2
$\begingroup$

$f(x,y)=ce^{-(x+y)}$ when $0< x< y+1$ if $x$ didn't belong to $0< x< y+1$ then $f(x,y)=0$

The questions are :

Find $c$

Determine the distribution function $F(y)$

  • 0
    @SiongThyeGoh the first question : i started with : $c\int_{0}^{y+1}\int_{x-1}^{+\infty }e^{-x-y}dx.dy=1$ – 2017-01-01
  • 0
    and found $c=2e^{-1}$ – 2017-01-01
  • 0
    Are you sure $c=\frac{2}{e}$ ? – 2017-01-01
  • 0
    yes positive . if u think i made a mistake please show me – 2017-01-01
  • 0
    @AymanErroutabi You can't get it from what you put in the comment above, so I assume its a typo. But indeed, $c \int_0^{\color{purple}\infty} \int_{x-1}^\infty e^{-x-y}\operatorname d y\operatorname d x=1 \implies c=2/e\quad\checkmark$ – 2017-01-02

1 Answers 1

1

You have the right idea, it's just that the integral you set up (as I read from your comment) appears to be wrong. So you recognize that integrating $f(x,y)$ over $\Bbb R^2$ should equal $1$ since it is a probability distribution, and since $f(x,y)$ is only nonzero whenever $(x,y)\in A$, you only need integrate over $A$. From the question we know that $A=\{(x,y)\in\Bbb R^2:0