Will the limit of integration for joint probability evaluation change when the random variables are correlated? I was having some confusion regarding the evaluation of a joint probability when the joint pdf describe random variables that are correlated.
For concreteness, I will describe my situation and though process as follow:
Suppose we have two independent random variables $X$, $Y$ (with positive supporting set) with the corresponding pdf ${f_X}\left( x \right) = {\lambda _1}\exp \left( { - {\lambda _1}x} \right)$ and ${f_Y}\left( y \right) = {\lambda _2}\exp \left( { - {\lambda _2}y} \right)$.
I know that the joint probability $\Pr \left[ {X + Y < 4} \right]$ can be easily evaluated without much effort by the following double integration. More specifically, I integrate the joint pdf ${{f_{X,Y}}\left( {x,y} \right)}$ with respect to $Y$ first and $X$ second.
$\Pr \left[ {X + Y < 4} \right] = \int\limits_{x = 0}^{x = 4} {\int\limits_{y = 0}^{y = 4 - x} {{f_{X,Y}}\left( {x,y} \right)dydx} }$
This is quite a standard problem and there is not much to said about it. Since I am an engineer guy so I really prefer thing to be as concrete as possible.
Now if  $X$ and $Y$ are correlated, the joint pdf will not be ${f_{X,Y}}\left( {x,y} \right) = {\lambda _1}\exp \left( { - {\lambda _1}x} \right) \times {\lambda _2}\exp \left( { - {\lambda _2}y} \right)$ but something else since there must be some other factor that describe the correlation.
What I want to know is that if I was to be given a correct joint pdf ${{f_{X,Y}}\left( {x,y} \right)}$ in the case $X$ and $Y$ are correlated and I want to evaluate the joint probability $\Pr \left[ {X + Y < 4} \right]$.
Will  the limit of integration changes given that we keep the order of integration as before (y first and x second) ?
I have a gut feeling that the limit of integration do not change but I am not really sure about it.
Also, ${f_{X,Y}}\left( {x,y} \right)$ could be chosen as correlated Gaussian joint pdf instead of exponential. Or more specifically
${f_{X,Y}}\left( {x,y} \right) = \frac{1}{{2\pi {\sigma _X}{\sigma _Y}\sqrt {1 - {\rho ^2}} }}\exp \left( { - \frac{1}{{2\left( {1 - {\rho ^2}} \right)}}\left[ {{{\left( {\frac{{x - {\mu _X}}}{{{\sigma _X}}}} \right)}^2} - 2\rho \left( {\frac{{x - {\mu _X}}}{{{\sigma _X}}}} \right)\left( {\frac{{y - {\mu _Y}}}{{{\sigma _Y}}}} \right) + {{\left( {\frac{{y - {\mu _Y}}}{{{\sigma _Y}}}} \right)}^2}} \right]} \right)$
Please correct me if my arbitrary choice of random variable types (exponential at the beginning of the post) cause any problem regarding the thinking process.
Thank you for your enthusiasm !
 A: If the variables are independent, then
$$
\begin{array}{l}
 \Pr \left( {(X,Y) \in \left( {\left[ {x,x + dx} \right) \times \left[ {y,y + dy} \right)} \right)} \right) =  \\ 
  = p_{X,Y} (x,y)dxdy = p_X (x)dx\;p_Y (y)dy \\ 
 \end{array}
$$
That also means that
$$
\Pr \left( {\left( {Y\left| X \right.} \right) \in \left[ {y,y + dy} \right)} \right) = p_Y (y)dy
$$
Then, intuitively speaking,
when you plot $ p_{X,Y} (x,y) = p_X (x)p_Y (y)$ you will see
some "preferential orientation" of the graph along the $x,y$ axes.
Such "preference" depends on the $x$ and $y$ support and on the
$p_x , p_y$ functions, and might be signaled by symmetry axes ,
asymptotes, or other geometric characters.
But if there is some probabilistic relation between the variables, i.e. they are correlated, then
$$
\Pr \left( {\left( {Y\left| X \right.} \right) \in \left[ {y,y + dy} \right)} \right)
 = p_{Y|X} (x,y)dy \ne p_Y (y)dy
$$
and this will reflect in a "distortion" of the "preferential" lines.
To put it more rigorous, we shall recall that the probability is a measure, thus the "preferential" lines
would be seen in terms of geodesics, Jordan determinant ...
But keeping it simple, considering the example of the 2D Gaussian you give, the correlation
is in fact "inside" the formula of the Gaussian, the region of integration in $x,y$ is unaffected, and it remains that
$$
\Pr \left( {X + Y < 4} \right) = \int\limits_{x + y < 4} {p_{X,Y} (x,y)dxdy}
  = \int_{x =  -  - \infty }^\infty  {dx\int_{y =  - \infty }^{4 - x} {p_{X,Y} (x,y)dy} } 
$$
whether or not the integral be easy to compute.
