Determine the value of c such that the function f(x, y)=c x y for 0<x<3 and 0<y<3 satisfies the

Determine the value of c such that the function f(x, y)=c x...

Key Concepts

Computing Probabilities in Continuous Distributions

Calculating probabilities for events involving continuous random variables requires integrating the pdf over the region specified by the event. This approach allows the determination of the probability that a variable falls within a particular range, and is fundamental to the study of continuous random variables.

Expectation in Continuous Distributions

The expectation (or mean) of a continuous random variable is a measure of its central tendency, calculated by integrating the product of the variable and its probability density function over its support. In the context of multivariate distributions, expectations are often computed by integrating out one variable while weighing by the joint density.

Conditional Distribution

The conditional distribution of one random variable given the value of another is obtained by dividing the joint pdf by the marginal pdf of the given variable. This represents the probability distribution of one variable under the condition that the other variable is known, and is an important concept in statistical inference and decision theory.

Marginal Distribution

The marginal distribution of a random variable is derived by integrating the joint pdf over the other variable(s). This gives a single-variable probability distribution, providing insight into the behavior of one random variable irrespective of the value of the other. Marginal distributions are critical for understanding components of multivariate distributions.

Double Integration in Multivariate Distributions

Double integration is a method used to calculate probabilities and expectations for joint continuous distributions by integrating the joint pdf over the relevant region. This process allows one to aggregate the probabilities of outcomes over two dimensions, making it fundamental in solving problems involving bivariate distributions.

Normalization Constant

In continuous probability distributions, the normalization constant ensures that the total probability over the support sums (or integrates) to one. When a function is proposed as a pdf, a constant multiplier is often determined by solving an integral equation to satisfy this requirement. This step is essential for constructing valid probability models.

Support of the Distribution

The support of a probability distribution refers to the set of all possible values that the random variables can take with non-zero probability. For continuous distributions, it is typically defined as an interval (or a region) where the pdf is non-zero. Understanding the support is crucial for setting up the limits during integration for probability computations.

Joint Probability Density Function

A joint probability density function (pdf) describes the likelihood of two continuous random variables occurring simultaneously. It must satisfy two main properties: the function is non-negative over its entire domain, and its integral over the complete support equals 1. This concept is central in multivariate probability and statistics.

Transcript

00:01 Hi guys and this problem is a joint probability density function must satisfy the condition of double integration from negative infinity to infinity.

00:10 Okay for x and y which is f of x and y d x and y d x x x equals 1.

00:22 Okay so we have integration from 0 to 3 then integration from 0 to 3 for c xy d x dy x dy y equals 1 so after completing the integration we get 84 81 over 4 c equals 1 so we get the value of c such as 4 over 81 okay okay now let's find the probability that x less than 2 and y less than 2 so it's probability of x less than 2 and y less than 3 not 2 okay so it's integration from 0 to 3 then from 0 to 3 for 4 over 8 1 x y d y d x x okay so after completing the integration let's start by the inner integral okay so it's a over 4 over 81 times integration from 0 to 2 for x times y squared over 2 then by its limits 3 and 0 d x okay so this is 4 over 81 times 9 over 2 integration 4 x d x from 0 to 2 okay okay so this probability is 0 .444 .4.

02:18 Okay.

02:19 Now let's find the probability of x, where x less than 2 .5.

02:26 So it's integration from 0 to 2 .5.

02:30 Then for f of x.

02:32 So f of x, it's just integration from 0 to 3.

02:36 For 4 over 81 x y, d y, d x, okay? so this is 4 over 81 times 9 over 2 times integration from 0 to 2 .5, okay, 2 .5 for x dx, okay.

03:11 So this is 0 .69444.

03:18 Okay, now let's find the probability such as y, where y between 1 and less than 2 .5.

03:28 So it's integration from 1 to 2 .5 for f of x.

03:34 Okay, so f of x, it's just integration from 0 to 3.

03:40 For f of y sorry it's not f of x 4 over 81 okay x y d x then integration relative with respect to d y okay so this is 2 over 9 times y squared over 2 and 2 and then substituted by its limits which is 2 .5 and 1 so it's 0 .5833.

04:21 Okay...

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