# Thread: Catch my mistake: verifying joint pdf.

1. ## Catch my mistake: verifying joint pdf.

The joint pdf of $\displaystyle X$ and $\displaystyle Y$ is given by:

$\displaystyle f_{X,Y}(x,y) = \frac{16}{15}(x^{2} + \frac{xy}{2}), 0<x<1, 0<y<x+1$

Verify that this is a legitimate joint pdf. So I try to integrate the pdf and see if it equals one.

$\displaystyle \frac{16}{15}\int_0^1 \int_0^{x+1}(x^{2} + \frac{xy}{2})dydx = \frac{16}{15}\int_0^1 [x^{2}y + \frac{xy^{2}}{4}]_0^{x+1}dx =$$\displaystyle \frac{16}{15}\int_0^1 (x^3 + x^2 + \frac{x^3 + 2x^2 + x}{4})dx =$$\displaystyle \frac{16}{15}[\frac{x^4}{4} + \frac{x^3}{3} + \frac{x^4}{16} + \frac{x^2}{4} +$$\displaystyle \frac{x^2}{8}]_0^1=\frac{49}{45}$

As you can see, the integral of the pdf does not equal one. But it has to since it asked me to verify that it is legitimate! Where is my mistake? Thanks.

2. Originally Posted by jfz23
[snip]

$\displaystyle \frac{16}{15} \int_0^1 \left( x^3 + x^2 + \frac{x^3 + 2x^2 + x}{4} \right) \, dx$

$\displaystyle = \frac{16}{15} \left[ \frac{x^4}{4} + \frac{x^3}{3} + \frac{x^4}{16} + \frac{x^2}{4} + \frac{x^2}{8}\right]_0^1$ Mr F says: The mistake is in this line.

As you can see, the integral of the pdf does not equal one. But it has to since it asked me to verify that it is legitimate! Where is my mistake? Thanks.
..

3. Ahh I found it! Such a simple calculation error...