1
$\begingroup$

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?

Thanks.

$\endgroup$
  • $\begingroup$ 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
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.