I said that $|\Psi(x_1,x_2,t)|^2 = |\Psi_1(x_1,t)\Psi_2(x_2,t)|^2$ and I think that the pdf of the product of two independent random variables is just the product of the pdfs, so $|\Psi_1(x_1,t)\Psi_2(x_2,t)|^2 = |\Psi_1(x_1,t)|^2|\Psi_2(x_2,t)|^2$ hence $|\Psi(x_1,x_2,t)|^2 = |\Psi_1(x_1,t)|^2|\Psi_2(x_2,t)|^2$