prob distribution and fundamental theorm of calculus over only one parameter of function

Question

Given:

$$z=x+y$$

$$\frac{d}{dz}F_z(z) =\frac{d}{dz}F_{X,Y}(x,y)= f_Z(z)=f_{X,Y}(x,y)$$

$$F_{X,Y}(x,-\infty) = 0$$

Why is this integral true?

$$\frac{d}{dz}\Bigg(\int \limits_{-\infty}^{z-x}f_{X,Y}(x,y)~ dy\Bigg) = f_{X,Y}(x,z-x)$$

I was wondering if somebody could write this out in more steps so that I could understand what is happening, to provide the rule that is being invoked.

I'm thinking this has something to do with the fundamental theorem of calculus:

$$\int \limits^b_a f(t)~ dt = F(b) - F(a)$$

where

$$F'(x) = f(x)$$

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One of the things that bothers me is when you integrate over y that means you get a CDF on the y variable, but the x variable is still a pdf... I want to call the integral $F_{X,Y}(x,y)$, but i think that's not right...becuase the capital letter F only applies to the Y parameter, the X is still undercase f... then we take the derivative of this mess relative to z... and it gives us back $f_X,Y(x,y)$

I think it makes perfect sense if you treat X and Y are independent variables, because $F_{XY}(x,y) = F_X(X) F_Y(y)$, and that problem disappears... but the example says we are suppose to assume that they are not independent first...to get this result $f_{X,Y}(x,z-x)$... and that's where I get stuck... because I don't understand how this integral works...

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here's from the book:

Answer 1

Your first inequality is wrong because the link between the variables is missing. How you wrote it, $F_{X,Y}(x,y)$ does not have a $z$.

Leaving that aside, by definition you have

$$F_Z(z) = \mathbb P(Z\le z) = \mathbb P(X+Y\le z) = \iint_{\{x+y\le z\}}f_{X,Y}(x,y)dxdy.$$

The region described in Fig. 4-6 is exactly $\{x+y\le z\}$. The remaining is calculus: Fubini, exchanging derivatives with integrals, and FTC.

Answer 2

$\int_a^b f(t) \, dt = F(b) - F(a)$ is the second fundamental theorem of calculus, but what you really want is the first fundamental theorem of calculus: $$\frac{d}{dx} \int_a^x f(t) \, dt = f(x).$$

Combined with the chain rule, this immediately implies $$\frac{d}{dz} \int_{-\infty}^{z-x} g(y) \, dy = g(z-x) \cdot \frac{d}{dz}(z-x) = g(z-x)$$ for any $g$.

Finally, if for a fixed $x$ we define $g(y) := f_{X,Y}(x, y)$, then the above statement can be rewritten as $$\frac{d}{dz} \int_{-\infty}^{z-x} f_{X,Y}(x, y) \, dy = f_{X,Y}(x, z-x).$$ If it helps, just think of $x$ as a fixed number, like $2$ or something, so $f_{X,Y}(2, y)$ is just a function of $y$.