Answer to Question #249600 in Statistics and Probability for aditi

Question

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.

Answer 1

a)

$P(X_1<2X_2)=\int^{\infin}_0\int^{\infin}_{0}\int^{2x_2}_0e^{-(x_1+x_2+x_3)}dx_1dx_2dx_3=$

$=-\int^{\infin}_0\int^{\infin}_{0}(e^{-(3x_2+x_3)}-e^{-(x_2+x_3)})dx_2dx_3=$

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

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

$P(X_1=2X_2)=\int^{\infin}_0\int^{\infin}_{0}\int^{2x_2}_{2x_2}e^{-(x_1+x_2+x_3)}dx_1dx_2dx_3=0$

mgf of X1+2X2+X3:

$M_{X_1+2X_2+X_3}(t)=\int^{\infin}_0\int^{\infin}_0\int^{\infin}_0e^{(t-1)(x_1+2x_2+x_3)}dx_1dx_2dx_3=\infin$

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

We have:

$f_{X_1}=f_{X_2}=f_{X_3}=-e^{-(x_1+x_2+x_3)}$

$f_{X_1}\cdot f_{X_2}\cdot f_{X_3}=-e^{-3(x_1+x_2+x_3)}\neq f(X_1, X_2,X_3)$

So, the variables are dependent.