5.2 - bivariate continuous distributions

Joint CDF

  • Recall definition of CDF.
  • For any random variables \(X\) and \(Y\), the joint bivariate distribution function \(F\) is:

\[F(x,y) = P(X\leq x, Y \leq y); x,y \in \mathbb{R}\]

Properties:

  1. \(\lim_{x\rightarrow -\infty}F(x,y) =\lim_{y\rightarrow -\infty}F(x,y) = 0\)
  2. \(\lim_{x\rightarrow \infty}\lim_{y\rightarrow \infty}F(x,y) =1\)
  3. Bivariate non-decreasing property: If \(x < x^*\) and \(y < y^*\), then

\[F(x^*, y^*) - F(x^*,y)-F(x,y^*)+F(x,y)= P(x < X \leq x^*, y < Y \leq y^*) \geq 0\]

Visual depiction of cumulative probability over a rectangular region

Joint continuous definitions

  • \(X\) and \(Y\) are said to be jointly continuous if \(F(x,y)\) is continuous in both arguments.

  • If there exists \(f(x,y)\) such that:

\[F(x,y) = \int_{-\infty}^x\int_{-\infty}^y f(t_1, t_2) dt_1 dt_2,\]

then \(f(x,y)\) is called the joint probability density function or joint pdf.

Properties of joint pdf

If \((X,Y)\) are jointly continuous random variables with joint pdf \(f(x,y)\), then:

  1. \(f(x,y) \geq 0\ \forall\ x,y\)
  2. \(\int_{-\infty}^\infty \int_{-\infty}^\infty f(x,y)\ dx\ dy = 1\)

Example: joint uniform

Suppose \(X\) and \(Y\) have joint pdf given by:

\[f(x,y) = \begin{cases} 1 & 0 < x < 1, 0 < y < 1 \\ 0 & otherwise \end{cases}\]

TASKS:

  1. Show that this is a valid joint pdf.
  2. Find the joint CDF.
  3. Find \(P(Y < X^2)\).

Interactive: https://www.desmos.com/3d/un0eouqhgq

1. Show valid pdf

Support (always draw!):

Support of jointly uniform(0,1) random variables

\[\int_0^1 \int_0^1 1\ dx\ dy = \int_0^1 \left(x\big|_0^1\right)\ dy\] \[\int_0^1 1\ dy = y\big|_0^1 =1\]

2. Find \(F(x,y)\)

Always draw/shade the region of integration!

\((x,y) \in (0,1)^2\)

Region of integration for finding F(x,y) when x and y are in support

\[\tiny F(x,y) = \int_0^x\int_0^y 1\ dv\ dt\] \[\tiny = \int_0^x\left(u \big|_0^y\right)dt = \int_0^x y\ dt = yt\big|_{t=0}^x = xy\]

\(0<x<1, y \geq 1\)

Region of integration for finding F(x,y) when 0<x<1 and y is greater than 1

\[\tiny F(x,y) = \int_0^x\int_0^1 1\ dv\ dt\] \[\tiny= \int_0^x \left(y\big|_0^1\right)\ dt =\int_0^x 1 \ dt = x\]

\(x\geq 1, 0 < y <1\)

Region of integration for finding F(x,y) when 0<y<1 and x is greater than 1

\[\tiny F(x,y) = \int_0^1\int_0^y 1\ dv\ dt\] \[\tiny= \int_0^1 \left(u\big|_0^y\right)\ d = \int_0^1 y\ dt = y\]

All cases

\[F(x,y) = \begin{cases} 0 & x < 0\ \ OR\ \ y < 0 \\ xy & 0 < x < 1, 0 < y < 1 \\ x & 0 < x < 1, y \geq 1 \\ y & x \geq 1, 0 < y < 1\\ 1 & x\geq 1, y \geq 1 \end{cases}\]

3. Find \(P(Y<X^2)\)

First step: Draw region!

