AskSia

Plus

Let \[ f(x, y)=\left(\frac{1}{2 \pi}\right) e^{-\left(x^{2}+y^{2}\right) / 2}\...

Apr 10, 2024

Let f(x, y)=\left(\frac{1}{2 \pi}\right) e^{-\left(x^{2}+y^{2}\right) / 2}\left[1+x y e^{-\left(x^{2}+y^{2}-2\right) / 2}\right], \quad-\infty<x<\infty,-\infty<y<\infty Show that

f(x, y)

is a joint pdf and the two marginal pdfs are each normal. Note that

X

and

Y

can each be normal, but their joint pdf is not bivariate normal. Figure 4.5-I Conditional pdf of

Y

, given that

x=5,10,15

Solution by Steps

step 1

To show that

f(x, y)

is a joint pdf, we need to show that it is non-negative and that its integral over the entire space is 1

step 2

First, we check if

f(x, y)

is non-negative for all

x

and

y

. Since

e^{-\left(x^{2}+y^{2}\right) / 2}

is always positive, and

1+x y e^{-\left(x^{2}+y^{2}-2\right) / 2}

is positive because

e^{-\left(x^{2}+y^{2}-2\right) / 2}

is always less than or equal to 1 for all real

x

and

y

f(x, y)

is non-negative

step 3

Next, we integrate

f(x, y)

over the entire space to check if it sums to 1. We need to evaluate the integral

\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} f(x, y) \, dy \, dx

step 4

The integral of

f(x, y)

can be split into two parts due to the addition inside the brackets. The first part is the integral of the Gaussian function, which is known to integrate to 1 over its domain. The second part involves the product of

x

y

, and the Gaussian function, which integrates to 0 due to symmetry

step 5

Therefore, the integral of

f(x, y)

over the entire space is 1, and

f(x, y)

is a valid joint pdf

step 6

To find the marginal pdfs, we integrate

f(x, y)

over

y

to find the marginal pdf of

X

, and over

x

to find the marginal pdf of

Y

step 7

The marginal pdf of

X

\int_{-\infty}^{\infty} f(x, y) \, dy

, which simplifies to

\frac{1}{\sqrt{2\pi}} e^{-x^2/2}

, a standard normal distribution

step 8

Similarly, the marginal pdf of

Y

\int_{-\infty}^{\infty} f(x, y) \, dx

, which also simplifies to

\frac{1}{\sqrt{2\pi}} e^{-y^2/2}

, another standard normal distribution

step 9

Although

X

and

Y

are each normally distributed, the joint pdf

f(x, y)

is not bivariate normal because it includes the term

x y e^{-\left(x^{2}+y^{2}-2\right) / 2}

, which introduces a dependency between

X

and

Y

that is not present in a bivariate normal distribution

Answer

f(x, y)

is a joint pdf, and the marginal pdfs of

X

and

Y

are each normal.

Key Concept

Joint Probability Density Function (pdf) and Marginal pdfs

Explanation

A joint pdf is valid if it is non-negative and integrates to 1 over the entire space. Marginal pdfs are obtained by integrating the joint pdf over the other variable, and they can be normal even if the joint distribution is not bivariate normal.