Is there a way of determining the joint probability density function of two random variables? If we have two independent random variables, $X$ and $Y$ that both are uniform on [0,1], then how do one calculate the joint probability density function, knowing that the two PDFs are 1 each? It wouldn't simply be the product of the two PDFs, right?
[Math] Finding the joint probability density function of two independent random variables
density functionprobability distributions
Related Solutions
Yes, if you know the joint probability distribution of $X,Y$ then the distributions of $U$ and $V$ are, respectively:
$$\begin{align} \mathsf P(U=u) & = \sum_\overbrace{x\in\Omega_X, u-x\in\Omega_Y} \mathsf P(X=x, Y=u-x) \\ \mathsf P(V=v) & = \sum_\overbrace{x\in\Omega_X, v/x\in\Omega_Y} \mathsf P(X=x, Y=v/x) \end{align}$$
The joint distribution on $U,V$, would similarly be $$\mathsf P(U=u, V=v) = \sum_\overbrace{x\in\Omega_X, u-x\in\Omega_Y, v/x\in\Omega_Y} \mathsf P(X=x, Y=u-x, Y=v/x)$$
However, for any given pair of $U=u,V=v$ there is in fact only two corresponding pair of $X=\Box,Y=\Box$.
$\begin{align} (U=X+Y) \wedge (V= XY) & \iff \left(X= \dfrac{U\pm\sqrt{U^2-4V}}{2}\right) \wedge \left(Y=\dfrac{U\mp\sqrt{U^2-4V}}{2}\right) \\[4ex] \therefore \mathsf P(U=u, V=v) & = \mathsf P\left(X= \dfrac{u\pm\sqrt{u^2-4v}}{2}, Y=\dfrac{u\mp\sqrt{u^2-4v}}{2}\right) \\[1ex] & = {\mathsf P\left(X= \tfrac{u+\sqrt{u^2-4v}}{2}, Y=\tfrac{u-\sqrt{u^2-4v}}{2}\right) + \mathsf P\left(X= \tfrac{u-\sqrt{u^2-4v}}{2}, Y=\tfrac{u+\sqrt{u^2-4v}}{2}\right)} \end{align}$
If I were you, I would use transformations.
$$ Y1 = X1 - X2 = u_1(x_1,x_2)$$ $$ Y2 = X1 + X2 = u_2(x_1,x_2)$$ $$ X_1 = \frac{Y_1+Y_2}{2}=w_1(y_1,y_2)$$ $$ X_2 = \frac{Y_1-Y_2}{2}=w_2(y_1,y_2)$$
$$ f(y_1,y_2) = f(w_1(y1, y2),w_2(y1,y2)|J|$$
j is jaconian
I solved same problem with standard normal distributions. I added picture of my process. or You can solve this problem by using distribution method.
good luck :)
Best Answer
It would simply be the product of the two pdf's.
If $X$ and $Y$ have densities $f_X$ and $f_Y$ respectively then independence of $X$ and $Y$ is exactly the statement that $(X,Y)$ has density $g(x,y)=f_X(x)f_Y(y)$.