Second step: determine order of integration

Vertical strip; \(dy\ dx\)

Region of integration for finding P(Y is less than X squared), vertical strip

\[\tiny P(Y<X^2) = \int_0^1 \int_0^{x^2} 1\ dy\ dx\] \[\tiny P(Y<X^2) = \int_0^1 x^2\ dx = \frac{x^3}{3}\big|_0^1 = \frac{1}{3}\]

Horizontal strip; \(dx\ dy\)

Region of integration for finding P(Y is less than X squared), horizontal strip

\[\tiny P(Y<X^2) = \int_0^1 \int_{\sqrt{y}}^{1} 1\ dx\ dy\] \[\tiny P(Y<X^2) = \int_0^1 (1-\sqrt{y})\ dy = \left(y-2\frac{y^{3/2}}{3}\right)\big|_0^1 = 1-\frac{2}{3}\]

Example: triangular support

Suppose \(X\) and \(Y\) are jointly continuous with joint pdf:

\[f(x,y) = \begin{cases}3x & 0 < y < x < 1 \\ 0 & otherwise \end{cases}\]

  1. Show that this is a valid joint pdf
  2. Find \(F(x,y)\).
  3. Find \(P(0\leq X \leq 0.5, Y > 0.25)\).

Interactive: https://www.desmos.com/3d/bw7gleh7jj

1. Valid pdf

Draw the support!

Horizontal strip; \(dx\ dy\):

Figure of triangular support with horizontal integration strip

\[\scriptsize \int_0^1 \int_y^1 3x\ dx\ dy=\int_0^1 \left(\frac{3x^2}{2}\big|_y^1\right)\ dy\] \[\scriptsize =\int_0^1 \frac{3}{2}(1-y^2)\ dy = \frac{3}{2}\left(y-\frac{y^3}{3}\right)\big|_0^1 = \frac{3}{2}\frac{2}{3}=1\]

Vertical strip; \(dy\ dx\):

Figure of triangular support with vertical integration strip

\[\scriptsize \int_0^1 \int_0^x 3x\ dy\ dx=\int_0^1 \left(3xy\big|_{y=0}^x\right)\ dy\] \[\scriptsize =\int_0^1 3x^2\ dx = x^3\big|_0^1 = 1\]

2. CDF cases

\(F(x,y)\) cases:

\(0 < y < x < 1\):

Region of integration when x and y are in the support

\(y \geq x, 0 < x < 1\):

Region of integration when y > x and 0 < x < 1

\(0 < y < 1, x \geq 1\):

Region of integration when 0<y<1 x> 1

Strategy: Find CDF for case 1; let \(x, y \rightarrow \infty\) for other two cases.

CDF case 1

Vertical strips:

Image representing limits of integration when using vertical strips

\[\scriptsize F(x,y) = \int_0^y\int_0^t 3t\ dv\ dt + \int_y^x \int_0^y 3t\ dv\ dt\]

Horizontal strip:

Image representing limits of integration when using horizontal strip

\[\scriptsize F(x,y) = \int_0^y\int_v^x 3t\ dt\ dv = \int_0^y\left( \frac{3t^2}{2}\big|_v^x\right) dv\] \[\scriptsize = \int_0^y\frac{3}{2}\left(x^2-v^2\right) dv = \frac{3}{2}\left(x^2v-\frac{v^3}{3}\right)\big|_0^y \] \[\scriptsize = \frac{3x^2y-y^3}{2}\]

CDF cases 2-3

\(0 < y < x < 1\):

Region of integration when x and y are in the support

\[\scriptsize F(x,y)= \frac{3x^2y-y^3}{2}\]

\(y \geq x, 0 < x < 1\):

Region of integration when y > x and 0 < x < 1

\[\scriptsize \lim_{y\rightarrow \infty}F(x,y)=F(x,x)\] \[ \scriptsize= \frac{3x^3-x^3}{2}=x^3\]

\(0 < y < 1, x \geq 1\):

