The joint density of $Y_1, Y_2$ is given by

$$f_{Y_1Y_2}(y_1, y_2) = \begin{cases}\frac{1}{2} e^{-\frac{1}{2}(y_1+y_2)} & 0\leq y_2 \leq y_1 < \infty\\ 0 & \text{otherwise}. \end{cases}$$

The marginal densities of $Y_1, Y_2$ are given in the book as:

$$f_{Y_1}(y_1) = \begin{cases} e^{-y_1} & y_1 \geq 0\\ 0 & \text{otherwise}, \end{cases}\qquad f_{Y_2}(y_2) = \begin{cases} e^{-\frac{1}{2}y_2}(1-e^{-\frac{1}{2}y_2}) & y_2 \geq 0\\ 0 & \text{otherwise}. \end{cases}$$

But I get it the other way, i.e.,

$$f_{Y_1}(y_1) = \begin{cases} e^{-\frac{1}{2}y_1}(1-e^{-\frac{1}{2}y_1}) & y_1 \geq 0\\ 0 & \text{otherwise}, \end{cases}\qquad f_{Y_2}(y_2) = \begin{cases} e^{-y_2} & y_2 \geq 0\\ 0 & \text{otherwise}. \end{cases}$$

Could someone tell me which is correct?


  • $\begingroup$ The condition for the first case of $f_{Y_1Y_2}(y_1, y_2)$ is probably $0\leq y_1 \leq y_2 < \infty$, not $0\leq y_2 \leq y_1 < \infty$. Then the marginal densities would be correct. $\endgroup$ – Did May 25 '14 at 9:17

Your answer is correct. In order to find the density function of $Y_2$, we "integrate out" $y_1$, and therefore $$f_{Y_2}(y_2)=\int_{y_1=y_2}^\infty \frac{1}{2}e^{-y_2/2}e^{-y_1/2}\,dy_1.$$ We get $e^{-y_2}$ (for $y_2\ge 0$).

The density function of $Y_1$ is calculated in a similar way, but in integrating out $y_2$, we integrate from $0$ to $y_1$. The expression is marginally more complicated because of the evaluation at $0$ term.

Remark: For two random variables, indices are not really a good idea. They look alike, and carry no clear geometry. You didn't get caught.


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