Say I have two independent random variables $X$ and $Y$ both having the exponential distribution. I.e.

$f_X(x) = \lambda_1 e^{-\lambda_1 x}, \ x \ge 0, 0$ elsewhere

$f_Y(y) = \lambda_2 e^{-\lambda_2 y}, \ x \ge 0, 0$ elsewhere

Does this mean that the joint distribution is the following

$f_{X, Y}(x, y) = \lambda_1 \lambda_2 e^{-\lambda_1 x - \lambda_2 y}, \ x, y \ge 0, 0$ elsewhere

I.e. you can simply multiply the functions?

  • $\begingroup$ Yes, independent iff the joint pdf is a product of marginal pdf $\endgroup$
    – Lost1
    Jun 30, 2014 at 10:52

1 Answer 1


Yes, it does. To see why, let $A,B\in\mathcal{B}(\mathbb{R})$ be measurable sets. Then $$ P((X,Y)\in A\times B)=P(X\in A)P(Y\in B)=\Big(\int_A f_X(x)\,\mathrm dx\Big)\Big(\int_B f_Y(y)\,\mathrm dy\Big) $$ by independence. Now, the product of the two integrals can be written as a double-integral: $$ P((X,Y)\in A\times B)=\iint_{A\times B} f_X(x)f_Y(y)\,\mathrm dx\,\mathrm dy. $$ Since this holds for all sets of the form $A\times B$ where $A,B\in\mathcal{B}(\mathbb{R})$, a standard argument (Dynkin) shows that it holds for all $C\in\mathcal{B}(\mathbb{R}^2)$, i.e. $$ P((X,Y)\in C)=\iint_C f_X(x)f_Y(y)\,\mathrm dx\,\mathrm dy, \quad C\in\mathcal{B}(\mathbb{R}^2). $$ Since $(x,y)\mapsto f_X(x)f_Y(y)$ is non-negative and measurable, we conclude that it is density of $(X,Y)$.


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