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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!

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    $\begingroup$ 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 $\{(x,y);x-3y=0\}$? $\endgroup$
    – Xi'an
    Mar 16, 2015 at 21:45
  • $\begingroup$ @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? $\endgroup$
    – pit_
    Mar 16, 2015 at 22:12
  • $\begingroup$ what is the surface of a line? $\endgroup$
    – Xi'an
    Mar 17, 2015 at 6:52

1 Answer 1

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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.

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    $\begingroup$ Wouldn't $P(X=0.5) = P(0.5 \le X \le 0.5) = \text{ therefore, } 0$ ? $\endgroup$
    – pit_
    Mar 16, 2015 at 22:51
  • $\begingroup$ 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. $\endgroup$ Mar 16, 2015 at 23:48
  • $\begingroup$ The above equations that I have written is all I have. $\endgroup$
    – pit_
    Mar 17, 2015 at 0:32
  • $\begingroup$ 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. $\endgroup$ Mar 17, 2015 at 1:10
  • $\begingroup$ 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. $\endgroup$
    – pit_
    Mar 17, 2015 at 2:28

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