Question #249600

Let f(x1,x2,x3)=exp[−(x1 + x2 + x3)], 0 < x1 < ∞, 0 < x2 < ∞, 0 < x3 < ∞, = zero elsewhere, be the joint pdf of X1, X2, X3. (a) Compute P (X1 < 2X2) and P (X1 = 2X2). (b) Determine the mgf of X1+2X2+X3. Are X1, X2, and X3 random variables independent? Give reasoning.


1
Expert's answer
2021-10-12T11:08:58-0400

a)

P(X1<2X2)=0002x2e(x1+x2+x3)dx1dx2dx3=P(X_1<2X_2)=\int^{\infin}_0\int^{\infin}_{0}\int^{2x_2}_0e^{-(x_1+x_2+x_3)}dx_1dx_2dx_3=


=00(e(3x2+x3)e(x2+x3))dx2dx3==-\int^{\infin}_0\int^{\infin}_{0}(e^{-(3x_2+x_3)}-e^{-(x_2+x_3)})dx_2dx_3=


=0(e(3x2+x3)/3+e(x2+x3))0dx3==-\int^{\infin}_{0}(-e^{-(3x_2+x_3)}/3+e^{-(x_2+x_3)})|^{\infin}_0dx_3=


=0(ex3/3ex3)dx3=11/3=2/3=-\int^{\infin}_{0}(e^{-x_3}/3-e^{-x_3})dx_3=1-1/3=2/3


P(X1=2X2)=002x22x2e(x1+x2+x3)dx1dx2dx3=0P(X_1=2X_2)=\int^{\infin}_0\int^{\infin}_{0}\int^{2x_2}_{2x_2}e^{-(x_1+x_2+x_3)}dx_1dx_2dx_3=0


mgf of X1+2X2+X3:


MX1+2X2+X3(t)=000e(t1)(x1+2x2+x3)dx1dx2dx3=M_{X_1+2X_2+X_3}(t)=\int^{\infin}_0\int^{\infin}_0\int^{\infin}_0e^{(t-1)(x_1+2x_2+x_3)}dx_1dx_2dx_3=\infin


X1, X2,X3 are independent if the joint p.d.f. is the product of the individual p.d.f.’s

We have:


fX1=fX2=fX3=e(x1+x2+x3)f_{X_1}=f_{X_2}=f_{X_3}=-e^{-(x_1+x_2+x_3)}

fX1fX2fX3=e3(x1+x2+x3)f(X1,X2,X3)f_{X_1}\cdot f_{X_2}\cdot f_{X_3}=-e^{-3(x_1+x_2+x_3)}\neq f(X_1, X_2,X_3)

So, the variables are dependent.


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