Suppose that X and Y have the joint density function f(x, y)=c √(1-x^2-y^2), x^2+y^2 ≤1 a. Fin

Suppose that X and Y have the joint density function f(x,...

Key Concepts

Independence of Random Variables

Two random variables are said to be independent if their joint pdf can be factored into the product of their marginal pdfs. This concept is important when determining the relationship between variables; if the joint density does not split into the product of marginals, the variables are dependent.

Marginal Density Functions

Marginal density functions are obtained by integrating the joint pdf with respect to one variable, effectively 'summing out' that variable. This process gives the pdf for the remaining variable alone and is key in analyzing the individual behavior of each random variable in a multivariate distribution.

Conditional Density Functions

Conditional density functions provide the probability of one variable given the value of another. They are calculated by dividing the joint pdf by the marginal pdf of the given variable. This concept is essential in understanding how the distribution of one variable changes when additional information about another variable is known.

Normalization of a Density Function

Normalization refers to the process of finding the constant that makes a probability density function integrate to one over its support. This ensures that the pdf is a valid representation of probabilities. The constant is typically determined by setting up and evaluating an integral over the entire region where the pdf is defined.

Joint Density Function

A joint probability density function (pdf) describes the likelihood of random variables taking on a set of values simultaneously. It is defined over a specific domain, and its integral over the entire support must equal one. In problems involving joint densities, we often need to integrate over the defined region to verify normalization and to compute probabilities related to combined events.

Polar Coordinate Transformation

When the region of integration is circular or exhibits radial symmetry, converting to polar coordinates simplifies the integration process. In polar coordinates, the variables are expressed in terms of radius and angle, and the differential area element dx dy transforms to r dr d?. This is often used to evaluate integrals over circular regions.

Transcript

00:01 So this question we have a joint density, f of x and y, is c root 1 minus x squared minus y squared, for x squared plus y squared less than or equal to 1.

00:12 So first of all, let's find c.

00:15 So the integral from minus root 1 minus y squared to root 1 minus y squared, and from minus 1 to 1 of c root 1 minus x squared minus y squared, dx, d, d, y, must be equal to 1.

00:34 But what we have to notice here is that we can make a change of variables.

00:38 So let's change to polar coordinates.

00:46 So that x squared plus y squared equals r squared, and dx, d, y, is equal to r, d, r, d, theta.

00:57 Then we have 1 is equal to the integral from 0 to 2 pi d theta, because we're integrating over a disk of radius 1 centered on 0, and the integral from 0 to 1, dr, we've got an r, and then a root 1 minus r squared, dr, and we've got the dr d theta.

01:21 So the theta part just gives us 2 pi.

01:26 The r part gives us 1 minus r squared to the three halves, but then we're going to get a 2 thirds, then we need to divide by minus 2, and we go between 0 and 1.

01:42 So that gives us minus 1 third.

01:48 So the three halves, yeah, and then the minus 2, all good.

01:51 Right.

01:52 So then at r equals 1, we don't get anything.

01:54 At r equals 0, we get 1 to the 3 halves, so we get 2 pi over 3.

02:00 But then we actually have a c involved here.

02:04 So there's a c, there's a c here.

02:07 So that tells us that c is 3 over 2 pi.

02:13 So we have f of x and y is 3 over 2 pi root 1 minus x squared minus y squared for x squared plus y squared less than or equal to 1.

02:27 Okay, part b, let's sketch the drawing density.

02:32 So this is going to be, so let's go x, y, f of x and y.

02:42 Now what we can see is that this goes as the square root of 1 minus r squared.