Finding the density function from joint density function

232 Views Asked by At

I'm reading the conditional distributions section of Probability and Random Processes by Grimmett and Stirzaker and I've come across a brief exercise I can't seem to figure out.

We're given earlier that $f_{X}(x)=\int_{-\infty}^{\infty}f(x,y)dy$ and the example states that X and Y have joint density function

$f_{X,Y}(x,y)=\frac{1}{x}$ for $0 \leq y \leq x \leq 1$

And then it says: show for yourself (exercise) that $f_{X}(x)=1$ if $0 \leq x \leq 1$.

I think I must be being stupid, but I keep getting:

$f_{X}(x)=\int_{-\infty}^{\infty}f(x,y)dy=\int_{0}^{1}\frac{1}{x}dy=\left [ \frac{y}{x} \right ]_{0}^{1}=\frac{1}{x}$. Why is this wrong?

Thanks for your help in advance.

1

There are 1 best solutions below

2
On BEST ANSWER

We want to "integrate out" $y$. Note that the joint density is $\frac{1}{x}$ in the part of the unit square below the line $y=x$. So we integrate the constant $\frac{1}{x}$ from $y=0$ to $y=x$. Not from $y=0$ to $y=1$. Above the line $y=x$, we have $f(x,y)=0$.