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One problem with a bivariate pdf f(x, y) is that you can’t really “see” the dist...

Sep 25, 2024

One problem with a bivariate pdf f(x, y) is that you can’t really “see” the distribu􏰁ons of the individual variables unless you integrate out a variable. For example, to get f(x) you’d have to integrate out y from f(x,y). A gentleman called Sarmanov came up with a very clever bivariate density in which the pdfs of the individual variables are directly visible. He defined: 𝑓(𝑥, 𝑦) = 𝑓􏰂(𝑥)𝑓􏰀(𝑦)[1 + 𝜔𝑔􏰂(𝑥)𝑔􏰀(𝑦)] where the func􏰁ons 𝑔􏰃 (. ) have the property that: 􏰅 􏰄 𝑓􏰃(𝑢)𝑔􏰃(𝑢)𝑑𝑢 = 0 􏰆􏰅 for i = 1, 2 and the parameter ω is such that 1 + 𝜔𝑔􏰂(𝑥)𝑔􏰀(𝑦) ≥ 0 for all x and y. Clearly for ω=0, x and y are independent so ω has something to do with dependence between x and y. The subscripts indicate that these can all be different func􏰁ons, i.e. x can have a Normal distribu􏰁on and y a Cauchy distribu􏰁on. 1.Show that f(x,y) is nonnega􏰁ve and integrates to 1 when you integrate over both x and y. Conclusion: this is indeed a proper bivariate pdf. 2. What is f(y|x)? 3. Give an expression for E(y | x). 4. If x and y have Normal distribu􏰁ons, then their pdf is 𝑓(𝑥) = 􏰂 exp (− 􏰂 𝑥􏰀). Can you √􏰀􏰇 􏰀 find a func􏰁on g(x) that has the required property noted above? If so, you’ve just found a bivariate pdf with Normal pdfs for the individual variables. Note: this is not an easy problem because of the demand that 1 + 𝜔𝑔􏰂(𝑥)𝑔􏰀(𝑦) ≥ 0 for at least some values of ω...

Solution by Steps

step 1

To show that

f(x,y)

is nonnegative, we note that

f(x,y) = f_1(x)f_2(y)[1 + \omega g_1(x)g_2(y)]

. Since

f_1(x)

and

f_2(y)

are probability density functions, they are nonnegative. The term

1 + \omega g_1(x)g_2(y)

is nonnegative if

\omega

is chosen such that

1 + \omega g_1(x)g_2(y) \geq 0

. Thus,

f(x,y) \geq 0

step 2

To show that

f(x,y)

integrates to 1, we compute the double integral:

\int \int f(x,y) \, dx \, dy = \int \int f_1(x)f_2(y)[1 + \omega g_1(x)g_2(y)] \, dx \, dy.

Using the property of integrals, we can separate this into two parts, leading to:

\int f_1(x) \, dx \cdot \int f_2(y) \, dy + \omega \int f_1(x)g_1(x) \, dx \cdot \int f_2(y)g_2(y) \, dy.

The first part equals 1, and the second part equals 0 due to the property of

g

. Thus, the total integral equals 1

step 3

The conditional density

f(y|x)

can be expressed using the joint density:

f(y|x) = \frac{f(x,y)}{f(x)}.

Substituting

f(x,y)

gives:

f(y|x) = \frac{f_1(x)f_2(y)[1 + \omega g_1(x)g_2(y)]}{f_1(x)} = f_2(y)[1 + \omega g_1(x)g_2(y)].

step 4

To find

E(y|x)

, we use the formula:

E(y|x) = \int y f(y|x) \, dy.

Substituting

f(y|x)

gives:

E(y|x) = \int y f_2(y)[1 + \omega g_1(x)g_2(y)] \, dy.

step 5

For

g(x)

to satisfy the property

\int f_1(u)g_1(u) \, du = 0

, we can choose

g(x) = x - \mu

where

\mu

is the mean of the Normal distribution. This leads to:

\int f_1(u)(u - \mu) \, du = 0.

Thus,

g(x)

can be defined as

g(x) = x - \mu

for the Normal distribution

Answer

f(x,y)

is nonnegative and integrates to 1, confirming it is a proper bivariate pdf. 2.

f(y|x) = f_2(y)[1 + \omega g_1(x)g_2(y)]

. 3.

E(y|x) = \int y f_2(y)[1 + \omega g_1(x)g_2(y)] \, dy

. 4.

g(x) = x - \mu

satisfies the required property.

Key Concept

Bivariate probability density functions and their properties

Explanation

The solution demonstrates how to verify the properties of a bivariate pdf and derive conditional expectations and functions.