exponential distributionprobability distributions
Let $X_1$ and $X_2$ be independent random variables each having a exponential distribution with mean $\lambda = 1$.
(a) Find the joint density of $Y_1 = X_1$ and $Y_2 = X_1 + X_2$.
(b) Get the marginal density of $f_1(y_1)$ and $f_2(y_2)$.
$f(x_1)=e^{-x_1}$, $f(x_2)=e^{-x_2}$
Since $X_1$ and $X_2$ are independent, $f(x_1, x_2)=f(x_1)f(x_2)=e^{-x_1-x_2}$
However, how could I find the joint density?
I mean, it is quite easy to find marginal ones from the joint one, but how can I do the opposite?
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 :)
The map $g:(x,y) \mapsto (x-2y,x)$ is a differentiable and invertible function between $(0,\infty)\times (0,\infty)$ and $R=\{(z,w) | z< w \text{ and } w>0\}$, so first of all we get that the support for $(Z,W)=(X-2Y,X)$ must be $R$.
The transformation theorem for probability densities states that:
$$f_{Z,W}(z,w) = f_{X,Y}(g^{-1}(z,w)) |det(\frac{dg^{-1}}{d(z,w)}(z,w))|,$$ where $\frac{dg^{-1}}{d(z,w)}(z,w)$ is the jacobian of $g^{-1}$.
(see https://en.wikipedia.org/wiki/Probability_density_function#Vector_to_vector)
We first compute $g^{-1}(z,w)= (w,\frac{w-z}{2})$ and the jacobian $$ \frac{dg^{-1}}{d(z,w)}(z,w) = \begin{pmatrix}0 & 1 \\ -\frac12 & \frac12 \end{pmatrix},$$ which has determinant $\frac12$ for all $z,w$. We now plugin, and get $$ f_{Z,W}(z,w) = \frac12 f_{X,Y} ((w,\frac{w-z}{2})) = e^{-w}e^{-2\frac{w-z}{2}}=e^{z-2w}.$$ for all $(z,w) \in \{(z,w) | z< w \text{ and } w>0\}$. Just to verify, that this is in fact a valid density we compute $$ \int_0^\infty \int_{-\infty}^w e^{z-2w} dzdw = 1$$
Best Answer
Hint: Use the transformation theorem.
If $Y=(Y_1,Y_2),$ where $Y_1$ and $Y_2$ are as described by you, $Y=g(X)$, $X=(X_1,X_2)$, then:-
$f_Y(y)=|J(y)|f_X(g^{-1}(y))$, if $y$ is on g's range (and zero otherwise).
Note that $J$ is the jacobian of the inverse transformation, $g^{-1}(Y)$.