If X and Y have joint density function f(x, y)=(1)/(x^2 y^2) x ≥1, y ≥1 (a) Compute the joint

If X and Y have joint density function f(x, y)=(1)/(x^2 y^2)...

Key Concepts

Support in Transformation

Understanding the support of the transformed variables is essential because it defines the region over which the new joint density function is valid. Determining the appropriate limits for the new variables ensures the correctness of further probability calculations.

Marginal Density

Marginal density is the probability distribution of a single variable obtained by integrating the joint density function over the range of the other variable. It simplifies a multivariate distribution to a univariate one, enabling the analysis of individual variable behavior.

Joint Density Function

The joint density function defines the probability distribution of two continuous random variables simultaneously. It specifies how probability mass is distributed over the two-dimensional space and is foundational for understanding multivariate probability problems.

Transformation of Random Variables

This concept involves finding the distribution of new variables defined as functions of the original random variables. It requires substituting the expressions relating the new variables to the original ones, allowing us to study different aspects of the distribution.

Jacobian Transformation

When performing a change of variables in multivariate distributions, the Jacobian determinant is used to adjust the density function for the transformation's effect on area or volume elements. This determinant accounts for the scaling introduced and is crucial for correctly obtaining the new density function.

Transcript

00:01 The joint probability density function is given by 1 by x square, y square, x is greater than or equal to 1, y is also greater than or equal to 1.

00:14 First let us find the marginal density of x.

00:17 It is y going from 1 to infinity 1 by x square y square, d y, because marginal density we should do integration with respect to y only.

00:27 What is integration of 1 by y square? it is negative 1 by y.

00:33 Integration of 1 by y square is negative 1 by y.

00:36 Substitute the limits 1 9 infinity.

00:38 So the marginal density of the random level x is 1 by x square x greater than equal to 1 .0 else.

00:47 So this is the marginal pdf of x by symmetry.

00:51 By symmetry, the marginal pdf of y is given by 1 by y square, y greater than equal to 1 .0 else.

01:02 So we got the individual density functions.

01:05 Now next thing is we need to find the density function of the random variable z y, sorry xy.

01:12 So what we do is well, let's start with a cdf.

01:15 Fz of z is probability that xy is less than or equal to z.

01:20 If z is less than one, obviously this probability will be zero because product of two random variables which are x and y, x random variable is greater than a equal to one, y also greater is equal to that the product should also be greater than equal to 1.

01:34 So when z is less than 1, this probability will be an impossible event.

01:38 The probability of impossible even.

01:40 Now what happens when z is greater than 1? so case 2, when z is greater than 1, our hyperbola x y is equal to z looks like this.

01:51 This is my hyperbola.

01:51 I'll draw a nice sketch using gizmos, so it will look like this.

01:56 So this is my hyperbola.

01:59 And if you can see, notice that this is my x is equal to 1 line, this is my y equal 1 line.

02:04 So tell me in this region, this is my region, x goes from 1 to which point and y goes from where to where.

02:12 If you see this is my line, x, y, curve xy is equal to z, where does y is equal to 1 meets when y is equal to 1, it x is 5 is 5.

02:25 It is 5 is 5.

02:26 It is x is x.

02:27 So, x is z.

02:31 So that means x belongs to 1 to z.

02:35 Why belongs to, y belongs to 1 to z by x.

02:44 So, that means my probability that xy is less than equal to z is given by integration x going from 1 to z, integration y going from 1 to z by x, 1 by x, 1 by square, and then into 1 by x square d x...

Please give Ace some feedback