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This question is from DeGroot's "Probability and Statistics"(Second Edition).

Suppose that $X$ and $Y$ are random variables that can only take values in the interval $0\leq X\leq2$ and $0\leq Y\leq2$. Suppose also that the joint CDF of $X$ and $Y$ for, $0\leq x\leq2$ and $0\leq y\leq2$ is as follows:$$F(x,y)=\frac{1}{16}xy(x+y)\tag1$$ Determine the CDF $F_1$ of just the random variable $X$.

Solution The value of $F(x,y)$ at any point $(x,y)$ in the $xy$-plane that does not represent a pair of possible values of $X$ and $Y$ can be calculated from Eq.$(1)$ and the fact that $F(x,y)=Pr(X\leq x \text{ and } Y\leq y )$.Thus, if either $x<0$ or $y<0$, then $F(x,y)=0$. If both $x>2$ and $y>2$, then $F(x,y)=1$. If $0\leq x\leq2$ and $y>2$, then $F(x,y)=F(x,2)$ and it follows from it follows from Eq.$(1)$ that $F(x,y)=\frac{1}{8}x(x+2).$Similarly,if $0\leq y\leq2$ and $x>2$, then $F(x,y)=\frac{1}{8}y(y+2).$The function $F(x,y)$ has now been specified for every point in the $xy$-plane.By letting $y\rightarrow\infty$, we find that the CDF of just the random variable $X$ is$$F_1(x) = \begin{cases} 0 & \text{for $x < 0$,} \\ \frac{1}{8}x(x+2) & \text{for $0 \leq x \leq 2,$ } \\ 1 & \text{for $x>2.$} \end{cases}$$

Now I can't understand how did the author conclude these:

  1. If either $x<0$ or $y<0$, then $F(x,y)=0$.
  2. If both $x>2$ and $y>2$, then $F(x,y)=1$.
  3. If $0\leq x\leq2$ and $y>2$, then $F(x,y)=F(x,2)$.
  4. if $0\leq y\leq2$ and $x>2$, then $F(x,y)=F(2,y)$.
Silent
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2 Answers2

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That $X$ only takes values in $[0,2]$ has the implication that $F_X(x)=0$ for $x<0$ and $F_X(x)=1$ for $x>2$, where $F_X(x)=P(X\leq x)$ is the CDF of $X$. Since $Y$ also only takes values in $[0,2]$ we have the same thing for $F_Y$.

The definition of $F$ is $F(x,y)=P(X\leq x,Y\leq y)$, where $P(X\leq x,Y\leq y)$ is short-hand notation for $P(\{X\leq x\}\cap\{Y\leq y\})$, i.e. the probability of the intersection between $A_x=\{X\leq x\}$ and $B_y=\{Y\leq y\}$.

  1. Assume for simplicity that $x<0$. Use the fact that $A_x\cap B_y\subseteq A_x$ to conclude.

  2. If both $x>2$ and $y>2$, then $P(A_x)=P(B_y)=1$. This is enough to ensure that $P(A_x\cap B_y)=1$ (why?).

  3. If $0\leq x\leq 2$ and $y>2$, then $P(B_y)=1$. Hence $P(A_x\cap B_y)=P(A_x)=P(A_x\cap B_2)$ since also $P(B_2)=1$.

  4. Similar argument as 3.

Stefan Hansen
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  • Thanks a lot, Sir, for your informative explanation. Will you just answer the "why", please?(I tried to find a convincing reason, but failed.) – Silent Oct 14 '13 at 12:05
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    In general, if $P(A)=P(B)=1$ then also $P(A\cap B)=1$. It's easiest to argue that $P((A\cap B)^c)=0$, but $(A\cap B)^c=A^c\cup B^c$ and hence $P((A\cap B)^c)\leq P(A^c)+P(B^c)=0$. – Stefan Hansen Oct 14 '13 at 12:07
  • Oh, thank you for so quick reply and satisfying me. – Silent Oct 14 '13 at 12:09
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Here I am not exactly answering your question (allready done by Stefan) but give you another route to the CDF of $X$.

Based on $F\left(x,y\right)=\frac{1}{16}xy(x+y)$ on $\left(0,2\right)^{2}$ we find that

$f\left(x,y\right)=$ $\frac{\partial^{2}}{\partial x\partial y}F(x,y)=\frac{1}{8}x+\frac{1}{8}y$

serves as density of $\left(X,Y\right)$ on $\left(0,2\right)^{2}$. Then

$f_{X}\left(x\right)=\int_{0}^{2}\left(\frac{1}{8}x+\frac{1}{8}y\right)dy=\frac{1}{4}x+\frac{1}{4}$

serves as density for $X$ on $\left(0,2\right)$. Here $f_{X}\left(x\right)=0$ if $x\notin\left(0,2\right)$. Then

$F_{X}\left(x\right)=\int_{0}^{x}\left(\frac{1}{4}u+\frac{1}{4}\right)du=\frac{1}{8}x\left(x+2\right)$ for $x\in\left(0,2\right)$.

Finally $F_{X}\left(x\right)=1$ if $x\geq2$ and $F_{X}\left(x\right)=0$ if $x\leq0$.

drhab
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