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I'm trying to understand this problem:

question http://puu.sh/5Qrh7.png

The solution is:

question http://puu.sh/5QrhP.png

But I have no idea how he got from the end of the first line to the second... can anyone help explain it?

Thanks!

Richard
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  • This would be much easier to read if it was done in two steps rather than one, finding $P(Y \le y) =P(X(2-X)\le y)$ and only then taking the derivative. – Henry Dec 18 '13 at 08:10

2 Answers2

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It seems that they solved the quadratic equation in X and defined the conditions for which
2 X - X^2 - y < 0

Claude Leibovici
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If $0<y<1$, then the polynomial $p(x)=-x^2+2x-y$ has zeros at $$ x_1=1-\sqrt{1-y},\quad\text{ and }\quad x_2=1+\sqrt{1-y}. $$ Note that, since $0<y<1$ we have $0<x_1<1<x_2$. So let us find out when $p$ is non-positive. Since $p(0)=-y<0$ then $p(x)<0$ for all $x\in (-\infty, x_1)$. Since $p(1)=1-y>0$ then $p(x)>0$ for all $x\in (x_1,x_2)$. Since $p(100)<0$ then $p(x)<0$ for $(x_2,\infty)$.

In conclusion $p(x)\leq 0$ if and only if $x\in (-\infty,x_1]\cup [x_2,\infty)$.

Stefan Hansen
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