# Marginal Probability Density Function of Joint Distribution

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!

• Your distribution is absolutely continuous with respect to the Lebesgue measure on the square $(0,2)x(0,1)$. What is the Lebesgue measure of the line $\{(x,y);x-3y=0\}$? Mar 16, 2015 at 21:45
• @Xi'an, thank you for your comment. I haven't had a chance to learn Lebesgue measure yet in my area of study. After looking up and trying to understand the concept, it simply represents volume/area/length of a set. Therefore, the Lebesgue measure of the line would be simply an integration of that equation with boundaries you have set. However, I don't understand where the range of y went. Am I understanding this correctly?
– pit_
Mar 16, 2015 at 22:12
• what is the surface of a line? Mar 17, 2015 at 6:52

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.

• Wouldn't $P(X=0.5) = P(0.5 \le X \le 0.5) = \text{ therefore, } 0$ ?
– pit_
Mar 16, 2015 at 22:51
• Yes. Define g(x)=1 if x=0.5, and 0 otherwise. This is called the indicator function for the set {0.5}. You arrive at zero by integrating $f(x) \cdot g(x)$ over your sample space, which in my example is the unit interval. What is g(x, y) in your example? And what is the entire sample space in your example. Mar 16, 2015 at 23:48
• The above equations that I have written is all I have.
– pit_
Mar 17, 2015 at 0:32
• I disagree. You also have your natural abilities of generalization and pattern matching. I asked you for P(X=0.5). The set of solutions to the equation X = 0.5 is {0.5}. So we computed the integral of fg where g was the "indicator" function of that set, to get the probability. In you example, what is the equation we want a set of solutions to, what is the indicator function g, and what is the integral of fg. Mar 17, 2015 at 1:10
• I revisited my Calculus book and various Wikipedia pages before coming back. The indicator function I have is $g(x,y) = x-3y$ for the entire square region, x:{0,2} y:{0,1}. Then I will need to integrate f(x,y) * g(x,y) dydx. With regions given from original piecewise function.
– pit_
Mar 17, 2015 at 2:28