I have this question regarding marginal probability density function of joint distribution. Following is the equation I have.
$$f(x,y) = \begin{cases} \frac{3}{2} y^2 & 0 \le x \le 2 \text{ and } 0 \le y \le 1 \\
0 & \text{otherwise} \end{cases}$$
I am trying to find this probability:
$$P(X=3Y)$$
I have tried calculating $f_x(x)$ which equates to:
$$\int_0^\frac{x}{3} \frac{3}{2}y^2~dy$$
$$=\frac{3}{2} * \frac{\frac{x^3}{27}}{3}$$
$$=\frac{x^3}{54} \text{ if } 0 \le x \le 2$$
I am unsure whether what I did above is correct or not.
Furthermore, if it is calculating the probability of $X=3Y$, then why would I need to integrate this piecewise function?
Any insight on this would be greatly appreciated!
Best Answer
Given what you said about the nature of the problem (marginal distributions) I'm wondering if the problem actually asked for P(X | X = 3Y).
But if this is not the case The comment above applies. Consider a 1D example with PDF, f(x) = 1 for 0 < x < 1. What is P(X = 0.5)? How do you reach that answer? It should involve integrating across the entire sample space.