Joint density for exponential distribution

Question

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?

Answer 1

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, $g^{-1}(Y)$.

Answer 2

You needn't get the joint distribution from marginals. Unitarity is the result $\int_{[0,\,\infty]^2} e^{-x_1-x_2}dx_1 dx_2=1$. The Jacobian is $1$, so we can restate that as $\int_S e^{-y_2}dy_1 dy_2=1$, where $S$ is the subset of $[0,\,\infty]^2$ satisfying $y_1\le y_2$. Thus $e^{-y_2}1_S(y_1,\,y_2)$ is the joint pdf. I'll leave you to get the marginals from that.