Is my answer sheet wrong?

**MathSucker** · December 10th 2009, 12:02 PM

pdf:

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

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

Answer:

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

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

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

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

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

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

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

**matheagle** · December 10th 2009, 02:43 PM

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.

**mr fantastic** · December 10th 2009, 02:49 PM

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 ....

**MathSucker** · December 10th 2009, 03:10 PM

oops. fixed.

**MathSucker** · December 11th 2009, 09:50 AM

Originally Posted by MathSucker

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

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

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?

**mr fantastic** · December 11th 2009, 12:20 PM

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 ....

**MathSucker** · December 11th 2009, 03:51 PM

So from the beginning do I just integrate $f(x,y)$
to get $F(x,y)$ . And then find:

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

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

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

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

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

Correct?

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