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I'm having trouble understanding exactly why when you have a two-dimensional random variable $(X,Y)$ and you want to find the marginal probability distribution of $X$, for example, you integrate the joint probability density function with respect to y. That is,

$f_x=\int_{-\infty}^{\infty}f(x,y) \ dy$

I understand its analogous form in the discrete case:

$p(X=x_i)=\sum_j p(x_i,y_j)$

because here you're just adding up the different ways $X$ can be $x_i$ (i.e., it can be $x_i$ when $Y$ is $y_1$ or $y_2$ etc.) but I don't get exactly what is being done by integrating with respect to $y$. What is being summed up? And how does it work when there are no distinct probabilities but only a probability density, in terms of adding up possibilities to "marginalize out" a variable?

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The joint density $f_{XY}(x,y)$ is defined in terms of probability as $$\mathbb{P}((X,Y)\in A)=\int_A f_{XY}(x,y)\,\text{d}x\text{d}y$$ for all measurable sets $A$. When $A$ is of the special form $A=[a,b]\times[c,d]$, this translates as $$\mathbb{P}((X,Y)\in A)=\int_c^d\int_a^b f_{XY}(x,y)\,\text{d}x\text{d}y=\int_c^d\left\{\int_a^b f_{XY}(x,y)\,\text{d}x\right\}\text{d}y$$that is, as the integral in $y$ of a function of $y$ defined as$$g(y)=\int_a^b f_{XY}(x,y)\,\text{d}x$$In particular, when $a=-\infty$ and $b=+\infty$ you get that$$\mathbb{P}((X,Y)\in \mathbb{R}\times[c,d])=\int_c^d\int_{-\infty}^{+\infty} f_{XY}(x,y)\,\text{d}x\text{d}y$$which shows that$$\int_{-\infty}^{+\infty} f_{XY}(x,y)\,\text{d}x$$operates like a density for $Y$.

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  • $\begingroup$ I'm trying to understand the integral in terms of Riemann sums...could you help me out with that $\endgroup$ – Gabriel Mar 18 '16 at 20:26
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    $\begingroup$ @Gabriel, probability integrals do not always work with riemannian definition of an integral. If you still want to cast them as Riemannian, then imagine that x and y are cut into bins. In this case the integrals become sums. $\endgroup$ – Aksakal Mar 18 '16 at 20:32
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    $\begingroup$ There is not much difference with Riemann sums when $f$ is continuous or continuous by parts and when $A$ is a union or intersection of a collection of cubes $[a,b]\times[c,d]$, which is why I wrote this explanation for a single cube. $\endgroup$ – Xi'an Mar 18 '16 at 20:33
  • $\begingroup$ So then how would I interpret $\sum f(x,y) \Delta y $ $\endgroup$ – Gabriel Mar 18 '16 at 20:35
  • $\begingroup$ I do not understand what this sum means in terms of the original problem. $\endgroup$ – Xi'an Mar 18 '16 at 20:42

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