XY XY 23X4YSolutiongiven X and Y are independent variable an

X+Y     X-Y    2-3X+4Y

Solution

given

X and Y are independent variable and X ~ N(2,1)     and Y~N(3,2)

`then we have to find probability distriobution of each of the folowing

X+Y ,X-Y ,2-3X+4X

since moment generating pof each family of pdf is unique so we find these distribution using moment generating function\'

remember that for any normal distribution W that is W~N( mean , variance)

then Mgf of W is given by     e^{mean*t +(variance* t-square)/2}

now

let

Z=X +Y

so

M_z(t) =E(e^tz ) =E(e^t(X+Y)) =E(e^tX e^tY )

                         =E(e^tX) * E(e^tY)            (since X and Y are independent)

                          =M_X(t) *M_Y(t)

                           =e^{2t +(1*t-square)/2}   * e^{3t +(2*t square)/2}

                          =e^{5t+(3t-square)/2}

Since Mgf of Z is in the form of mgf of Normal distribution with mean =5 and variance =3

hence Z=X+Y ~ N(5 ,3)

now let

U=X -Y

then

M_z(t) =E(e^tz ) =E(e^t(X-Y)) =E(e^tX e^-tY )

                         =E(e^tX) * E(e^-tY)            (since X and Y are independent)

                          =M_X(t) *M_Y(-t)

                           =e^{2t +(1*t-square)/2}   * e^{-3t +(2*t square)/2}

                          =e^{-t+(3t-square)/2}

Since Mgf of U is in the form of mgf of Normal distribution with mean = -1 and variance =3

hence U=X-Y ~ N(-1 ,3)

now let

V=2 - 3X+4Y

so

M_z(t) =E(e^tz ) =E(e^t(2-3X+4Y)) =E(e^2te^-3tX e^4tY )

                         =e^2tE(e^-3tX) * E(e^4tY)            (since X and Y are independent)

                          =e^2tM_X(-3t) *M_Y(4t)

                           =e^2te^{-6t +(9*t-square)/2}   * e^{12t +(32*t square)/2}

                          =e^{8t+(41t-square)/2}

Since Mgf of V is in the form of mgf of Normal distribution with mean =8and variance =41

hence V=2-3X+4Y ~ N(5 ,3)

then