Question

3 rmed in 3a4 s e Co s easy to show th EelxJ-EY

Do all parts.

Answer 1

$(a)~F_Y(t)=P(Y\leq t)=P(c-X\leq t)=P(X\geq c-t)=1-F_X(c-t)\\ F_X(x)=\int_0^x\frac{1}{c}dt=\frac{x}{c},~0\leq x\leq c\\ Now,~F_Y(t)=1-F_X(c-t)=1-\frac{c-t}{c}=\frac{t}{c}=F_X(t)\\ (b)~F_Y(t)=P(Y\leq t)=P(c-X\leq t)=P(X\geq c-t)=1-F_X(c-t)\\ Now,~take~Y=X^2~then~ f_{Y}(t)=\frac{1}{2\sqrt{t}}\left(f_X(\sqrt{t})+f_X(-\sqrt{t}) \right )\\ =\frac{1}{2\sqrt{t}}\times \frac{1}{c},~0\leq t\leq c^2\\ =0~otherwise\\ Since~f_X(x)=\frac{1}{c}~if~0\leq x\leq c\\ =0~otherwise\\ so,~f_X(-\sqrt{t})=0~and~f_{X}(\sqrt{t})=\frac{1}{c}\\ Now,~\frac{1}{2c}\int_0^{c^2}t^{-\frac{1}{2}}dt=1\\ E_X(X^2)=\frac{1}{c}\int_0^cx^2dx=\frac{c^2}{3}\\ E_{X^2}(X)=E_Y(\sqrt{Y})=\frac{1}{2c}\int_0^{c^2}t^{\frac{1}{2}}t^{-\frac{1}{2}}dt=\frac{1}{2c}\int_0^{c^2}dt=\frac{c}{2}\\ E_Y(Y)=\frac{1}{2c}\int_0^{c^2}t^{\frac{1}{2}}dt=\frac{1}{2c}\times \frac{2}{3}\times c^3=\frac{c^2}{3}\\ Hence~E_{X^2}(X)=E_Y(Y)$