# Joint density for exponential distribution

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?

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