Question

The continuous random variables X and Y have joint pdf f(x,y)=3/(2 x^3 y^2) 1/x <= y <= x , 1<=x<=∞

find
1)Find the marginal distribution of X and Y.
2) Find P(X > 2) and P(X > 2 and Y > 2).

Answer 1

\[f(x,y) = \frac{ 3 }{ 2x^{3}y^{2} } \]
\[\frac{ 1 }{ x } \le y \le x\] and \[1 \le x < \infty \]

Answer 2

\[
p_X(x)=\int \frac{3}{2x^3y^2}dy
\]Finding the limits is the issue I suppose.

Answer 3

Since the lowest \(1/x\) can go is \(1/\infty \to 0\) and the highest \(x\) can go is \(\infty\), I suppose those are the limits.

Answer 4

Actually you should probably just do this: \[
p_X(x) = \int\limits _{\frac 1x}^x \frac{3}{2x^3y^2}dy=-\frac{3}{2x^3y}\Bigg|_{\frac{1}{x}}^x=\frac{3}{2x^2}-\frac{3}{2x^4}=\frac{3x^2-3}{2x^4}
\]

Answer 5

I'm sure you know how to integrate, so it should be no problem for you.

Answer 6

thanks, it's really helpful for finding limits