Question #336840

The joint density function of the random variables X1, X2, X3 is given as


f(x1, x2, x3) = {


c(x1 + 2x2 + 3x3), 0 < xi < 1, i = 1, 2, 3

0, elsewhere


Determine

(i) the value of c

(ii) the joint marginal density function of X1 and X2.


1
Expert's answer
2022-05-08T14:43:28-0400

We assume that variables are continuous. We remind that for random variables X1X_1, X2X_2, X3X_3 we have: P(X1x1,X2x2,X3x3)=0x10x20x3f(x1,x2,x3)dx1dx2dx3,P(X_1\leq x_1, X_2\leq x_2,X_3\leq x_3)=\int_{0}^{x_1}\int_{0}^{x_2}\int_{0}^{x_3}f(x_1,x_2,x_3)dx_1dx_2dx_3, where 0<x1<1,0<x2<1,0<x3<10<x_1<1,0<x_2<1,0<x_3<1.

(i) The value cc can be determined from the following equality: 1=010101f(x1,x2,x3)dx1dx2dx31=\int_{0}^{1}\int_{0}^{1}\int_{0}^{1}f(x_1,x_2,x_3)dx_1dx_2dx_3. We receive: 1=010101c(x1+2x2+3x3)dx1dx2dx3=c0101(12x12+2x2x1+3x3x1)01dx2dx3=c0101(12+2x2+3x3)dx2dx3=c01(12x2+x22+3x3x2)01dx3=c01(32+3x3)dx3=c(32x3+32x32)01=3c1=\int_{0}^{1}\int_{0}^{1}\int_{0}^{1}c(x_1+2x_2+3x_3)dx_1dx_2dx_3=c\int_{0}^{1}\int_{0}^{1}(\frac12x^2_1+2x_2x_1+3x_3x_1)|_0^1dx_2dx_3=c\int_{0}^{1}\int_{0}^{1}(\frac12+2x_2+3x_3)dx_2dx_3=c\int_{0}^{1}(\frac12x_2+x_2^2+3x_3x_2)|_0^1dx_3=c\int_{0}^{1}(\frac32+3x_3)dx_3=c(\frac32x_3+\frac32x_3^2)|_0^1=3c

From the latter we receive that c=13c=\frac13.

(ii) The joint marginal density function can be received in the following way: f(x1,x2)=1301(x1+2x2+3x3)dx3=13(x1x3+2x2x3+32x32)01=13(x1+2x2+32)f(x_1,x_2)=\frac13\int_0^1(x_1+2x_2+3x_3)dx_3=\frac13(x_1x_3+2x_2x_3+\frac32x_3^2)|_0^1=\frac13(x_1+2x_2+\frac32).

Thus, we receive that f(x1,x2)=13(x1+2x2+32)f(x_1,x_2)=\frac13(x_1+2x_2+\frac32).

Answer: (i) c=13c=\frac13; (ii) f(x1,x2)=13(x1+2x2+32)f(x_1,x_2)=\frac13(x_1+2x_2+\frac32).


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