Let $X,Y$ have joint density
$f(x, y) = 3xy/1250$ for $0 ≤ x, 0 ≤ y, x + y ≤ 10$ and $f(x,y) = 0 $ otherwise
Find the density $f(x)$ of $X$.
My attempt: $f(x) = \int_0^{10-x} \cfrac{3xy}{1250}dy = \cfrac{30}{1250}x^2-\cfrac{3}{1250}x^3$ for $0\leq x \leq 10-y$
But this does not match the answer in the book: $f(x) = \cfrac{3}{2500}x(10 − x)^2$ for $0 ≤ x ≤ 10 − y$
What am I misinterpreting about the bounds of the integral here? Where did I make a mistake?
Thank You!
On
I think you have the bounds of integration correct, but there may be an issue in the execution of the integral. Consider this:
$f(x)=\int_{0}^{10-x}{\frac{3}{1250}xydy}$
$f(x)=\frac{3}{1250}\int_{0}^{10-x}{xydy}$
$f(x)=\frac{3}{1250}{\frac{xy^2}{2}}|_{0}^{10-x}=\frac{3}{1250}(x(10-x)^2-0)$
Which then simplifies to $f(x) =\frac{3}{2500}x(10-x)^2$ for $0 \leq x \leq 10-y $
Both solutions are incorrect: the marginal density must not contain the variable $y$. By definition: $$ f_X(x)=\int_{-\infty}^\infty f(x,y)dy $$
The density of $X$ is given as follows.
For $x\not\in (0,10)$, $f_X(x)=0$.
For $0\le x\le 10$, $$ f_X(x)=\int_{-\infty}^\infty f(x,y)dy =\int_0^{10-x}\frac{3xy}{1250}dy =\frac{3x}{1250}\cdot \frac12 (10-x)^2 $$
It is useful to draw a picture for the region where $f$ is nonzero.