Since
$$
\begin{array}{l}
 \left( {\frac{{x - \mu _x }}{{\sigma _x }}} \right)^2  - 2\rho \left( {\frac{{x - \mu _x }}
{{\sigma _x }}} \right)\left( {\frac{{y - \mu _y }}{{\sigma _y }}} \right) + \left( {\frac{{y - \mu _y }}{{\sigma _y }}} \right)^2  =  \\ 
  = \xi ^2  - 2\rho \xi \eta  + \eta ^2  =  \\ 
  = \xi ^2  - 2\left( {\rho \eta } \right)\xi  + \left( {\rho \eta } \right)^2  + \left( {1 - \rho ^2 } \right)\eta ^2  =  \\ 
  = \left( {\xi  - \rho \eta } \right)^2  + \left( {1 - \rho ^2 } \right)\eta ^2  = u^2  + v^2  \\ 
 \end{array}
$$
we can do a change of variables, and consequently of the integration boundary
$$
\begin{array}{l}
 \left\{ \begin{array}{l}
 \frac{{x - \mu _x }}{{\sigma _x }} = \xi  \\ 
 \frac{{y - \mu _y }}{{\sigma _y }} = \eta  \\ 
 \end{array} \right. \Rightarrow \left\{ \begin{array}{l}
 u = \frac{{x - \mu _x }}{{\sigma _x }} - \rho \frac{{y - \mu _y }}{{\sigma _y }} \\ 
 v = \sqrt {\left( {1 - \rho ^2 } \right)} \frac{{y - \mu _y }}{{\sigma _y }} \\ 
 \end{array} \right. \Rightarrow \left\{ \begin{array}{l}
 x = \sigma _x u + \frac{{\rho \sigma _x }}{{\sqrt {\left( {1 - \rho ^2 } \right)} }}v + \mu _x  \\ 
 y = \frac{{\sigma _y }}{{\sqrt {\left( {1 - \rho ^2 } \right)} }}v + \mu _y  \\ 
 \end{array} \right. \\ 
  \Rightarrow x + y = \sigma _x u + \frac{{\rho \sigma _x  + \sigma _y }}
{{\sqrt {\left( {1 - \rho ^2 } \right)} }}v + \mu _y  + \mu _x  < 4 \Rightarrow  \\ 
  \Rightarrow \sigma _x u + \frac{{\rho \sigma _x  + \sigma _y }}
{{\sqrt {\left( {1 - \rho ^2 } \right)} }}v +  < 4 - \mu _y  - \mu _x  \\ 
  \Rightarrow J_{\frac{{(u,v)}}{{(x,y)}}}  = \left( {\begin{array}{*{20}c}
   {1/\sigma _x } & { - \rho /\sigma _y }  \\
   0 & {\sqrt {\left( {1 - \rho ^2 } \right)} /\sigma _y }  \\
\end{array}} \right) \Rightarrow dxdy = \frac{{\sigma _x \sigma _y }}{{\sqrt {\left( {1 - \rho ^2 } \right)} }}dudv \\ 
 \end{array}
$$
And now that we reached to an isotropous Gaussian, we can apply a pure rotation of the variables, to bring
one of them orthogonal to the bound line, and the other parallel to it .
( I leave to you to concude)
A: If $X$ and $Y$ are jointly continuous random variables with joint probability density function $f_{X,Y}(x,y)$, the probability that $X+Y \leq \alpha$ can be expressed as
$$P(X+Y\leq \alpha) = \int_{-\infty}^\infty \int_{-\infty}^{\alpha-y}f_{X,Y}(x,y) \,\mathrm dx \, \mathrm dy \tag{1}$$
regardless of whether $X$ and $Y$ are independent or what the support of $f_{X,Y}(x,y)$ happens to be. In the special case when the support of $f_{X,Y}(x,y)$ is the first quadrant, Eq. $(1)$ can be simplified to
$$P(X+Y\leq \alpha) = \int_{0}^\alpha \int_{0}^{\alpha-y}f_{X,Y}(x,y) \,\mathrm dx \, \mathrm dy. \tag{2}$$
Note that either case, the region of integration does not change regardless of whether $X$ and $Y$ are independent or dependent random variables; what changes is the integrand which can be factored into $f_X(x)f_Y(y)$ when $X$ and $Y$ are independent.
For the case when $X$ and $Y$ are jointly Gaussian random variables, we must use Eq. $(1)$ if we want to write an integral formula for $P(X+Y\leq \alpha)$ but there is a much easier way to compute the value of $P(X+Y\leq \alpha)$ (or write a pretty formula for it) than the complicated calculation proposed in the answer by @G.Cab.  If $X$ and $Y$ are jointly Gaussian random variables, then $X+Y$ is  a Gaussian random variable with mean $E[X]+E[Y]$ and variance $\sigma_X^2+\sigma_Y^2+2\rho_{X,Y}\sigma_X\sigma_Y$, and so
$$P(X+Y\leq \alpha) = \Phi\left(\frac{\alpha -E[X]-E[Y]}{\sqrt{\sigma_X^2+\sigma_Y^2+2\rho_{X,Y}\sigma_X\sigma_Y}}\right)$$
where $\Phi(\cdot)$ denotes the standard Gaussian cumulative distribution function.
A: Let $B$ be any Borel set in the plane. You are considering the set $B = \{(x,y) \mid x+y < 4\}.$ Then,
$$
P(X + Y < 4) = P((X,Y) \in B) = \int\limits_B f_{X,Y}(s,t)\ d(s,t).
$$
The first equality is convention of  and the second equality is definition of what a density is. Note that I have not used any form of $f_{X,Y}.$ So, the "limits of integration" will be always the same for a fixed $B$ (regardless of $f_{X,Y}$). However, once you know the actual shape of $f_{X,Y},$ (say $f_{X,Y} > 0$ on the region $R$ and zero elsewhere, think of positive random variables where $R$ would be $[0, \infty)^2$), then you can further reduce the "limits of integration" to the region where $f_{X,Y} > 0$ and so $P((X,Y) \in B) = \int_\limits{R \cap B} f_{X,Y}.$ You can write this as an iterated integral if you needed to calculate it, for that you need $R \cap B$ to be of "nice shape" as in $R \cap B = \{x \in [a,b], u(x) \leq y \leq v(x)\}$ or anything else that allows a nice iterated integration.
