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Let's say, I have joint probability function as follows:

$ f(x,y) = 4xy $ for $ 0 \le x \le 1 $ and $ 0 \le y \le 1 $

I want to get the marginal probability distribution of the random variable $ X $ from the joint probability distribution. So I integrate out the other random variable $ Y $. Then I get

$ f_X(x) = \int_{0}^{1} 4xy dy = [2xy^2]_{0}^{1} = 2x $

How can I use the resulted probability distribution function to find a probability between an interval of the random variable $ X $? Do I need to integrate the resulted function for some specific probability density between an interval of the random variable $ X $?

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Yes, for example, $$P(0<X<1)=\int_0^1 f(x)dx=\int_0^1 2x=1$$ as expected. Note that $f(x)=0$ when $x<0$ or $x>1$.

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  • $\begingroup$ Thank you for the answer. If possible, could I ask why we integrate with respect to the random variable of $ Y $? I can understand the mathematical way to get the marginal probability distribution function. Mathematically we can get rid of the other random variable but I just can't visualize the way we integrate out the random variable of $ Y $. $\endgroup$
    – xabzakabecd
    Commented Aug 31, 2020 at 12:05
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    $\begingroup$ Think about the discrete version: $$P(X=x)=\sum_y P(X=x, Y=y)$$ where the probability of $X=x$ is equal to the probabilities of all the events that the other RV is equal to some value with $x$ being constant $\endgroup$
    – gunes
    Commented Aug 31, 2020 at 12:12
  • $\begingroup$ To me, for discrete cases with 2 random variables, it looks like that we can directly sum over all the values of the other variable to get the marginal probability or probability distribution but for continuous cases, we integrate out the nuisance variable and get the marginal probability distribution function of the interested random variable, then use integration to get specific probability density for a specific interval with the resulting marginal probability distribution function. Am I correct? $\endgroup$
    – xabzakabecd
    Commented Aug 31, 2020 at 13:23
  • $\begingroup$ yes, that’s correct (probability of a specific interval, not the prob. density of a specific interval) $\endgroup$
    – gunes
    Commented Aug 31, 2020 at 13:41

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