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I am trying to understand bivariate probability distribution functions and I am following all of Statistics: a concise course in statistical inference book. In this book the author give one example for joint mass function as mentioned below:

Suppose that $$f (x, y) = \begin{cases}x + y &\text{ if }0 \le x \le 1, 0 \le y \le 1 \\ 0&\text{ otherwise.}\end{cases}$$ If $x = 1$ and $y = 1$ then $f(x,y) = 2$ (which is wrong because probability can not be greater than $1$ ) what is the meaning of this?

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Venkatesh
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1 Answers1

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Actually, it makes sense for the same reason that if $x=y=0$, then $f(x,y) = 0$. You are forgetting that to verify that a function is a density, you should integrate over all values, $$\int_0^1\int_0^1f_{X,Y}(x,y)\,dxdy = \int_0^1\int_0^1 x+y\,dxdy = 1.$$

Analogously, if $X\sim\text{unif}(0,1/2)$, then if $x = 1/2$, then $$f_X(x) = 2.$$ But $f_X(x)$ is still a density since $$\int_0^{1/2}f_X(x)\,dx =\int_0^{1/2} 2\,dx = 1.$$

Em.
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  • so what is the interpretation to f(x) = 2 in your answer? it can not represent probability right?the probability of getting any number between 0 and 1/2 is 2 . – Venkatesh Dec 08 '15 at 08:35
  • Nooooo? The probability of getting any value between (0,.5) when $X\sim\text{unif}(0,.5)$ is 1. Ummmmmm, it might be helpful for you to review properites of discrete and continuous random variables. – Em. Dec 08 '15 at 08:59
  • sorry,i am totally confused can you please explain where i am wrong ! – Venkatesh Dec 08 '15 at 09:45
  • Actually, let me clarify my statement regarding the uniform $X$. What I meant was $P(X\leq .5) = 1$. However, $P(X = a) = 0$ for any constant $a$. Regarding the density, $f_X(a) = 2$ for any constant $0\leq a\leq .5$. This is simply the height at $x =a$. I'm not exactly sure about the "interpretation" of the height at a particular point. – Em. Dec 08 '15 at 09:54
  • Now i understood that 2 does not have any meaning except it reflects constant probability.when we consider continuous random variables we take intervals into consideration to calculate the probability .thank you so much :) – Venkatesh Dec 08 '15 at 16:29