Joint Probability Density Function

A Joint Probability Density Function $f(x,y)$ describes the Multivariate Distribution of two continuous random variables. Probability is volume under the surface $f$ : to find $P((X,Y) \in A)$ , integrate $f$ over the region $A$ .

$P\bigl((X,Y) \in A\bigr) = \iint_A f(x,y)\,dA$

$f_X(x) = \int f(x,y)\,dy$

$f_{Y \mid X}(y \mid x) = \frac{f(x,y)}{f_X(x)}$

A valid joint density satisfies $f(x,y) \geq 0$ and integrates to 1 over its support. Setting that double integral equal to 1 is how an unknown constant $c$ is found.
The marginal $f_X(x)$ integrates $y$ out — the limits on that inner integral depend on $x$ whenever the support is not a rectangle. The marginal is a function of $x$ alone; if $y$ survives, the limits were wrong.
$X$ and $Y$ are independent only if $f(x,y) = f_X(x)f_Y(y)$ and the support is a rectangle. A triangular support such as $0 < x < y < 1$ forces dependence no matter how the formula factors, because the range of $Y$ depends on $x$ .
$E[g(X,Y)] = \iint g(x,y) f(x,y)\,dx\,dy$ — see Moments for Joint Distributions and Covariance.

Setting up the limits

The integrand is rarely the hard part; the region is. Work in this order:

Sketch the support. Shade the set of $(x,y)$ where $f > 0$ .
Pick an outer variable and read off its full numeric range — those limits must be constants.
Slice. Holding the outer variable fixed, read the inner variable's range off the sketch — those limits may involve the outer variable.
Sanity check: the outer limits are always numbers; the inner limits may mention the outer variable but never the inner variable itself — the limits on $dy$ never contain $y$ .

$\int_{a}^{b}\!\!\int_{g_1(x)}^{g_2(x)} f(x,y)\,dy\,dx$

Reversing the order of integration is a legitimate move and is often far easier — but the limits must be re-derived from the sketch, not swapped mechanically.

Finding the Normalizing Constant on a Triangle $f(x,y) = c(x+y)$ on the region $0 < x < y < 1$ . Find $c$ .
Answer The support is the triangle above the line $y = x$ inside the unit square. Integrate $x$ from 0 to 1; for each $x$ , $y$ runs from $x$ up to 1:
 $\begin{align*} 1 &= \int_0^1\!\!\int_x^1 c(x+y)\,dy\,dx \\ &= c\int_0^1 \left[xy + \tfrac{y^2}{2}\right]_{y=x}^{y=1} dx \\ &= c\int_0^1 \left(x + \tfrac{1}{2} - x^2 - \tfrac{x^2}{2}\right) dx \\ &= c\int_0^1 \left(x + \tfrac{1}{2} - \tfrac{3x^2}{2}\right) dx \\ &= c\left[\tfrac{1}{2} + \tfrac{1}{2} - \tfrac{1}{2}\right] \\ &= \frac{c}{2} \end{align*}$ 
So $c = 2$ .

Marginal Density on a Non-Rectangular Support Losses on two coverages have $f(x,y) = 8xy$ for $0 < x < y < 1$ . Find the marginal density of $Y$ .
Answer To get $f_Y(y)$ , integrate $x$ out. On the sketch, fixing $y$ means $x$ runs from 0 up to $y$ — not from 0 to 1:
 $\begin{align*} f_Y(y) &= \int_0^y 8xy\,dx \\ &= 8y\left[\tfrac{x^2}{2}\right]_0^y \\ &= 4y^3, \quad 0 < y < 1 \end{align*}$ 
Check: $\int_0^1 4y^3\,dy = 1$ . Using limits 0 to 1 for $x$ would have given $4y$ , which integrates to 2 — an immediate signal the region was wrong.

Probability over a Region and a Conditional Density With $f(x,y) = 8xy$ on $0 < x < y < 1$ , find $P(Y > 2X)$ and the conditional density $f_{X \mid Y}(x \mid y)$ .
Answer The event $Y > 2X$ is the part of the triangle above the line $y = 2x$ . Integrating $x$ outermost, $y$ runs from $2x$ to 1, and $x$ can go no higher than $\tfrac{1}{2}$ :
 $\begin{align*} P(Y > 2X) &= \int_0^{1/2}\!\!\int_{2x}^{1} 8xy\,dy\,dx \\ &= \int_0^{1/2} 4x\left[y^2\right]_{2x}^{1} dx \\ &= \int_0^{1/2} \left(4x - 16x^3\right) dx \\ &= \left[2x^2 - 4x^4\right]_0^{1/2} \\ &= \tfrac{1}{2} - \tfrac{1}{4} \\ &= 0.25 \end{align*}$ 
Dividing the joint by the marginal $f_Y(y) = 4y^3$ from the previous example:
 $\begin{align*} f_{X \mid Y}(x \mid y) &= \frac{8xy}{4y^3} \\ &= \frac{2x}{y^2}, \quad 0 < x < y \end{align*}$ 
Given $Y = y$ , $X$ is confined to $(0, y)$ — the conditional support shrinks with $y$ , which is exactly why $X$ and $Y$ cannot be independent here even though $8xy$ factors as $(8x)(y)$ .