# Is my answer sheet wrong?

• Dec 10th 2009, 12:02 PM
MathSucker
pdf:

$\displaystyle f(x,y)=\left\{ \begin{array}{lr} \frac{2x+y}{4}&0\le x\le 1;0\le y\le 2\\ 0&otherwise \end{array} \right.$

Question: Find cdf (remember to consider all cases).

$\displaystyle x \le 0 \, , F(x,y)=0$
$\displaystyle y \le 0 \, , F(x,y)=0$

$\displaystyle x > 1\, , y > 2, \, F(x,y)=1$

$\displaystyle 0 \le x \le 1, 0 \le y \le 2$ :

$\displaystyle F(x,y)=\int_{0}^{x}\int_{0}^{y}\frac{2u+v}{4}\,dvd u = \frac{xy}{8}(2x+y)$

Shouldn't that be it? What's with these next two cases if x and y are never greater than 1 and 2 respectively?

$\displaystyle 0 \le x \le 1 \, , y \ge 2 \, : F(x,y)=\frac{x}{2}(x+1)$

$\displaystyle 0 \le y \le 2 \, , x \ge 1 \, , F(x,y)=\frac{y}{8}(2+y)$
• Dec 10th 2009, 02:43 PM
matheagle
Do you have a typo?
It doesn't make sense that x is bounded above by 1
and x is larger than y
and yet you have 0<y<2
clearly y cannot pass 1.
• Dec 10th 2009, 02:49 PM
mr fantastic
Quote:

Originally Posted by matheagle
Do you have a typo?
It doesn't make sense that x is bounded above by 1
and x is larger than y
and yet you have 0<y<2
clearly y cannot pass 1.

You beat me to it. Actually I think there are typos ....
• Dec 10th 2009, 03:10 PM
MathSucker
oops. fixed.
• Dec 11th 2009, 09:50 AM
MathSucker
Quote:

Originally Posted by MathSucker
$\displaystyle 0 \le x \le 1 \, , y \ge 2 \, : F(x,y)=\frac{x}{2}(x+1)$

$\displaystyle 0 \le y \le 2 \, , x \ge 1 \, , F(x,y)=\frac{y}{8}(2+y)$

Even with the correction, what is the point of these two cases. Why would I be finding equations for cases where x and y are greater than 1 and 2 if x only takes values between 0 and 1 and y only takes values between 0 and 2?

Do I fundamentally misunderstand this problem?
• Dec 11th 2009, 12:20 PM
mr fantastic
Quote:

Originally Posted by MathSucker
Even with the correction, what is the point of these two cases. Why would I be finding equations for cases where x and y are greater than 1 and 2 if x only takes values between 0 and 1 and y only takes values between 0 and 2?

Do I fundamentally misunderstand this problem?

The pdf is defined for x > 1 and y > 2 (just as it's defined for x < 0 and y < 0). The question said all cases. Therefore ....
• Dec 11th 2009, 03:51 PM
MathSucker
So from the beginning do I just integrate $\displaystyle f(x,y)$
to get $\displaystyle F(x,y)$. And then find:

$\displaystyle F(0,y)=F(x,0)=F(0,0)=0$ for $\displaystyle x>0$ and $\displaystyle y>0$

$\displaystyle F(2,1)=1$ for $\displaystyle x>1$ and $\displaystyle y>2$

$\displaystyle F(x,y)=F(x,y)$ for $\displaystyle 0 \le x \le 1 ; 0 \le y \le 2$

$\displaystyle F(1,y)=$ (given above) for $\displaystyle x>1 ; 0 \le y \le 2$

$\displaystyle F(x,2)=$ (given above) for $\displaystyle y>2 ; 0 \le x \le 1$

Correct?