Answer to Question #336840 in Statistics and Probability for alvin

Question

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.

Answer 1

We assume that variables are continuous. We remind that for random variables $X_1$, $X_2$, $X_3$ we have: $P(X_1\leq x_1, X_2\leq x_2,X_3\leq x_3)=\int_{0}^{x_1}\int_{0}^{x_2}\int_{0}^{x_3}f(x_1,x_2,x_3)dx_1dx_2dx_3,$ where $0<x_1<1,0<x_2<1,0<x_3<1$.

(i) The value $c$ can be determined from the following equality: $1=\int_{0}^{1}\int_{0}^{1}\int_{0}^{1}f(x_1,x_2,x_3)dx_1dx_2dx_3$. We receive: $1=\int_{0}^{1}\int_{0}^{1}\int_{0}^{1}c(x_1+2x_2+3x_3)dx_1dx_2dx_3=c\int_{0}^{1}\int_{0}^{1}(\frac12x^2_1+2x_2x_1+3x_3x_1)|_0^1dx_2dx_3=c\int_{0}^{1}\int_{0}^{1}(\frac12+2x_2+3x_3)dx_2dx_3=c\int_{0}^{1}(\frac12x_2+x_2^2+3x_3x_2)|_0^1dx_3=c\int_{0}^{1}(\frac32+3x_3)dx_3=c(\frac32x_3+\frac32x_3^2)|_0^1=3c$

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

(ii) The joint marginal density function can be received in the following way: $f(x_1,x_2)=\frac13\int_0^1(x_1+2x_2+3x_3)dx_3=\frac13(x_1x_3+2x_2x_3+\frac32x_3^2)|_0^1=\frac13(x_1+2x_2+\frac32)$.

Thus, we receive that $f(x_1,x_2)=\frac13(x_1+2x_2+\frac32)$.

Answer: (i) $c=\frac13$; (ii) $f(x_1,x_2)=\frac13(x_1+2x_2+\frac32)$.