I was doing a question and I came across the following logic:
Claim: If you have $X$ and $Y$ as two random variables and
$$f_{X,Y} (x,y) = \frac{1}{2\sqrt{3} \pi} e^{-\frac{1}{2}\left[\frac{(x - 4)^2}{3} + (y-2)^2 \right]}, \quad x \in \mathbb{R}, y \in \mathbb{R},$$
then,
$$f_{X,Y} (x,y) = \underbrace{\frac{1}{\sqrt{2\pi (3)}} e^{-\frac{1}{2} (x - 4)^2/3}}_{f_{X}(x)}\cdot \underbrace{\frac{1}{\sqrt{2\pi}} e^{-\frac{1}{2} (y - 2)^2}}_{f_Y (y)} \quad ,$$
which means $X,Y$ are independent and $ X \sim N(4,3)$ and $ Y \sim N(2,1). $
Is this true? I am unsure whether a joint density function always split uniquely into two marginal PDF's when the random variables are independent? What if it splits in two ways ($f_{X,Y} = f_{X_1} \cdot f_{Y_1}$ or $f_{X_2} \cdot f_{Y_2}$), where both are valid marginal PDF's? If this is possible, does that mean you can't always deduce the PDF of $X$ and $Y$ by splitting the joint density?
Thanks!
If $f_{X,Y}(x,y)=f_X(x)f_Y(y)$ then $X,Y$ are independent. Actually, we have $$P(X\in A,Y\in B)=\iint_{A\times B} f_{X,Y}(x,y)\,dxdy=\int_A f_X(x)\,dx \int_B f_Y(y)\,dy=P(X\in A)P(X\in B).$$ From the definition we know that $X,Y$ are independent.
If $f_{X,Y}(x,y)=f_{X_1}(x)f_{Y_1}(y)=f_{X_2}(x)f_{Y_2}(y)$, integrating with respect to $x$ gives $f_{Y_1}(y)=f_{Y_2}(y)$ so $Y_1$ and $Y_2$ are identically distributed. Hence you can deduce the PDF of $X$ and $Y$ by splitting the joint density if it really can (in which case they are independent).