题目
设 sim N(0,1) ,则 =(X)^2 的概率密度函数为-|||-A ) _(r)(y)=dfrac (1)(sqrt {2pi )}(e)^-dfrac (y{2)} gt 0 .-|||-bigcirc ._(Y)(y)=dfrac (1)(2sqrt {2pi )}(e)^-dfrac (y{2)}(y)^-dfrac (1{2)} gt 0-|||-C. . _(Y)(y)=dfrac (1)(2sqrt {2pi )}(e)^-dfrac (y{2)} ,gt 0-|||-bigcirc ._(Y)(y)=dfrac (1)(sqrt {2pi )}(e)^-dfrac (y{2)}(y)^-dfrac (1{2)} gt 0
题目解答
答案
B. ${f}_{Y}(y)=\dfrac {1}{2\sqrt {2\pi }}{e}^{-\dfrac {y}{2}}{y}^{-\dfrac {1}{2}}$ $y\gt 0$