0

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!

  • The density of $X$ should not depend on $y$. There might be a typo in your book. –  Jan 22 '21 at 21:08
  • If it doesn't depend on y, what should the bounds on $x$ be? –  Jan 22 '21 at 21:21
  • By definition of the marginal density $f_X(x)=\int_{-\infty}^\infty f(x,y)dy$, this is a function in $x$, not $y$. If there is any "bounds" at all, it must be a constant.// There must be a typo in the solution. What book are you reading? –  Jan 22 '21 at 21:24
  • This book is introduction to probability by Mark Ward I am looking at exercise $26.5$ on page $340$ –  Jan 22 '21 at 21:26

2 Answers2

0

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 $

sliberatore18
  • 11
0

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.

enter image description here

  • I see, this makes more sense to me since $x$ shouldn't depend on $y$, so they just messed up the bounds. thank you so much for your time and help! –  Jan 22 '21 at 21:34
  • @BillBillwater: you are welcome! –  Jan 22 '21 at 21:35