Region of integration when 0<y<1 x> 1

\[\scriptsize \lim_{x\rightarrow \infty}F(x,y)=F(1,y)\] \[ \scriptsize= \frac{3y-y^3}{2}\]

All cases

\[F(x,y) = \begin{cases} 0 & x < 0\ or\ y < 0 \\ \frac{3x^2y-y^3}{2} & 0 < y < x < 1 \\ x^3 & 0 < x < 1, y \geq x\\ \frac{3y-y^3}{2} & 0 < y < 1, x \geq 1\\ 1 & x\geq 1, y \geq 1 \end{cases}\]

\(P(0\leq X \leq 0.5, Y > 0.25)\)

Shade the region of integration!

Vertical strip:

Region of integration with vertical strip

\[\scriptsize P(0\leq X \leq 0.5, Y > 0.25) = \int_{0.25}^{0.5}\int_{0.25}^x 3x\ dy\ dx\]

Horizontal strip:

Region of integration with horizontal strip

\[\scriptsize P(0\leq X \leq 0.5, Y > 0.25) = \int_{0.25}^{0.5}\int_{y}^{0.25} 3x\ dx\ dy\]

Example: Infinite area support

Let \(X\) and \(Y\) be joint continuous random variables with joint pdf given by:

\[f(x,y) = \begin{cases} xe^{-y} &0 < x < y < \infty \\ 0 & otherwise \end{cases}\]

  1. Show that this is a valid joint pdf.
  2. Find \(P(X+Y<3)\).

Interactive: https://www.desmos.com/3d/efodrvofks

1. Valid pdf

Draw the support!

Vertical strip; \(dy\ dx\):

Region of integration with vertical strip

\[\scriptsize \int_0^\infty \int_x^\infty xe^{-y}\ dy\ dx\]

Horizontal strip; \(dx\ dy\):

Region of integration with horizontal strip

\[\scriptsize \int_0^\infty \int_0^y xe^{-y}\ dx\ dy=\int_0^\infty \left(\frac{x^2}{2}\big|_0^y\right)e^{-y}\ dy\] \[\scriptsize =\int_0^\infty \frac{1}{2}y^2e^{-y}\ dy=\frac{1}{2}\Gamma(3) = 1\]

\(P(X+Y < 3)\)

\(P(X+Y<3) = P(Y < 3-X)\); draw the region!

Vertical strip; \(dy\ dx\):

Region of integration with vertical strip

\[\scriptsize \int_0^{1.5} \int_x^{3-x} xe^{-y}\ dy\ dx\]

Horizontal strips; \(dx\ dy\):

Regions of integration with horizontal strips

\[\scriptsize \int_0^{1.5} \int_0^{y} xe^{-y}\ dx\ dy + \int_{1.5}^{3} \int_0^{3-y} xe^{-y}\ dx\ dy\]

Evaluating the first integral

\[\scriptsize P(X+Y<3) = \int_0^{1.5} \int_x^{3-x} xe^{-y}\ dy\ dx = \int_0^{1.5} \left(-xe^{-y}\big|_{y=x}^{3-x}\right)dx\]

\[\scriptsize =\int_0^{1.5} x \big(e^{-x} - e^{-(3 - x)}\big)\ dx= \underbrace{\int_{0}^{1.5} x e^{-x}\, dx}_{I_1} - \underbrace{\int_{0}^{1.5} x e^{-(3 - x)}\, dx}_{I_2}.\]

\[\scriptsize I_1 = \int_{0}^{1.5} x e^{-x}\, dx = -(x + 1)e^{-x}\Big|_{0}^{1.5} = 1 - 2.5 e^{-1.5}.\]

\[\scriptsize I_2 = \int_{0}^{1.5} x e^{-(3 - x)}\, dx = e^{-3} \int_{0}^{1.5} x e^{x}\, dx = e^{-3} (x - 1)e^{x}\big|_{0}^{1.5} = e^{-3}\big(0.5 e^{1.5} + 1\big) = 0.5 e^{-1.5} + e^{-3}.\]