A$, therefore
$$F_{X,Y}(x_0,y_0) = 1$$
STEP3:
Now, we write the CDF expression. There is several ways of doing this. the most simple, in my opinion, is to write it as a piecewise function.
$$ F_{X,Y}(x_0,y_0) = \begin{cases} 0 & y_0 < 0\\ \frac{1}{32}{x_0}^5 & x_0\le y_0 \text{ and } 0 \le x_0 \le 2, y_0 \ge 0\\ \frac{5}{64}{x_0}^2{y_0}^3 - \frac{3}{64}{y_0}^5 & x_0 > y_0 \text{ and } x_0 \le 2, y_0 \ge 0\\ \frac{5}{16}{y_0}^3 - \frac{3}{64}{y_0}^5 & x_0 > y_0 \text{ and } x_0 \ge 2, y_0 \ge 0 \\ 1 & \text{otherwise} \end{cases}$$
Alternatively, you can write as one single expression using indicator functions and/or $min$ and $max$ functions. One solution is the following
$$F_{X,Y}(x_0,y_0) = \left(\frac{5}{64}{x_0}^2{min(y_0,x_0)}^3 - \frac{3}{64}{min(y_0,x_0)}^5\right)I(0 \le x_0 \le 2)I(y_0 \ge 0) + \left(\frac{5}{16}{min(2,y_0)}^3 - \frac{3}{64}{min(2,y_0)}^5\right)I(x_0> 2)I(y_0\ge 0)$$