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I have random values $X,Y$ and

$f(x,y) = 2x+4y$ , where $0\le x\le 1$ , $0 \le y \le 1-x$

How to calculate distribution of

$Z = \min(X,Y)$ ?

As I calculated, $X$ and $Y$ are dependent, because

$f(x) \cdot f(y) \neq f(x,y)$

Ant
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Armyx
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1 Answers1

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Note that $X\geqslant0$, $Y\geqslant0$ and $X+Y\leqslant1$ almost surely hence $0\leqslant Z\leqslant1/2$ almost surely. For every $z$ in $(0,1/2)$, $[Z\geqslant z]=[X\geqslant z,Y\geqslant z]$ is $$ [X\geqslant z,1-X\geqslant Y\geqslant z]=[1-z\geqslant X\geqslant z,1-X\geqslant Y\geqslant z], $$ hence $$ P(Z\geqslant z)=\int_z^{1-z}\int_z^{1-x}f(x,y)\,\mathrm dy\,\mathrm dx. $$ Considering that you are able to compute the RHS as $G(z)$, the PDF $g$ of $Z$ is such that $g(z)=-G'(z)$ on $0\lt z\lt1/2$. Unless I am mistaken, and to let you check your computations, let me mention that one gets $$ g(z)=3(1-4z^2)\,\mathbf 1_{0\lt z\lt1/2}. $$

Did
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  • Why you use P(Z >= z) instad of P(Z < z) ? I didn't understand what you wrote about RHS and PDF, because I'm not familiar with this – Armyx May 24 '14 at 21:07
  • Events (Z>z) are easier to deal with than (Z<z) when Z is a minimum (to see why, read the post). RHS = right hand side. PDF = probability density function. – Did May 24 '14 at 21:12
  • ok thank you. When z isn't in range (0; 1/2) g(z) = 0, right ? – Armyx May 24 '14 at 21:22