Bivariate Random Variable

Let $(X, Y)$ be a bivariate random variable with support $$S = \{ (x, y) \mid 0 < x < 7, x < y < x + 2 \}$$ and its joint pdf $f(x, y) = 1/14$ for $(x, y) \in S$.

(A) Find the marginal probability density function for $X$.

I got $f_X(x) = 1/7$ for $0 < x < 7$.

(B) Find the marginal probability density function for $Y$.

I got $$f_Y(y) = \int_0^6\frac1{14}\mathsf dx = \frac12$$ for $x < y < x + 2$.

This part doesn't seem right to me. Can someone help me with this part?

• What are your calculations in (A) ? – callculus Jul 18 '15 at 14:12

Your marginal density for $X$ is correct. But indeed your computation for the marginal density of $Y$ is flawed; $Y$ takes values in $[0,9]$ (since $0<x<y$ and $x<y<x+2$). To compute the marginal density of $Y$, we need to integrate the joint density over $S$ with respect to $x$. For $0<y<2$, we have $0<x<y$ and $$f_Y(y)=\int_0^2 \frac1{14}\mathsf dx = \frac1{14}y$$ For $2<y<7$, we have $0<x<2$ and $$f_Y(y) = \int_0^2\frac1{14}\mathsf dx = \frac17.$$ (the bounds for that one are hard to see unless you draw a plot of $S$.) For $7<y<9$, we have $y<x<9$ and $$f_Y(y) = \int_y^9\frac1{14}\mathsf dx = \frac1{14}(9-y).$$

Hence $$f_Y(y) = \frac1{14}y 1_{(0,2)}(y) + \frac17 1_{(2,7)}(y) + \frac1{14}(9-y) 1_{(7,9)}(y).$$

• Thanks! I figured it out with your help. You are so right. I with we were able to post pics on here because being able to graph the Space to locate the bounds helps a ton. – Bill Jul 18 '15 at 16:19
• You can post images... – Math1000 Jul 18 '15 at 17:52

If you sketch the two-dimensional region, it would be easier to visualize what you need to do to find the marginal density for $Y$. Your answer for the marginal density for $X$ is correct. The one for $Y$ is more difficult because you need to consider three different regions, depending on the value of $y$ you are considering.

$$if \ \ \ 2<y<7 \ \ \ , f_Y(y)=\frac{1}{7}\ \ \$$ $$if \ \ \ 0<y<2 \ \ \ , f_Y(y)=\frac{y}{14}$$ $$if \ \ \ \ 7<y<9 \ \ \ , f_Y(y)=\frac{9-y}{14}$$ $$if \ \ \ y<0 \ \, \ \ \ \ f_Y(y)=0$$