Question

We start out with a couple of defintions and examples.
Definition: Let X and Y have joint pdf f(x,y). The conditional pdf of Y given X = x (resp. of X given Y = y) is defined by

h(y|x) = f (x, y) resp. g(x|y) = f (x, y) f1(x) f2(y)

If A is a subset of the real line, then

P(Y ∈A|X =x)= h(y|x)dy resp. P(X ∈A|Y =y)= g(x|y)dx . AA

Example 1 (seen in class) Consider the joint pdf of the random variables X and Y : 2, if 0 < x ≤ y < 1

Recall, the marginal pdf of X if f1(x) = 2(1−x), x ∈ (0,1) and the marginal pdf of Y is f2(y) = 2y, y ∈ (0,1). Both X and Y have the interval (0,1) as their support (i.e., their pdf’s are equal to 0 outside (0, 1)).

f (x, y) = 0, otherwise

Given y ∈ (0, 1), the conditional density of X , given Y = y is
g(x|y)= f(x,y) = 2 = 1. x∈(0,1).

f2(y) 2y y It should be clear that g(x|y) = 0 if x ̸∈ (0, 1). Therefore,

1/y, if 0 < x < y, y ∈ (0, 1)
g(x|y) = 0, otherwise (1)

Keep in mind that in equation (1), y is held constant while x varies, subject to the stated constraint. It is easy to see that given Y = y (y ∈ (0, 1)), the conditional distribution of X is the uniform distribution over the interval (0, y).

As a numerical application, let’s compute the conditional probability of the event X ≤ 1/4) given Y = 1/2. By definition,

1/4 1/4 P((X ≤ 1/4|Y = 1/2) = g(x|1/2)dx =

00

2dx = 1/2.

Your turn: We reverse the roles of X and Y . (use the back page if you run out of space)
(i) What is the conditional pdf, h(y|x), of Y given X = x, x ∈ (0, 1). Present your result in the form of

equation (1) above and name the distribution (it’s recognaizable). (ii) Compute P (Y > 5/9|X = 2/3).

Example 2 (also seen in class) Consider the joint pdf of the random variables X and Y : 2e−x−y, if 0 < x ≤ y < ∞

Recall, the marginal pdf of X if f1(x) = 2e−2x, x ∈ (0, ∞) and the marginal pdf of Y is f2(y) = 2e−y(1−e−y), y ∈ (0,∞). Both X and Y have the interval (0,∞) as their support (i.e., their pdf’s are equal to 0 outside (0, ∞)).

Given x ∈ (0, ∞), the conditional density of Y , given X = x is f (x, y) 2e−x−y

f (x, y) = 0, otherwise

h(y|x) = f1(x) = 2e−2x = ex−y. y > x. It should be clear that h(y|x) = 0 if y ∈ (0, x]. Therefore,

ex−y, if y > x, x > 0
h(y|x) = 0, otherwise (2)

Keep in mind that in equation (2,) x is held constant while y varies, subject to the stated constraint (h(y|x) is not a recognizable pdfl–it’s sometimes called the shifted exponential distribution).

As a numerical application, let’s compute the conditional probability of the event Y ≤ 3) given Y = 1. By definition,

3 3
P((Y ≤3|X=1)= h(y|1)dy= e1−ydy=1−e−2.

11

Your turn: We reverse the roles of X and Y . (use the back page if you run out of space)
(i) What is the conditional pdf, g(x|y), of X given Y = y, y ∈ (0, ∞). Present your result in the form of

equation (2) (the distribution is not recognaizable). (ii) Compute P (X > 1|Y = 2).

Answer 1