Mostrando
Si , encuentre la distribución de Y = 2 XX∼ C( 0 , 1 )X∼C(0 0,1)X\sim\mathcal C(0,1) .Y= 2 X1 -X2Y=2X1-X2Y=\frac{2X}{1-X^2} Tenemos FY( y) = P r ( Y≤y)FY(y)=PAGr(Y≤y)F_Y(y)=\mathrm{Pr}(Y\le y) = P r ( 2 X1 - X2≤ y)=PAGr(2X1-X2≤y)\qquad\qquad\qquad=\mathrm{Pr}\left(\frac{2X}{1-X^2}\le y\right) =⎧⎩⎨⎪⎪⎪⎪⎪⎪Pr(X∈(−∞,−1−1+y2√y])+Pr(X∈(−1,−1+1+y2√y]),ify>0Pr(X∈(−1,−1+1+y2√y])+Pr(X∈(1,−1−1+y2√y]),ify<0={Pr(X∈(−∞,−1−1+y2y])+Pr(X∈(−1,−1+1+y2y]),ify>0Pr(X∈(−1,−1+1+y2y])+Pr(X∈(1,−1−1+y2y]),ify<0\qquad\qquad=\begin{cases} \mathrm{Pr}\left(X\in\left(-\infty,\frac{-1-\sqrt{1+y^2}}{y}\right]\right)+\mathrm{Pr}\left(X\in\left(-1,\frac{-1+\sqrt{1+y^2}}{y}\right]\right),\text{if}\quad y>0\\ \mathrm{Pr}\left(X\in\left(-1,\frac{-1+\sqrt{1+y^2}}{y}\right]\right)+\mathrm{Pr}\left(X\in\left(1,\frac{-1-\sqrt{1+y^2}}{y}\right]\right),\text{if}\quad y<0 \end{cases} Me pregunto si la …