Answer to Question #339375 in Statistics and Probability for ken

Question

Suppose that X1 and X2 have the joint probability distribution

f(x1, x2) = kx1x2 x1 = 1, 2, 3 x2 = 2, 3, 4

0, otherwise

Find the value of k so that the two variables are independent of each other

Answer 1

The joint probability density has to satisfy: $\sum_{x_1,x_2}f(x_1,x_2)=1$. We will check that:

$\sum_{x_1,x_2}f(x_1,x_2)=\sum_{x_1=1,2,3;x_2=2,3,4}kx_1x_2=k(2+3+4+4+6+8+6+9+12)=54k$. We receive that $k=\frac{1}{54}$. Find the marginal densities: $f_{X_1}(x_1)=\sum_{x_2}f(x_1,x_2)=k(2x_1+3x_1+4x_1)=9kx_1$.

$f_{X_2}(x_2)=\sum_{x_1}f(x_1,x_2)=k(x_2+2x_2+3x_2)=6kx_2$.

As we can see: $f_{X_1}(x_1)f_{X_2}(x_2)=54k^2x_1x_2=kx_1x_2=f(x_1,x_2)$. The latter implies that $X_1$ and $X_2$ are independent.