Chapter 5 Continuous Joint Distributions

M241 Probability Class Discussion notes

For the joint distribution of a pair of continuous random variables X and Y, we are interested in P[(X,Y) is in some set B], where B is a subset of the plane.

Uniform distribution on region R in plane: (X,Y) has the uniform distribution on R if P[(X,Y) is in C] = Area (C)/Area(R)

The simplest example to consider is independent uniformly distributed random variables X and Y.
Suppose X and Y are uniform on (0,a) and (0,b) respectively.

This means that the joint distribution of independent uniformly distributed random variables X and Y is uniform on the on the rectangle (0,a) X (0,b). The same would be true for X and Y independent, uniformly distributed on arbitrary intervals (c, c + a) and (d , d + b).

We can use methods of plane geometry or simple calculus to calculate probabilities of functions g(X,Y) where X and Y are independent uniform random variables.

Here We have to calculate the are of the shaded region (see page 342 bottom) (of course divided by the area of the unit square - which is just 1).
This is the area under the curve y = x2 from 0 to 1 -- just using simple integratoin we get 1/3.

Example 2 page 342 and 343 gives some more examples which apply straight-forward calculations of areas to get the desired probabilities.

Example 3 page 343. Probability of two people meeting. Arrival times independent and uniformly distributed on 5:00 to 5:30. Person waits five minutes and leaves. (Map this half hour into unit interval - interval of length 1/6 represents 5 minutes.) Find probability that they meet.

X = arrival time of first person
Y = arrival time of second person.
Meet if and only if their arrival time differs by at most 5 minutes; i.e. |X-Y| <= 1/6 or equivalently X-1/6 <= Y <= X+1/6
See sketch bottom page 343.
Probability = area of shaded region = 1 - (5/6)²

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₁₀

¹⁰

The joint density function for a pair of random variables X and Y is the function f(x,y) such that

The geometric interpretation of this is that the probability that (X,Y) is in B is the volume under the surface f(x,y) over the region B. See the figures 1 and 2 page 346 and 347.
See page 348 and 349 for a review of the properties of discrete joint distributions and their equivalents for continuous distributions. Note that the properties are virtually the same, except that for continous distribtuions we replace sum with integral.
Especially note the definition of the marginal density functions:
The marginal density of X (i.e. the density function of X alone), f_X(x), and the marginal density of Y (i.e. the density function of Y alone),f_Y(y) are computed from the join density f(x,y) by "integrating out" the y (or x):

For independent random variables the joint density f(x,y) is just the product of the marginal densities f_X(x) f_Y(y)
The Expectation of a function g(X, Y) is calculated by

Example 1: Let (X,Y) be uniformly distributed on the triangle {(x,y): 0 < x < y < 1}

Problem 1: Find the density function f(x,y). Uniform means f(x,y) = c, a constant. To be a valid joint density, we must have volume under f(x,y) = c over the triangle = 1. So c * ½ = 1 meaning that c is 2.
Problem 2: Find the marginal densities.
Problem 3: Not independent since the joint density not the product of the marginals!
Problem 4: Find E(X) and E(Y). Just use formulas
Problem 5: Find E(XY) - again use formula.

Let X and Y be independent and exponentially distributed random variables with parameters m and l. Calculate P(X < Y).
We just need to integrate the joint density over the region {(x, y) : 0 < x < y}
By independence, the joint density is the product of the densities.