# A problem on bivariate random variables

Suppose we have absolutely continuous random vectors $$X=(X_1,X_2)$$ and $$Y=(Y_1,Y_2)$$. And we have $$Y_i=a_iX_i+b_i$$, and $$a_i>0, b_i\geq 0$$ $$i=1,2$$ . Let $${F}$$ be a distribution function such that $${F}_X(x_1,x_2)=P(X_1\leq x_1,X_2\leq x_2).$$ We know that $$\frac{\partial^2}{\partial x_1\partial x_2}F_X(x_1,x_2)=f_X(x_1,x_2).$$

Now, \begin{align} F_Y(y_1,y_2)&=P(Y_1\leq y_1,Y_2\leq y_2)\\ &=P(a_1X_1+b_1\leq y_1,a_2X_2+b_2\leq y_2)\\ &=P(X_1\leq (y_1-b_1)/a_1,X_2\leq (y_2-b_2)/a_2)\\ &=F_X((y_1-b_1)/a_1,(y_2-b_2)/a_2)\\ &=F_X(x_1,x_2) \end{align} Form here can we say that $$f(y_1,y_2)=f(x_1,x_2)$$? Thanks beforehand!

• Ask yourself whether the univariate version of this is true. It is not, but exploring why it is false will point you towards a correct formula.
– whuber
Commented Jan 31 at 14:59

Remember, when $$\mathbf X\mapsto \mathbf Y$$ via a one-one onto transformation, say $$y_i:= g_i(x_1, \ldots, x_n),$$ and $$h_i:=g_i^{-1},$$ then if $$\frac{\partial h_i}{\partial y_j}$$ are continuous for all $$i, j\in\{1, 2,\ldots, n\},$$ then $$f_\mathbf Y(\mathbf y) =f_\mathbf X(h_1(\mathbf y), \ldots, h_n(\mathbf y) ) |\mathrm J(\mathbf y) |,$$ where $$\mathrm J(\mathbf y) :=\det\left(\frac{\partial h_i}{\partial y_j}\right) .$$
• I think the general intuitive essence can be seen when dealing with linear transformation of Lebesgue measure on $\mathbb R^n,$ @Galen. Commented Jan 31 at 6:01